Notes for Exam with Cheat Sheet - Calculus and Analytic Geometry II | MATH 124, Study notes of Analytical Geometry and Calculus

Download Notes for Exam with Cheat Sheet - Calculus and Analytic Geometry II | MATH 124 and more Study notes Analytical Geometry and Calculus in PDF only on Docsity! Math 124 (1997—1998) by Neal Koblitz ii Send comments to: koblitz@math.washington.edu Printed on November 29, 2000 Preface – Advice for Students The overwhelming emphasis in this course is on applications – story problems. To master story problems one needs a tremendous amount of practice – doing homework, worksheet, and practice exam problems. For this reason, this course will require much more of your time than would a high school or community college calculus course that does not have such a focus on word problems. Here are some pointers for doing story problems: (1) Carefully read each problem twice before writing anything. (2) Assign letters to quantities that are described only in words; draw a dia- gram if appropriate. (3) Decide which letters are constants and which are variables. A letter stands for a constant if its value remains the same throughout the problem. (4) Using math notation, write down what you know and then write down what you want to Þnd. (5) Decide what category of problem it is (this might be obvious if the problem comes at the end of a particular chapter, but will not necessarily be so obvious if it comes on an exam covering several chapters). (6) Double check each step as you go along; don’t wait until the end to check your work. (7) Use common sense; if an answer is out of the range of practical possibilities, then check your work to see where you went wrong. Suggestions for Using These Notes (1) Read the example problems carefully, Þlling in any steps that are left out (ask someone if you can’t follow the solution to a worked example). (2) Later use the worked examples to study by covering the solutions, and seeing if you can solve the problems on your own. (3) Give yourself the practice midterms and Þnals in an exam situation (with- out access to material other than a single sheet of notes, and with the same time limit as on the exam). (4) The answers to homework and practice exam problems that are in the back of the notes should be used only as a Þnal check on your work, not as a crutch. Keep in mind that sometimes an answer could be expressed in various ways that are algebraically equivalent, so don’t assume that your answer is wrong just because it doesn’t have exactly the same form as the answer in the back. v Formulas Used in Math 124 and 125 Algebra Completing the square: X2 + bX + c = (X + b 2 )2 − b 2 4 + c . Quadratic formula: roots of aX2 + bX + c are −b±√b2 − 4ac 2a . Exponents: ab · ac = ab+c; a b ac = ab−c; (ab)c = abc; a1/b = b √ a . Geometry Circle: circumference = 2πr; area = πr2 . Sphere: vol = 4 3 πr3; surface area = 4πr2 . Cylinder: vol = πr2h; lateral area = 2πrh; total surface area = 2πrh+ 2πr2. Cone: vol = 1 3 πr2h; lateral area = πr p r2 + h2; total surface area = πr p r2 + h2 + πr2. Analytic geometry Point-slope formula for straight line: y = y0 +m(x− x0). Circle centered at (h, k): (x− h)2 + (y − k)2 = r2. Ellipse with semimajor axis along x-axis and semiminor axis along y-axis: x2 a2 + y2 b2 = 1. Trigonometry sin = opposite hypotenuse ; cos = adjacent hypotenuse ; tan = opposite adjacent ; sec = 1 cos ; csc = 1 sin ; cot = 1 tan ; tan = sin cos ; cot = cos sin ; sin x = cos ³π 2 − x ´ ; cosx = sin ³π 2 − x ´ ; sin(x+ π) = − sinx; cos(x+ π) = − cosx. Law of cosines: a2 = b2 + c2 − 2bc cosA. Law of sines: a sinA = b sinB = c sinC . Sine of sum of angles: sin(x+ y) = sinx cos y + cosx sin y. sin2 and cos2 formulas: sin2+cos2 = 1; sin2 x = 1− cos(2x) 2 ; cos2 x = 1 + cos(2x) 2 . CHAPTER 1 Straight Lines, Polygonal Lines, Slopes, Distance, Circles Much of the mathematics in this chapter will be review for you. However, the examples will be oriented toward applications (“story problems”), and so will take some thought. In the (x, y) coordinate system we normally write the x-axis horizontally, with positive numbers to the right of the origin, and the y-axis vertically, with positive numbers above the origin. That is, unless stated otherwise, we take “rightward” to be the positive x-direction and “upward” to be the positive y-direction. In a pure math situation, we normally choose the same scale for the x- and y-axes. For example, the line joining the origin to the point (a, a) makes an angle of 45◦ with the x-axis (and also with the y-axis). But in applications, often letters other than x and y are used, and often dif- ferent scales are chosen in the x- and y-directions. For example, suppose you drop something from a window, and you want to study how its height above the ground changes from second to second. It is natural to let the letter t denote the time (the number of seconds since the object was released) and to let the letter s denote the height. For each t (say, at one-second intervals) you have a corresponding height s. This information can be tabulated, and then plotted on the (t, s) coordinate plane. Here is an example: t (sec) s (meters) 0 80.0 1 75.1 2 60.4 3 35.9 4 1.6 1 2 3 4 20 40 60 80 We use the word “quadrant” for each of the four regions the plane is divided into: the 1st quadrant is where points have both coordinates positive, and the 2nd, 3rd, and 4th quadrants are counted off counterclockwise as follows: 2nd 1st 3rd 4th Suppose we have two points A(2, 1) and B(3, 3) in the (x, y)-plane. We often want to know the change in x-coordinate (also called the “horizontal distance”) in going from A to B. This is often written ∆x, where the meaning of ∆ is “change in” (thus, ∆x can be read as “change in x” – it denotes a single number, and should not be read as “delta times x”). In our example, ∆x = 3 − 2 = 1. Similarly, the “change in y” is written ∆y. In our example, ∆y = 3−1 = 2, the difference between 1 4 1. LINES,SLOPES,CIRCLES of velocity, which is 50 mph. To Þnd the equation of the line, we use the point-slope formula: y − 110 t− 1 = −50, so that y = −50(t− 1) + 110 = −50t+ 160. The meaning of the y-intercept 160 is that when t = 0 (when you started the trip) you were 160 miles from Seattle. To Þnd the t-intercept, set 0 = −50t + 160, so that t = 160/50 = 3.2. The meaning of the t-intercept is: the time when you’ll be in Seattle. After traveling 3 hrs 12 min, your distance y from Seattle will be 0. 1. Distance Between Two Points; Circles Given two points (x1, y1) and (x2, y2), recall that their horizontal distance from one another is ∆x = x2 − x1, and their vertical distance from one another is ∆y = y2−y1. The actual distance from one point to the other is the hypotenuse of a right triangle with legs ∆x and ∆y: ∆x ∆y (x2, y2) (x1, y1) The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides: distance = p (∆x)2 + (∆y)2 = p (x2 − x1)2 + (y2 − y1)2. For example, the distance between our two points A(2, 1) and B(3, 3) is equal top (3− 2)2 + (3− 1)2 = √5 = 2.236 · · · . As a special case of the distance formula, suppose we want to know the distance of a point (x, y) to the origin. According to the distance formula, this is equal top (x− 0)2 + (y − 0)2 = p x2 + y2. A point (x, y) is at a distance r from the origin if and only if p x2 + y2 = r, or, if we square both sides: x2 + y2 = r2. This is the equation of the circle of radius r centered at the origin. The special case r = 1 is called the unit circle; its equation is x2 + y2 = 1. Similarly, if C(h, k) is any Þxed point, then a point (x, y) is at a distance r from the point C if and only if p (x− h)2 + (y − k)2 = r, i.e., if and only if (x− h)2 + (y − k)2 = r2. This is the equation of the circle of radius r centered at the point (h, k). For example, the circle of radius 5 centered at the point (0, 5) on the y-axis has equation (x− 0)2+ (y− 5)2 = 25. If we expand (y− 5)2 = y2− 10y+25 and cancel the 25 on both sides, we can rewrite this as: x2 + y2 − 10y = 0. 1. DISTANCE BETWEEN TWO POINTS; CIRCLES 5 Homework 1. 1. For each pair of points A(x1, y1) and B(x2, y2) Þnd (i) ∆x and ∆y in going from A to B, (ii) the slope of the line joining A and B, (iii) the equation of the line joining A and B in the form y = mx+ b, and (iv) the distance from A to B: (a) A(2, 0), B(4, 3); (b) A(1,−1), B(0, 2); (c) A(0, 0), B(−2,−2); (d) A(−2, 3), B(4, 3); (e) A(−3,−2), B(2, 3); (f) A(0.01,−0.01), B(−0.01, 0.05). 2. Graph each of the following lines, after changing to the form y = mx + b, and also Þnd the y-intercept and x-intercept: (a) y − 2x = 2; (b) x+ y = 6; (c) x = 2y − 1; (d) 3 = 2y; (e) 2x+ 3y + 6 = 0. 3. Find the equation of the circle of radius 3 centered at: (a) (0, 0); (b) (5, 6); (c) (−5,−6); (d) (0, 3); (e) (0,−3); (f) (3, 0). 4. Graph the circles (a) x2 + y2 + 10y = 0; (b) x2 − 10x+ y2 = 24; (c) x2 − 6x+ y2 − 8y = 0. 5. Let x stand for temperature in degrees Celsius (centigrade), and let y stand for temperature in degrees Fahrenheit. A temperature of 0◦C (freezing of water) corresponds to 32◦F, and a temperature of 100◦C (boiling of water) corresponds to 212◦F. Find the equation of the line that relates temperature Fahrenheit y to temperature Celsius x. Graph the line, and Þnd the y- and x-intercepts. What is the practical meaning of the intercepts? 6. A car rental Þrm has the following charges for a certain type of car: $25 per day with 100 free miles included, $0.15 per mile for more than 100 miles. Suppose you want to rent a car for one day, and you know you’ll use it for more than 100 miles. What is the equation relating the cost y to the number of miles x that you drive the car? 7. An instructor gives a 100-point Þnal exam, and decides that a score 90 or above will be a grade of 4.0, a score of 40 or below will be a grade of 0.0, and between 40 and 90 the grading will be linear. Let x be the exam score, and let y be the corresponding grade. Find a formula of the form y = mx+ b which applies to scores x between 40 and 90. 8. A photocopy store advertises the following prices: 56c per copy for the Þrst 20 copies, 4 6c per copy for the 21st through 100th copy, and 36c per copy after the 100th copy. Let x be the number of copies, and let y be the total cost of photocopying. (a) Graph the cost as x goes from 0 to 200 copies. (b) Find the equation in the form y = mx+ b that tells you the cost of making x copies when x is more than 100. 9. In the Kingdom of Xyg the tax system works as follows. Someone who earns less than 100 gold coins per month pays no tax. Someone who earns between 100 and 1000 golds coins pays tax equal to 10% of the amount over 100 gold coins that he or she earns. Someone who earns over 1000 gold coins must hand over to the King all of the money earned over 1000 in addition to the tax on the Þrst 1000. (a) Draw a graph of the tax paid y versus the money earned x, and give formulas for y in terms of x in each of the regions 0 ≤ x ≤ 100, 100 ≤ x ≤ 1000, and x ≥ 1000. (b) Suppose that the King of Xyg decides to use the second of these line segments (for 100 ≤ x ≤ 1000) for x ≤ 100 as well. Explain in practical terms what the King is doing, and what the meaning is of the y-intercept. 6 1. LINES,SLOPES,CIRCLES 10. Market research tells you that if you set the price of an item at $1.50, you will be able to sell 5000 items; and for every 10 cents you lower the price below $1.50 you will be able to sell another 1000 items. Let x be the number of items you can sell, and let P be the price of an item. (a) Express P linearly in terms of x, in other words, express P in the form P = mx+ b. (b) Express x linearly in terms of P . 11. The tax for a single taxpayer is described in the Þgure below. Use this informa- tion to graph tax versus taxable income (i.e., x is the amount on Form 1040, line 37, and y is the amount on Form 1040, line 38). Find the slope and y-intercept of each line that makes up the polygonal graph. 1990 Tax Rate Schedules Caution: use ONLY if your taxable income (form 1040, line 37) is $50,000 or more. If less, use the Tax Table. (Even though you cannot use the tax rate schedules below if your taxable income is less than $50,000, we show all levels of taxable income so that taxpayers can see the tax rate that applies to each level.) Schedule X–Use if your filing status is Single Schedule Z–Use if your filing status is Head of household If the amount on Form 1040, line 37, is: Over– But not over– Enter on Form 1040, line 38 of the amount over– If the amount on Form 1040, line 37, is: Over– But not over– Enter on Form 1040, line 38 of the amount over– $0 $19,450 . . . . . . . . . . . . . . . 15% $0 $0 $26,050 . . . . . . . . . . . . . . . 15% $0 19,450 47,050 $2,917.50+28% 19,450 26,050 67,200 $3,907.50+28% 26,050 47,050 97,620 $10,645.50+33% 47,050 67,200 134,930 $15,429.50+33% 67,200 97,620 . . . . . . . . . Use Worksheet below to Þgure your tax. 134,930 . . . . . . . . . Use Worksheet below to Þgure your tax. 2. FUNCTIONS 9 In interval notation, we write: the domain is the interval (0, 12 min(a, b)). We now give more examples of the domain of a pure-math function. Example 2.2. (Circle of radius r centered at the origin) The equation for this circle is usually given in the form x2 + y2 = r2. To write the equation in the form y = f(x) we solve for y, obtaining y = ±√r2 − x2. But this is not a function, because when we have an x it does not give us a single value of y but rather two values (provided that x is between r and −r). To get a function, we must choose one of the two signs in front of the square root. If we choose the positive sign, for example, we get the upper semicircle y = f(x) = √ r2 − x2. The domain of this function is the interval [−r, r], i.e., x must be between −r and r (including the endpoints). If x is outside of that interval, then r2 − x2 is negative, and we cannot take the square root. In terms of the graph, this just means that there are no points on the curve whose x-coordinate is greater than r or less than −r. -1 1 1 y = √ 1− x2 Example 2.3. Find the domain of y = f(x) = 1√ 4x− x2 . To answer this question, we must rule out the x-values that make 4x−x2 negative (because we cannot take the square root of a negative) and also the x-values that make 4x − x2 zero (because if 4x− x2 = 0, then when we take the square root we get 0, and we cannot divide by 0). In other words, the domain consists of all x for which 4x− x2 is strictly positive. We give two different methods to Þnd out when 4x− x2 > 0. First method. Factor 4x − x2 as x(4 − x). The product of two numbers is positive when either both are positive or both are negative, i.e., if either x > 0 and 4 − x > 0, or else x < 0 and 4 − x < 0. The latter alternative is impossible, since if x is negative, then 4− x is greater than 4, and so cannot be negative. As for the Þrst alternative, the condition 4− x > 0 can be rewritten (adding x to both sides) as 4 > x, so we need: x > 0 and 4 > x (this is sometimes combined in the form 4 > x > 0, or, equivalently, 0 < x < 4). In interval notation, this says that the domain is the interval (0, 4). Second method. Write 4x − x2 as −(x2 − 4x), and then complete the square, obtaining − ³ (x−2)2−4 ´ = 4−(x−2)2. For this to be positive we need (x−2)2 < 4, which means that x− 2 must be less than 2 and greater than −2: −2 < x− 2 < 2. 10 2. FUNCTIONS Adding 2 to everything gives 0 < x < 4. Both of these methods are equally correct; you may use either in a problem of this type. A function does not always have to be given by a single formula. For example, suppose that y = v(t) is the velocity function for a car which starts out from rest (zero velocity) at time t = 0; then increases its speed steadily to 20 m/sec, taking 10 seconds to do this; then travels at constant speed 20 m/sec for 15 seconds; and Þnally applies the brakes to decrease speed steadily to 0, taking 5 seconds to do this. The formula for y = v(t) is different in each of the three time intervals. The graph of this function is shown below, along with the three formulas: y = v(t) =  2t, if 0 ≤ t ≤ 10; 20, if 10 ≤ t ≤ 25; 120− 4t, if 25 ≤ t ≤ 30. 10 20 30 10 20 y = v(t) Not all functions are given by formulas at all. A function can be given by an experimentally determined table of values. For example, the population y of the U.S. is a function of the time t: we can write y = f(t). This is a perfectly good function – you could graph it if you had data for various t– but you couldn’t Þnd an algebraic formula for it. 1. Shifts and Dilations Many functions in applications are built up from simple functions by inserting constants in various places. It is important to understand the effect such constants have on the appearance of the graph. Horizontal shifts. If you replace x by x − C everywhere it occurs in the formula for f(x), then the graph shifts over C to the right. (If C is negative, then this means that the graph shifts over |C| to the left.) For example, the graph of y = (x − 2)2 is the x2-parabola shifted over to have its vertex at the point 2 on the x-axis. The graph of y = (x+ 1)2 is the same parabola shifted over to the left so as to have its vertex at −1 on the x-axis. Vertical shifts. If you replace y by y−D, then the graph moves up D units. (If D is negative, then this means that the graph moves down |D| units.) If the formula is written in the form y = f(x) and if y is replaced by y −D to get y −D = f (x), we can equivalently move D to the other side of the equation and write y = f(x) +D. Thus, this principal can be stated: to get the graph of y = f (x)+D, take the graph of y = f(x) and move it D units up. For example, the function y = x2−4x = (x−2)2−4 can be obtained from y = (x− 2)2 (see the last paragraph) by moving the graph 4 units down. The result is the x2-parabola shifted 2 units to the right and 4 units down so as to have its vertex at the point (2,−4). 1. SHIFTS AND DILATIONS 11 Warning. Do not confuse f (x) +D and f(x+D). For example, if f(x) is the function x2, then f (x) + 2 is the function x2 + 2, while f(x + 2) is the function (x+ 2)2 = x2 + 4x+ 4. Example 2.4. (Circles) An important example of the above two principles is the circle x2+ y2 = r2. This is the circle of radius r centered at the origin. (As we saw, this is not a single function y = f(x), but rather two functions y = ±√r2 − x2 put together; in any case, the two shifting principles apply to equations like this one which are not in the function form y = f (x).) If we replace x by x−C and replace y by y −D – getting the equation (x − C)2 + (y −D)2 = r2 – the effect on the circle is to move it C to the right and D up, thereby obtaining the circle of radius r centered at the point (C,D). This tells us how to write the equation of any circle, not necessarily centered at the origin. We will later want to use two more principles concerning the effects of constants on the appearance of the graph of a function. Horizontal dilation. If x is replaced by x/A in a formula, then the effect on the graph is to expand it by a factor of A in the x-direction (away from the y-axis). This wording supposes that A > 1. If A is between 0 and 1 then the effect on the graph is to contract by a factor of 1/A (towards the y-axis). We use the word “dilate” in both cases. For example, replacing x by x/0.5 = 2x has the effect of contracting toward the y-axis by a factor of 2. If A is negative, then we dilate by a factor of |A| and then ßip about the y-axis. Thus, replacing x by −x has the effect of taking the mirror image of the graph with respect to the y-axis. For example, the function y = √−x, which has domain x ≤ 0, is obtained by taking the graph of √x and ßipping it around the y-axis into the second quadrant. Vertical dilation. If y is replaced by y/B in a formula and B > 0, then the effect on the graph is to dilate it by a factor of B in the vertical direction. Note that if we have a function y = f(x), replacing y by y/B is equivalent to multiplying the function on the right by B: y = Bf(x). The effect on the graph is to expand the picture away from the x-axis by a factor of B if B > 1, to contract it toward the x-axis by a factor of 1/B if 0 < B < 1, and to dilate by |B| and then ßip about the x-axis if B is negative. Example 2.5. (Ellipses) A basic example of the two expansion principles is given by an ellipse of semimajor axis a and semiminor axis b. We get such an ellipse by starting with the unit circle – the circle of radius 1 centered at the origin, the equation of which is x2 + y2 = 1 – and dilating by a factor of a horizontally and by a factor of b vertically. To get the equation of the resulting ellipse, which crosses the x-axis at ±a and crosses the y-axis at ±b, we replace x by x/a and y by y/b in the equation for the unit circle. This gives x2 a2 + y2 b2 = 1. Finally, if you want to analyze a function that involves both shifts and dilations, it is usually simplest to work with the dilations Þrst, and then the shifts. For instance, if you want to dilate a function by a factor of A in the x-direction and then shift C to the right, you do this by replacing x Þrst by x/A and then by (x − C)/A in the formula. As an example, suppose that, after dilating our unit circle by a in the x-direction and by b in the y-direction to get the ellipse in the last paragraph, we CHAPTER 3 Instantaneous Rate Of Change: The Derivative Suppose that y is a function of x, say y = f (x). It is often necessary to know how sensitive the value of y is to small changes in x about a Þxed value. Example 3.1. Take, for example, y = f(x) = √ 625− x2 (the upper semicircle of radius 25 centered at the origin). When x = 7, we Þnd that y = √ 625− 49 = 24. Suppose we want to know how much y changes when x increases a little, say to 7.1 or 7.01. In the case of a straight line y = mx + b, the slope m = ∆y∆x measures the change in y per unit change in x. Let us look at the same ratio ∆y∆x for our function y = f (x) = √ 625− x2 when x changes from 7 to 7.1. Here ∆x = 7.1 − 7 = 0.1 is the change in x, and ∆y = f(x+∆x)− f(x) = f(7.1)− f(7) = p 625− 7.12 − p 625− 72 = 23.9706− 24 = −0.0294. Thus, ∆y/∆x = −0.0294/0.1 = −0.294. Geometrically, this means that the chord of the circle drawn from the point (7, 24) to the point (7.1, 23.9706) has slope equal to −0.294. In general, if we draw the chord from the point (7, f(7)) to a nearby point on the semicircle (7 +∆x, f (7 +∆x)), the slope of this chord is the so-called difference quotient slope of chord = f (7 +∆x)− f(7) ∆x = p 625− (7 +∆x)2 − 24 ∆x . For example, if x changes only from 7 to 7.01, then the difference quotient (slope of the chord) is equal to (23.997081− 24)/0.01 = −0.2919. As our second x value 7+∆x moves in towards 7, the chord joining (7, f(7)) to (7 +∆x, f(7 +∆x)) shifts slightly. As can be seen in the picture below, as ∆x gets smaller and smaller, the chord joining (7, 24) to (7+∆x, f(7+∆x))x gets closer and closer to the tangent line to the circle at the point (7, 24). Recall that the tangent line is the line that just grazes the circle at that point, i.e., it doesn’t meet the circle at any second point. Thus, as ∆x gets smaller and smaller, the slope ∆y/∆x of the chord gets closer and closer to the slope of the tangent line. 15 16 3. RATE OF CHANGE 5 10 15 20 25 30 0 5 10 15 20 25 30 tangent line chord 5 10 15 20 25 30 0 5 10 15 20 25 30 tangent line chord 5 10 15 20 25 30 0 5 10 15 20 25 30 tangent line chord ∆x = 8 ∆x = 1 ∆x = 0.5 The slope of the tangent line to the circle at (7, 24) is called the derivative of our function f(x) = √ 625− x2 at 7. It is denoted f 0(7) (we say “f prime of 7,” or equivalently, “the derivative of f at 7”). The slope of the chord joining (7, f(7)) to (7 +∆x, f(7 +∆x)), namely, ∆y/∆x = (f(7 +∆x)− f (7))/∆x, gets closer and closer to this value f 0(7). We write f 0(7) = lim ∆x−→0 f (7 +∆x)− f(7) ∆x and we say that f 0(7) is the limiting value or simply the limit as ∆x approaches zero of the difference quotient (f(7 +∆x)− f (7))/∆x. In the particular case of a circle, there’s a simple way to Þnd the derivative. Namely, the tangent to a circle at a point is perpendicular to the radius drawn to the point of contact, and so its slope is the negative reciprocal of the slope of the radius. The radius joining (0, 0) to (7, 24) has slope 24/7. Hence, the tangent line has slope −7/24 = −0.29166 · · · . We write: f 0(7) = −0.29166 · · · . Notice that when ∆x is small, such as 0.01, the slope of the chord joining (7, f(7)) to (7+∆x, f(7+∆x)) is a good approximation to the value f 0(7). For example, when ∆x = 0.01, we saw that this difference quotient is −0.2919, which is close to f 0(7). Now suppose that we choose a different x, say x = 15, and we want to know how fast y is changing as x increases a little from 15. As an approximation we could take the chord joining (15, f(15)) to (15 +∆x, f(15 +∆x)) when ∆x = 0.01: ∆y ∆x = f(15.01)− f(15) 0.01 = 19.992476− 20 0.01 = −0.7504. The limiting value of this difference quotient as ∆x becomes smaller and smaller is what we mean by f 0(15). The value f 0(15) is the slope of the tangent to the circle at the point (15, 20). As before, we can Þnd this slope exactly, since the tangent to a circle is perpendicular to the radius. The answer is f 0(15) = −15/20 = −0.75. Again we see that, if ∆x is small (like 0.01), then the slope of the chord joining (x, f(x)) to (x+∆x, f(x+∆x)) is very close to this value f 0(x) (here x = 15). 3. RATE OF CHANGE 19 Homework 3. 1. Draw the graph of the function y = f(x) = √ 169− x2 between x = 0 and x = 13. Find the slope ∆y/∆x of the chord between the points of the circle lying over (a) x = 12 and x = 13, (b) x = 12 and x = 12.1, (c) x = 12 and x = 12.01, (d) x = 12 and x = 12.001. Now use the geometry of tangent lines on a circle to Þnd (e) the exact value of the derivative f 0(12). Your answers to (a)—(d) should be getting closer and closer to your answer to (e). 2. Use geometry to Þnd the derivative f 0(x) of the function f(x) = √ 625− x2 in the text for each of the following x: (a) 20, (b) 24, (c) −7, (d) −15. Draw a graph of the upper semicircle, and draw the tangent line at each of these four points. 3. Let y = f(t) = t2, where t is the time in seconds and y is the distance in meters that an object falls on a certain airless planet. Draw a graph of this function between t = 0 and t = 3. Make a table of the average velocity of the falling object between (a) 2 sec and 3 sec, (b) 2 sec and 2.1 sec, (c) 2 sec and 2.01 sec, (d) 2 sec and 2.001 sec. Then use algebra to Þnd a simple formula for the average velocity between time 2 and time 2 + ∆t. (If you substitute ∆t = 1, 0.1, 0.01, 0.001 in this formula you should again get the answers to parts (a)—(d).) Next, in your formula for average velocity (which should be in simpliÞed form) determine what happens as ∆t approaches zero. This is the instantaneous velocity. Finally, in your graph of y = t2 draw the straight line through the point (2, 4) whose slope is the instantaneous velocity you just computed. 4. If an object is dropped from an 80-meter high window, its height y above the ground at time t sec is given by the formula y = f(t) = 80 − 4.9t2. (Here we are neglecting air resistance; the graph of this function was shown at the beginning of the Þrst section.) Find the average velocity of the falling object between (a) 1 sec and 1.1 sec, (b) 1 sec and 1.01 sec, (c) 1 sec and 1.001 sec. Now use algebra to Þnd a simple formula for the average velocity of the falling object between 1 sec and 1 +∆t sec. Determine what happens to this average velocity as ∆t approaches 0. That is the instantaneous velocity at time t = 1 sec (it’ll be negative, because the object is falling). 5. Draw the graph of the function y = f(x) = 1/x between x = 1/2 and x = 4. Find the slope of the chord between (a) x = 3 and x = 3.1, (b) x = 3 and x = 3.01, (c) x = 3 and x = 3.001. Now use algebra to Þnd a simple formula for the slope of the chord between (3, f (3)) and (3 +∆x, f (3 +∆x)). Determine what happens when ∆x approaches 0. In your graph of y = 1/x, draw the straight line through the point (3, 1/3) whose slope is this limiting value of the difference quotient as ∆x approaches 0. 6. Find an algebraic expression for the difference quotient ¡ f(1 +∆x)− f(1)¢/∆x when f (x) = x2 − (1/x). Simplify the expression as much as possible. Then deter- mine what happens as ∆x approaches 0. That value is f 0(1). 7. Draw the graph of y = f(x) = x3 between x = 0 and x = 1.5. Find the slope of the chord between (a) x = 1 and x = 1.1, (b) x = 1 and x = 1.001, (c) x = 1 and x = 1.00001. Then use algebra to Þnd a simple formula for the slope of the chord between 1 and 1 +∆x. (Use the expansion (A+ B)3 = A3 + 3A2B + 3AB2 +B3.) Determine the limit as ∆x approaches 0, and in your graph of y = x3 draw the straight line through the point (1, 1) whose slope is equal to the limit you just found. CHAPTER 4 The Derivative Function In the last section we saw how, given a function y = f (x) and some particular value of x, we can speak of the derivative of the function at x, denoted f 0(x). In geometrical terms, f 0(x) is the slope of the tangent line which grazes the curve at the point (x, f(x)). In this section, instead of taking just one particular value of x (or a few particular values, as in Problem 2 of the last homework), we let x vary. That is, we are interested in the rule which for every x in the domain of our function f(x) determines the value of the derivative f 0(x). That gives us a new function f 0, called “f prime” or “the derivative of f .” Example 4.1. Find the derivative function f 0(x) for the Þrst function discussed in the last section: f(x) = √ 625− x2. The point on the upper semicircle above x has coordinates (x, √ 625− x2). The radius out to that point has slope ∆y/∆x = ( √ 625− x2 − 0)/(x − 0) = ( √ 625− x2)/x. Since the tangent to the circle at that point is perpendicular to this radius, its slope is the negative reciprocal, i.e., −x/√625− x2. Thus, f 0(x) = −x/√625− x2. For example, if you plug x = 20, 24, −7, or −15 into this formula, you get the answers to Problem 2 of the last homework. Notation. There are various other ways to denote the derivative f 0(x). If y = f(x), we can also write y0, or else dy/dx, or else ddxf(x). For example, using the different types of notation in Example 1, where f(x) = √ 625− x2, we can write: y0 = f 0(x) = dy dx = d dx ³p 625− x2 ´ = − x√ 625− x2 . Example 4.2. Find the derivative function for the second function in the last sec- tion: y = f (t) = t2. The derivative f 0(t) is the instantaneous velocity, which we Þnd by taking the limit as ∆t approaches zero of the average velocity during the time interval between t and t+∆t. That average velocity is the difference quotient distance traveled time = (t+∆t)2 − t2 ∆t = t2 + 2t(∆t) + (∆t)2 − t2 ∆t = 2t(∆t) + (∆t)2 ∆t = 2t+∆t. The limit of this as ∆t approaches zero is 2t. Thus, f 0(t) = d dt ¡ t2 ¢ = 2t. In the case of a distance function, the derivative (the instantaneous velocity) is often denoted with a dot rather than a prime; so we can write: úy = d dt ¡ t2 ¢ = 2t. 21 24 4. THE DERIVATIVE FUNCTION Example 4.6. Use graphical differentiation to Þnd the graph of f 0(x) for the curve y = f(x) pictured below. -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 y = f(x) Following the 6-step procedure, we make a table of values of f 0(x) for x at intervals of 0.2 from −1 to 1: x -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 f 0(x) 0.3 0.4 0.5 0.6 1.0 inÞnite 1.0 0.6 0.5 0.4 0.3 (Note that this is a very approximate procedure, so our values of f 0(x) can be relied upon only to 1 or 2 signiÞcant Þgures.) Based on this table of values, we obtain an approximate graph of the derivative function f 0(x). Note that the derivative becomes inÞnite as we get closer and closer to x = 0, because the tangent to the curve there is vertical (has inÞnite slope). We say that f 0(0) “does not exist” (or “is inÞnite”). -1 -0.5 0.5 1 0.5 1 1.5 f 0(x) 2. GRAPHICAL DIFFERENTIATION 25 Homework 4. 1. Using the procedure used above in Example 4.1, Þnd the derivative function for y = f(x) = √ 169− x2. 2. Using the procedure that we used in Examples 4.2 and 4.3, Þnd the derivative function for: (a) y = f(t) = 80− 4.9t2, (b) y = f(x) = x2 − (1/x), (c) y = f(x) = ax2+ bx+ c (where a, b, and c are constants), (d) y = f(x) = x3. Please show every step clearly. 3. For each of the following two graphs y = f(x), use graphical differentiation to sketch the graph of the derivative function. For x at regular intervals from one end of the domain to the other, estimate the derivative f 0(x) using a ruler, as explained in the text. Show clearly the points where the derivative is zero and where the derivative does not exist (or is inÞnite). -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1 2 3 4 5 0 1 2 3 4 26 4. THE DERIVATIVE FUNCTION 4. The 24 graphs below are labeled by letters from (a) to (x). For each of the following graphs of f(x), give the letter of the graph that looks most like it could be the graph of the derivative function f 0(x): (1) b, (2) c, (3) e, (4) g, (5) h, (6) j, (7) l, (8) p, (9) r, (10) s, (11) t, (12) u, (13) w, (14) x. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s) (t) (u) (v) (w) (x) 2. THE TANGENT LINE APPROXIMATION 29 to know f 0(x). Now ddxx 3 = 3x2 (see Problem 2(d) in the last homework set, or else use the Power Rule above). Hence f 0(x) = f 0(2) = 3 · 22 = 12. We are now ready to use the tangent line approximation formula: f(2 + 0.1) ≈ f(2) + (0.1)f 0(2) = 8 + 0.1 · 12 = 9.2. Of course, we could easily Þnd (2.1)3 exactly by calculator (or by multiplying by hand): the result is 9.261. 2 8 9.261 1.9 2 2.1 2.2 7 7.5 8 8.5 9 9.5 10 Since the tangent line approximation does not give the exact answer, and since we could have found 2.13 easily enough without it, you might wonder why one bothers with the tangent line approximation. There are several answers: (1) In easy examples, with experience you can perform the tangent line approximation in your head, thereby getting a better “feel” for how the function changes when you make a small change from x to x + ∆x. For instance, in Example 5.1 we can say that “the cube of a number near 2 is going to be near 8, and the difference between the cube and 8 is going to be about 12 times the difference between our number and 2.” (2) In many practical applications, the tangent line approximation gives simple answers to questions about experimental error, especially percent error. (3) Later on, in the section on implicit functions, we will encounter examples where y cannot be written in terms of x, and so there is no way to evaluate the y-coordinate of the point on the curve above x+∆x except through an approximation method, such as the tangent line approximation. Example 5.2. Find 1/4.99. Here we’re interested in evaluating the reciprocal function very near to an x- value whose reciprocal we can evaluate in our heads, namely: f(5) = 1/5 = 0.2. That is, our function is f(x) = 1/x, our x is 5, and x+∆x = 4.99, i.e., ∆x = −0.01. Since f 0(x) = −1/x2 = −1/25 = −0.04, we have: 1/4.99 = f(4.99) = f(5 + (−0.01)) ≈ f(5) + (−0.01)f 0(5) = 0.2− 0.01 · (−0.04) = 0.2004. Using a calculator, we can Þnd the exact value: 1/4.99 = 0.200400801 · · · . Thus, in this case the tangent line approximation is accurate to 5 decimal places. 30 5. THE TANGENT LINE APPROXIMATION Example 5.3. On the circle of radius 10 centered at the origin Þnd a point in the Þrst quadrant whose x-coordinate is 6.01. Here we know the “easy point” (6, 8) on the circle, and we want to know the y-coordinate of the nearby point (6.01, ?). The function is y = f (x) = √ 100− x2, our x is 6, and our x + ∆x = 6.01, i.e., ∆x = 0.01. Using the technique in the last section (e.g., Problem 1 on the homework), we can Þnd the derivative: f 0(x) = −x/√100− x2 = −6/√100− 62 = −0.75. Thus, the tangent line approximation tells us that f(6.01) ≈ f(6) + 0.01 · f 0(6) = 8 + 0.01(−0.75) = 7.9925. (The exact value, computed by calculator, is 7.992490225 · · · .) Example 5.4. If 1 quart equals 0.9464 liter, and 1 liter of water occupies exactly 1000 cc (cubic centimeters), i.e., a cubic container 10 cm on a side, then what are the dimensions (in cm) of a cubic container that holds 1 quart of water? In a story problem like this, we must Þrst analyze the operation that is being performed. The question can be reworded as follows: if a cube holding 1000 cc has side 10 cm, then what is the side of a cube holding 0.9464 as much? The process of going from the volume to the side is the cube-root function. Since “0.9464 as much as 1000 cc” obviously means 946.4 cc, we can rephrase the question once again: if the cube-root of 1000 is 10, then what is the cube-root of 946.4? Let s be the length of the side of the cube and V its volume, then s = f(V ) = 3 √ V . In the tangent line approximation we take V = 1000 and V + ∆V = 946.4, i.e., ∆V = −53.6. By the Power Rule, we have f 0(V ) = 13V −2/3. When V = 1000, we have 1000−2/3 = 1/100, and so f 0(1000) = 1/(3 · 100) = 1/300. So the tangent line approximation says that 3 √ 946.4 = f(946.4) ≈ f (1000) + (−53.6) · f 0(1000) = 10− 53.6/300 = 9.821 cm. (Note: the exact value obtained by calculator is 9.81804 · · · .) Alternate Method. Write V = s3. Then by the tangent line approximation ∆V ≈ 3s2∆s. Since s = 10 and ∆V = −53.6, we have −53.6 ≈ 300∆s. Solving for ∆s gives ∆s ≈ −0.179, so (new s) = 10 +∆s ≈ 10− 0.179 = 9.821 cm. The next example involves percent error. If you measure a value x, then the error (meaning the maximum possible error, or perhaps the maximum error that is likely) is the amount by which the true value could possibly differ from your measured value. We shall denote the error ∆x. In most cases the true value could be on either side of the measured value, so the error is actually ±∆x. By the percent error we mean the percent that this error is of the measured value, namely: percent error = 100 ∆x x . For example, if you measure the speed of a car to be 50 mph to within 1 mph on either side, that means you have a percent error of ±2%. If you know the percent error, say ±p%, then to convert this to absolute error ∆x, use the formula: ∆x = p100x. Example 5.5. If you determine the length of a side of a cubic container by measuring the volume of liquid it holds, then an error of ±1% in measuring the volume will lead to what percent error in your value for the length of a side? An error of ±p% will lead to what percent error for the side? This is a lot like Example 5.4. Again our process is: starting with the volume of a cube, determine its side. That is, our function is again the cube-root function. 2. THE TANGENT LINE APPROXIMATION 31 But we’re not given any concrete values – and to answer the question we don’t need any. So let V stand for the measured value of the volume, and let s = f(V ) = 3 √ V be the length of a side. We want to know the error ∆s that is caused if V is replaced by V ± ∆V , where ∆V is a 1% error, i.e., ∆V = 0.01V . Actually, what we want to know is the percent error in the side s, i.e., 100(∆s)/s. According to the tangent line approximation, ∆s ≈ f 0(V )∆V = 1 3 V −2/3 · 0.01V = 1 3 · 0.01 3 √ V , where the last equality comes from the rule of exponents: V −2/3 ·V = V −2/3 ·V 1 = V −2/3+1 = V 1/3 = 3 √ V . But 3 √ V is precisely the value s = f(V ). So our percent error in s is 100 ∆s s ≈ 100 µ 1 3 ¶ 0.01 3 √ V s = 1 3 %. If we take −∆V in place of ∆V , we get −1/3 %. Thus, the percent error in s is ±13%. This answers our Þrst question. If the error in V is ±p% instead of ±1%, everything would go through the same way, except with a factor of p everywhere: 100 ∆s s ≈ 100f 0(V )∆V s = 100 (1/3)V −2/3 · 0.01pV s = 1 3 p, so the error in s is ±13p%. That is, a certain percent error in measuring the volume of a cube leads to a percent error one-third as great in the value computed for the side. Note. Notice where the 1/3 came from in the last example: the exponent in x1/3 that comes down in the rule for the derivative of xn. In the same way, it turns out that whenever your function is of the form y = f(x) = Cxn (where C is any constant), there is a simple relationship between the percent error in your measurement of x and the percent error in y obtained using the tangent line approximation. More examples of this will show up in the homework. CHAPTER 6 Rules For Finding Derivatives 1. Linearity of the Derivative This means two things: (1) If you know the derivative f 0(x) of a function f(x), then the derivative of a constant multiple cf(x) is simply cf 0(x). The derivative of a constant c times a function is equal to c times the derivative of the function: d dx ¡ cf(x) ¢ = c d dx f(x). This principle is obvious if you think of the derivative as a rate of change. For example, it says that if f (t) is the distance function for a bicycle, and if the distance function for a car is always 5 times the distance for the bicycle, then the car must be going at 5 times the velocity of the bicycle: d dt ¡ 5f (t) ¢ = 5 d dt f(t) . (2) If you know the derivatives f 0(x) and g0(x) of two functions f (x) and g(x), then the derivative of the sum f (x) + g(x) is simply f 0(x) + g0(x). The derivative of the sum of two functions is equal to the sum of the derivatives of the functions: d dx ¡ f(x) + g(x) ¢ = d dx f (x) + d dx g(x). This principle is also obvious when one thinks of rates. It says that the rate at which the sum of two distances is changing is equal to the sum of the rates at which each distance is changing. Principle (2) can be used repeatedly, if we have a sum of more than two functions. It can also be used in combination with the Þrst principle. When combined in this way, the two principles together tell us that it is easy to Þnd the derivative of any linear combination of functions whose derivatives we know. By a “linear combination” we mean any function that is constructed using the process of addition and multiplying by constants. For example, a polynomial is a linear combination of power functions: c0 + c1x+ c2x 2 + · · ·+ cnxn. Example 6.1. In the section on the derivative function, we found the derivative of t2 (or x2), ofmx+b, and of the reciprocal function. Use that, together with linearity of the derivative, to do Problem 2(a)-(c) of that section much more quickly. 35 36 6. RULES FOR FINDING DERIVATIVES We have: d dt ¡ 80− 4.9t2¢ = d dt 80− 4.9 d dt t2 = 0− 4.9(2t) = −9.8t. d dx µ x2 − 1 x ¶ = d dx x2 − d dx x−1 = 2x− ¡−x−2¢ = 2x+ 1 x2 . d dx ¡ ax2 + bx + c ¢ = a d dx x2 + d dx (bx+ c) = 2ax+ b. Example 6.2. Find the derivative of y = ¡ 1− 5x3¢2. Since ¡ 1− 5x3¢2 = 1− 10x3 + 25x6, the derivative is d dx ¡ 1− 10x3 + 25x6¢ = d dx 1− 10 d dx x3 + 25 d dx x6 =0− 10 · 3x2 + 25 · 6x5 = −30x2 + 150x5. 2. The Product Rule Suppose that we want to Þnd the derivative of a function that is written as a product f (x) = u(x)v(x), where each of the two factors u(x) and v(x) is a function whose derivative we know. For example, f(x) = x3 √ 25− x2 is the product of a function u(x) = x3 whose derivative was already computed (namely, u0(x) = 3x2) and a function v(x) = √ 25− x2 whose derivative we also know (namely, v0(x) = −x/√25− x2). We derive the product rule using the tangent line approximation formula for u(x) and v(x): u(x+∆x) ≈ u(x) + (∆x)u0(x); v(x+∆x) ≈ v(x) + (∆x)v0(x). We now look at the difference quotient for f(x) = u(x)v(x), whose limit as ∆x approaches zero is the derivative f 0(x): f(x+∆x)− f (x) ∆x = u(x+∆x)v(x+∆x)− u(x)v(x) ∆x ≈ ³ u(x) + (∆x)u0(x) ´³ v(x) + (∆x)v0(x) ´ − u(x)v(x) ∆x = u(x)v(x) + (∆x) ³ u(x)v0(x) + u0(x)v(x) ´ + (∆x)2u0(x)v0(x)− u(x)v(x) ∆x = ³ u(x)v0(x) + u0(x)v(x) ´ + (∆x)u0(x)v0(x). As ∆x approaches zero, the last term (∆x)u0(x)v0(x) drops out, and we are left with u(x)v0(x) + u0(x)v(x). Thus, if f(x) = u(x)v(x), then f 0(x) = u(x)v0(x) + u0(x)v(x). This rule can be written in shorthand as follows: (Product rule) d dx (uv) = uv0 + u0v. 4. THE SECOND DERIVATIVE AND ACCELERATION 39 Example 6.7. Find the second derivative of (a) y = xn, (b) y = 1/x, (c) ax2+bx+c (where a, b, c are constants). (a) We have d dx µ d dx xn ¶ = d dx ¡ nxn−1 ¢ = n(n− 1)xn−2. Then (b) is the special case n = −1, i.e., the second derivative of x−1 is 2x−3. In part (c) we have d dx ¡ d dx (ax2 + bx+ c) ¢ = d dx (2ax+ b) = 2a. Dot Notation. If the time t is playing the role of the x-variable, i.e., if y = f(t), then we often use dots rather than primes for the derivatives: úy = d dt f(t) , ÿ = d2 dt2 f(t) . If the y-variable is denoted by a letter other than y, we still use the dot notation. For example, if the distance function is denoted s = s(t), then we write s̈(t) for the second derivative of the distance function. In the case of a distance function s = s(t), both the Þrst derivative function ús(t) and the second derivative function s̈(t) have a familiar practical meaning. The Þrst derivative ús(t) is, as we’ve seen, the velocity function, which is often denoted v(t): v(t) = ús(t) = d dt s(t). The derivative of that – the instantaneous rate at which the velocity is changing – is what is called the acceleration, often denoted a(t): a(t) = d dt v(t) = d2 dt2 s(t) = s̈(t). Earlier we saw that the velocity at a certain instant t is the slope of the tangent line to the graph of the distance function at the point (t, s(t)). Similarly, the acceleration at a certain instant t is the slope of the tangent line to the graph of the velocity function at the point (t, v(t)). Example 6.8. A bicycle is traveling along a road (which we take to be the x-axis). Its distance in meters from the reference point (0 on the x-axis) at time t seconds is given by the formula x(t) = 50 + 5t + t2. Find the derivative function úx(t) and the second derivative function ẍ(t), graph all three functions (distance, velocity, and acceleration), and explain in words what is going on. First v(t) = úx(t) = d dt (50 + 5t + t2) = 5 + 2t. Next, a(t) = ẍ(t) = d dt v(t) = d dt (5 + 2t) = 2. The distance function x(t) tells us that the bicycle is at x(0) = 50 meters from the reference point at time t = 0 and is moving to the right at a certain velocity v(t). The formula for this velocity function v(t) = 5 + 2t says that the speed of the bicycle is 5 m/sec at Þrst (t = 0), but it increases steadily. The rate at which the velocity increases is the acceleration function a(t) = 2, which happens to be a constant function in this example. The acceleration is measured in “meters per second per second” or “meters per second squared,” written 2 m/sec2. Here are the graphs of the distance, velocity, and acceleration functions: 40 6. RULES FOR FINDING DERIVATIVES 1 2 3 4 5 0 20 40 60 80 100 1 2 3 4 5 0 2 4 6 8 10 1 2 3 4 5 0 1 2 3 4 5 s(t) v(t) = ús(t) a(t) = s̈(t) m m/sec m/sec2 Example 6.9. (Free fall without air resistance) This is a good place to discuss the vertical motion of an object moving without air resistance under the inßuence of gravity. Let y denote the height (say in meters) of the object above ground level and let t be the time (in seconds). Suppose that at time t = 0 an object is located at height y = y0 and has velocity v0. (Notice that v0 positive means that the object is initially moving upward.) Then y and t are related by the formula (Falling Body Formula) y = y(t) = −1 2 gt2 + v0t+ y0 . Taking derivatives gives the formula for the velocity v(t) = d dt y(t) = −gt+ v0 , and taking one more derivative gives the acceleration a(t) = d dt v(t) = −g . In particular, the acceleration of a falling body is constant, it is called the accel- eration due to gravity. On the surface of the earth g = 9.8m/sec2. (Notice that the acceleration is negative because the force of gravity points down.) 5. The Chain Rule The chain rule is the way to Þnd the derivative of a function that is built up from simpler functions by the composition of two functions, i.e. by taking a “function of a function.” For example, the function √ 1 + x4 is built up from the square root function and the function 1 + x4. The function 1/ p 5−√x is built up from the reciprocal square root function (i.e., the (−1/2)-power function) and the function 5−√x. We shall write y = f(x) = g(u(x)) to indicate that y is obtained by taking a function g(u) (which we sometimes call the “outside function”) and applying it to another function u(x) (which we call the “inside function”). In the two examples just mentioned: g(u) = √ u and u(x) = 1 + x4 =⇒ y = f(x) = g(u(x)) = p 1 + x4; g(u) = 1/ √ u and u(x) = 5−√x =⇒ y = f(x) = g(u(x)) = 1/ q 5−√x. 5. THE CHAIN RULE 41 The chain rule enables us to Þnd dy/dx once we know the derivative of the outside function dy/du = g0(u) and the derivative of the inside function du/dx = u0(x). We shall derive the chain rule using the tangent line approximation. The tangent line approximation formula for the function y = g(u) can be written in the form ∆y ≈ g0(u)∆u. Similarly, the tangent line approximation formula for the function u(x) can be written ∆u ≈ u0(x)∆x. Putting these two facts together, we see that a small change ∆x produces a change u0(x) times as much in u, which, in turn, produces a change g0(u) times as much in y, i.e., ∆y ≈ g0(u)∆u ≈ g0(u)u0(x)∆x. Since we also have ∆y ≈ f 0(x)∆x, it follows that the two “proportionality factors” between ∆x and ∆y must be equal, i.e., f 0(x) = g0(u)u0(x). This is the chain rule. It is especially easy to remember when it is written in the dy/dx notation: (Chain Rule) dy dx = dy du · du dx . It is as if the “dy,” “du,” and “dx” can be treated individually as algebraic quantities, where we cancel the two “du” on the right. Schematically, we can regard a function of a function as a 2-step procedure, Þrst going from x to u and then from u to y. The chain rule says that the derivative from x to u must be multiplied by the derivative from u to y to get the derivative from x to y. Before giving some computational examples, we shall give a story problem to show that the chain rule makes sense from a practical point of view. Example 6.10. Suppose your income is in the 28% tax bracket, and it is increasing at the rate of $1000 per year. At what rate is your tax increasing? As before, let us use x to denote income and y to denote the income tax. Recall that “28% tax bracket” means that y as a function of x has slope 0.28 over the range of values of x (your income) in question. In this problem y depends on x, which in turn depends on the time t. We could write that y = g(x) and x = x(t) together give y = f (t) = g(x(t)). (Thus, x is playing the role of u in the chain rule, and t is playing the role of x.) The chain rule says that dy dt = dy dx · dx dt = (28%) · (1000 dollars/ year) = 0.28 · 1000 dollars/year = 280 dollars/year. That is, tax is increasing at the rate of $0.28 for every dollar increase in income, and income is increasing at the rate of $1000 for every year increase in time; hence, tax is increasing at the rate 0.28 · 1000 = 280 dollars every year. Example 6.11. (a) Use the chain rule to Þnd the derivative of the upper semicircle function y = √ r2 − x2 (where r is a constant), and see that the answer agrees with our earlier formula for the derivative obtained using geometry. Also Þnd the deriv- ative of the two examples mentioned above: (b) y = √ 1 + x4, (c) y = 1/ p 5−√x. (a) We take y = g(u) = √ u for the outside function and u = r2 − x2 for the inside function. Then, since du/dx = −2x, we have dy dx = dy du · du dx = 1 2 √ u · (−2x) = − x√ u = − x√ r2 − x2 , as expected. 44 6. RULES FOR FINDING DERIVATIVES for the distance from the person’s eye to the object. (b) Find a formula in terms of t for the rate at which that distance is changing. Practice First Midterms Practice First Midterm #1 (50 points in all, time = 1 hour) 1. (12 points) The graph of y = f(x) = 2x−x2 is given at the right. Find the domain of each of the following functions. You do not need to use any algebra in this problem. (a) p f(x) = √ 2x− x2 (b) p f(−x) = √−2x− x2 (c) 1f(x−2) (d) 1√ f(x+1) -3 -2 -1 1 2 3 4 -3 -2 -1 1 2 2. (12 points) At time t = 0 an object starts from 0 moving to the right at 2 m/sec. Its velocity changes, as shown in the graph at the right. This graph shows the velocity with which the object is moving to the right as a function of time. Give the letters of ALL labeled points where (a) the object is actually moving right- ward; (b) the object is actually moving leftward; (c) the object has positive (rightward) ac- celeration; (d) the object has zero acceleration; (e) the object’s speed heading to the right is maximum; (f) the object’s speed heading to the left is maximum. -2 -1 1 2 A B C D E F G H I J 3. (13 points) A box with square top and bottom and rectangular sides has surface made up of two s × s sides and four s × t sides. Thus, if your box has dimensions 5 ft × 5 ft × 4 ft, then the total surface area A is 130 sq ft. If you want to increase s by 1 inch to 5 ft 1 in, how should t change so that the surface area stays the same? Use the tangent line approximation, and give your answer in feet. 4. (13 points) A cubic container with side 9 cm holds 93 = 729 cubic cm. Suppose you have a bottle containing 750 ml of liquor. (A milliliter (ml) is the same thing as a cubic cm.) Use the tangent line approximation to Þnd the side of a cubic container needed to hold the 750 ml of liquor. 45 46 PRACTICE FIRST MIDTERMS Practice First Midterm #2 (50 points in all, time = 1 hour) 1. (8 points) Sketch the graph of the functions (a) y = (x+ 2)2 − 4, (b) y = − p 25− (5x)2 2. (6 points) The graph of the function g(x) on the interval [0, 2π] is pictured below to the left. Find the domain of y = f (x) = 1/ p g(x+ (π/2)). (This function g(x) is actually the sine function restricted to the interval [0, 2π].) -1.5 -1 -0.5 0.5 1 1.5 -2 -1 1 2 A B C D E F G H I J 1 2 π π 3 2 π 2π Graph for Problem 2 Graph for Problem 3 3. (12 points) At time t = 0 an object starts from 0 moving to the left at 1 m/sec. Its velocity changes, as shown in the graph above at the right. This graph shows the velocity with which the object is moving to the right as a function of time. Give the letters of ALL labeled points where (a) the object is actually moving rightward; (b) the object is actually moving leftward; (c) the object has positive (rightward) acceleration; (d) the object has zero acceleration; (e) the object’s speed heading to the right is maximum; (f) the object’s speed heading to the left is maximum. 4. (12 points) From a point exactly 75 m above the ground you throw a stone upward at v0 m/sec, and you time how long it takes to hit the ground. If we neglect air resistance, the height of the stone at time t is given by the falling body formula s = −12gt2 + v0t+ s0, where g = 9.8 m/sec2. (a) If you time 5 sec for the stone to reach the ground, at what upward speed did you throw the stone? (b) If your stopwatch is accurate only to ±0.1 sec, how accurate is your answer to part (a)? Use the tangent line approximation. 5. (12 points) You have a 1000-square-foot apartment which you’re renting for the summer to someone from Europe, so you have to convert its area to square meters in order to explain its size. You know that there are about 10 square feet in one square meter. More precisely, 1 square meter = 10.77 square feet. Use the tangent line approximation to Þnd the area of your apartment in square meters. You do NOT need a calculator to do this problem correctly. CHAPTER 7 Implicit Functions and Implicit Differentiation 1. Implicit Functions We have seen that not all functions y = f(x) are given by formulas. Sometimes a function is determined by interpolating a table of experimentally determined data (drawing a smooth curve between the points). Another way a function can be determined is implicitly. That means that we have a formula which x and y to- gether must satisfy. The equation may take the form F (x, y) =constant, or perhaps F (x, y) = G(x, y). Example 7.1. (a) The circle can be written x2 + y2 = r2, which is actually the implicit equation of two functions at the same time: the upper semicircle function y = √ r2 − x2 and the lower semicircle function y = −√r2 − x2. (b) We can avoid fractional powers y = xl/m by raising both sides to the m-th power and writing the implicit equation ym = xl. (c) We can avoid negative powers by writing y = x−l/m in the form ymxl = 1. (d) The points (x, y) on an ellipse with foci at (x1, y1) and (x2, y2) and semimajor axis a satisfy the geometrical property: ( distance from (x, y) to (x1, y1)) + ( distance from (x, y) to (x2, y2)) = 2a, which leads to the following implicit equation for the ellipse:p (x− x1)2 + (y − y1)2 + p (x− x2)2 + (y − y2)2 = 2a. (e) In the case when the ellipse’s axes of symmetry are parallel to the x- and y-axes, a simpler implicit equation can be given, namely (x−h)2/a2+(y−k)2/b2 = 1. With many implicit equations, it is possible to solve for y in terms of x. For example, we can do this for the circle (Example 7.1(a)). In that case we have the choice of whether to use the explicit form y = f (x) or the implicit form F (x, y) =constant. However, there are other examples where it is impossible to solve algebraically for y in terms of x, and we have no choice but to work with the implicit equation. 2. Implicit Differentiation Suppose we want to Þnd the derivative y0 at a point (x, y) of an implicitly deÞned curve. What you do is: (1) take d/dx of both sides of the equation; (2) separate the terms involving y0 to one side of the equation and the terms without y0 to the other side; and (3) solve for y0. In step (1), it is important to remember that you are differentiating with respect to x (or whatever letter is playing the role of the x-variable), and so the chain rule must be used for expressions like y2. In other words, the letter y stands for a certain function of x, and so taking d/dx of y2 is just like taking d/dx of u(x)2: ddxy 2 = 2y dydx = 2y y 0. 49 50 7. IMPLICIT FUNCTIONS AND IMPLICIT DIFFERENTIATION Example 7.2. Find y0 at the point (x, y) for each implicitly deÞned function in Example 7.1. (a) When we take d/dx of both sides of x2 + y2 = r2, remember that (1) the derivative of y2 is not 2y but rather 2yy0, and (2) the derivative of r2 is not 2r but rather 0, because r2 is just a constant. Thus, taking d/dx of both sides gives: 2x + 2yy0 = 0. We next move the 2x to the right, and divide both sides by 2y to solve for y0: y0 = −2x/(2y) = −x/y. This answer agrees with our earlier formula for the derivative of y = √ r2 − x2, namely, y0 = −x/√r2 − x2, since y and √r2 − x2 are the same. Notice that when you differentiate an explicit function y = f(x) your answer is a formula in terms of x, whereas implicit differentiation generally leads to a formula involving both y and x. Parts (b) and (c) of this example will show us how to derive the nxn−1 rule when the power of x is a fraction n = l/m or n = −l/m, where l and m are positive integers. (In the section on the product rule, we already saw how to get the nxn−1 rule for any positive integer n.) (b) We take d/dx of both sides of ym = xl, and then solve for y0: mym−1y0 = lxl−1, so that y0 = l m xl−1 ym−1 . The last ratio simpliÞes algebraically, if we use the fact that ym−1 = ym/y = xl/xl/m. That is, y0 = l m xl−1 xl/xl/m = l m xl/m−1 = nxn−1 (here n = l m ). (c) We take d/dx of both sides of ymxl = 1 (using the product rule), and then solve for y0: ym ³ lxl−1 ´ +mym−1y0xl = 0, so that y0 = − ly mxl−1 mym−1xl . Canceling ymxl from the numerator and denominator in the last expression, we obtain: − l m x−1 y−1 = − l m y x = − l m x−l/m x = − l m x− l m −1 = nxn−1, where n = − l m . (d) When we take d/dx of both sides, we must use the chain rule several times, Þrst with the square root as the outside function and then when we Þnd the derivative of (x− x1)2, (y − y2)2, etc. The result is: (x− x1) + (y − y1)y0p (x− x1)2 + (y − y1)2 + (x− x2) + (y − y2)y0p (x− x2)2 + (y − y2)2 = 0. Solving for y0 is not difficult, but the expression we end up with is rather messy, so we shall omit it. If we had speciÞc values for the foci (x1, y1) and (x2, y2), and if we wanted to Þnd the slope y0 at a particular point (x, y), then we could substitute all of these numerical values in place of x, y, x1, y1, x2, y2, ending up with an equation of the form α+βy0 = 0 (where β comes from bringing together all of the terms with y0 and α comes from bringing together all the terms without y0). Such an equation has solution simply y0 = −α/β. We will see such situations later in the story problems. 2. IMPLICIT DIFFERENTIATION 51 (e) Here we get: 2(x− h)/a2 + 2(y − k)y0/b2 = 0. Solving for y0, we obtain y0 = − b 2 a2 x− h y − k . Example 7.3. If you know that the curve (xy)4 + x−2 + y−1 = C (where C is a constant) passes through the point (1, 2), Þnd the y-coordinate of the point (1.01, ?) on the curve near (1, 2). Use the tangent line approximation. (Note: The constant C is not given. We could easily compute it from the fact that the point (1, 2) satisÞes the equation. However, we will never need to know the value of C, so we won’t bother to compute it.) In this problem there is no way you can Þnd an exact expression for the y- coordinate corresponding to x = 1.01, because when you substitute x = 1.01 you get an equation for y that cannot be solved algebraically. The equation for y could be solved by an approximation method that we’ll learn later (Newton’s method). In the meantime, the best we can do is to use the tangent line approximation to Þnd the y-coordinate. In this problem we know that the implicitly deÞned y = f (x) has value 2 when x = 1 (this is what it means to say that the curve passes through the point (1,2)). We want to know f (1.01). According to the tangent line approximation, f(1.01) ≈ f(1) + 0.01f 0(1) = 2 + 0.01y0, where y0 denotes the derivative at the point (1, 2). To Þnd that derivative we Þrst take d/dx of both sides of our implicit equation. We use the chain rule and the product rule to Þnd ddx(xy) 4, and on the right we get 0 (the derivative of any constant is 0): 4(xy)3(xy0 + 1 · y) + (−2x−3) + (−y−2y0) = 0. Once we’ve taken d/dx we can substitute the particular x and y values of the point where we want the derivative. After that our equation will look a lot simpler – just some numbers together with the unknown y0: 4(1 · 2)3(y0 + 2) + (−2 · 1) + (−2−2y0) = 0, i.e., y0(32− 0.25) + 64− 2 = 0, i.e., y0 = −62/31.75 = −1.953. Finally, substituting this value of y0 = f 0(1) in the tangent line approximation gives the y-coodinate as 2 + 0.01 · (−1.953) = 1.98047. Example 7.4. Suppose that the ellipse in Example 7.1(d) has foci at the points (0,1) and (3,0), and passes through the point (2,3). Find the y-coordinate of the point (1.99, ?) through which it passes. Use the tangent line approximation. The tangent line approximation tells us that f(1.99) ≈ f(2) − 0.01f 0(2) = 3 − 0.01y0, where y0 is the derivative at the point (2, 3). The implicit differentiation of the equation of this ellipse has already been carried out, so it remains just to substitute the particular values we’re given: x1 = 0, y1 = 1, x2 = 3, y2 = 0, x = 2, y = 3: 2 + 2y0√ 8 + −1 + 3y0√ 10 = 0, so that µ 2√ 8 − 1√ 10 ¶ + y0 µ 2√ 8 + 3√ 10 ¶ = 0, i.e., 0.39088 + 1.65579y0 = 0. Thus, y0 = −0.39088/1.65579 = −0.23607, and so f(1.99) ≈ 3− 0.01y0 = 3.00236. 54 7. IMPLICIT FUNCTIONS AND IMPLICIT DIFFERENTIATION as Þeld equipment ages, its maximum operating settings are generally decreased. Find the equation of the line that best approximates the way in which the holding time would have to be increased as the maximum temperature rating falls slightly below the usual operating temperature. 8. One end of a string (the point A) is attached to the y-axis at a distance 20 meters from the origin; the string is threaded through a hole (the point B) at the end of a pole situated along the line x = 10; and the other end of the string (the point C) is attached to a hook that is free to slide along the x-axis and is being pulled by a force to the right that is sufficient to keep the string taut. See the diagram at the right. At time t = 0, when the top of the pole is at the point (10,10), the pole starts lowering at the steady rate of 30 cm/sec (=0.30 m/sec). The sliding end of the string reaches the point x = 20 at time t = 10. A B CD x (a) Write an equation relating t and x (the time is an implicit function of the location of the sliding end of the string). (b) Use the tangent line approximation to Þnd the time when the sliding end of the string has moved another 20 cm to the left from the point x = 20. CHAPTER 8 Trig Functions and Sinusoidal Functions We start by reviewing trig functions of an angle measured in radians. Because this is review material, we will go through it rapidly. We Þrst draw the unit circle in the xy-plane. The standard way of measuring an angle is to start with the positive x-axis and rotate counterclockwise. As we go through the angle, we sweep around the circumference of the unit circle. The radian measure of the angle is deÞned to be the distance around the circumference that we travel. That is, a full revolution of 360◦ is equal to 2π radians, half a revolution is 180◦ = π radians, and so on. Other frequently used equivalences are: 90◦ = π/2 rad, 60◦ = π/3 rad, 45◦ = π/4 rad, 30◦ = π/6 rad. All angles in this course will be assumed to be measured in radians unless otherwise stated. The designation “radian” will often be omitted, e.g., if we write “an angle of π/6,” we mean a 30◦ angle. By a negative angle we mean a clockwise angle. For example, an angle of −π/2 (i.e., −90◦) brings us to the same position as an angle of 3π/2 (which is 270◦). In general, if we have a circle of radius r and sweep out an angle of θ radians, then the distance around the circumference through which we have traveled is simply θr. Of course, one full circumference is 2πr, corresponding to rotation through 2π radians. Returning to the unit circle (see the drawing below), after rotating through an angle of θ radians let (x, y) be the point where we end up on the circumference. Then the six trig functions are deÞned as follows: sin θ = y cosecant θ = 1 sin θ = 1 y cos θ = x secant θ = 1 cos θ = 1 x tan θ = sin θ cos θ = y x cot θ = 1 tan θ = cos θ sin θ = x y (x, y) θ From the deÞnition of sin θ we see that as θ increases from 0 to 2π (i.e., as we go once around the unit circle), sin θ goes from 0 to 1 to 0 to −1 and back to 0. Then as θ goes from 2π to 4π (i.e., as we go once again around the circle), the same values of sin θ are repeated. We say that the function sin θ has period 2π, i.e., we have sin(θ + 2π) = sin θ. Here are some other relations for the trig functions that follow easily from the above deÞnitions: cos(θ + 2π) = cos θ, sin(θ + π) = − sin θ, cos(θ + π) = − cos θ, tan(θ + π) = tan θ, sin(−θ) = − sin θ, cos(−θ) = cos θ, 55 56 8. TRIG FUNCTIONS AND SINUSOIDAL FUNCTIONS tan(−θ) = − tan θ, sin2 θ + cos2 θ = 1, and cos θ = sin(π2 − θ). The last of these relations is familiar from high school trigonometry: the cosine of an angle is equal to the sine of the complementary angle. It can also be seen from our deÞnitions as follows: the x-coordinate of the point on the unit circle you arrive at when you rotate through θ counterclockwise starting at the positive x-axis is the same as the y-coordinate of the point you arrive at when you rotate through θ clockwise starting at the positive y-axis. Sometimes one also needs the following formulas for the trig functions of a sum of two angles: sin(α + β) = sinα cosβ + cosα sin β, and cos(α + β) = cosα cosβ − sinα sin β. As a special case we have the double angle formulas sin(2α) = 2 sinα cosα and cos(2α) = cos2 α− sin2 α. Also notice the sign of the trig functions in the different quadrants: sinx is positive in the 1st and 2nd quadrants (for angles between 0 and π), cosx is positive in the 1st and 4th, and tanx is positive in the 1st and 3rd. 1. Graphs of Trig Functions The most important graphs to be familiar with are the graphs of the functions y = sinx, y = cosx, and y = tanx. (We’re now using x for the angle in radians; x no longer denotes a coordinate of a point on the unit circle.) For certain x the trig functions have easily stated exact values, namely, when x is a multiple of π/6 or of π/4. Here it is useful to draw the 30◦ − 60◦ − 90◦ triangle with hypotenuse 1 and the 45◦ − 45◦ − 90◦ triangle with hypotenuse 1. Using these triangles, we arrive at a table of values of the trig functions, from which we can sketch the graphs on the next page. 2. Sinusoidal Functions A sinusoidal function is a function made up from the function y = sin x (or y = sin t, if time is playing the role of the x-variable) by inserting constants in various places. The reason for inserting these constants is that the pure sine function y = sin x almost never occurs in the real world, but curves with a sine-like appearance (this is what the word “sinusoidal” means) arise frequently. At this point you should review the material at the end of the section on func- tions: horizontal and vertical shifts, horizontal and vertical dilations, and the last batch of homework problems in that section. Example 8.1. Write the cosine function as a shift of the sine function. The graphs show that the cosine function is obtained by shifting the sine function π/2 to the left. Thus, cosx = sin(x − (−π2 )) = sin(x + π2 ). This relation can also be obtained using the identities listed above: cos x = cos(−x) = sin(π2 − (−x)) = sin(x+ π2 ). We now give the general form for a sinusoidal function y = f(x). We start with y = sin x. Step 1. Dilation. Vertically, we dilate by a factor of A, which is called the amplitude. We do this by replacing sin x by A sinx. This makes the function go to a maximum of A and a minimum of −A (rather than a maximum of 1 and a minimum of −1 as was the case for sin x). Next, we dilate horizontally in such a way that the function repeats its pattern from x to x + B (rather than from x to x+2π as with sin x). B is called the period. That is, we want to dilate horizontally by a factor of B/2π. We do this by replacing x by x/(B/2π) = 2πB x inside the sine. 2. SINUSOIDAL FUNCTIONS 59 We graph y(t) by Þrst plotting the two points t = 30, y = 37 and t = 210 (i.e., July 30), y = 65. We then draw a sine wave in such a way that its minimum is at the Þrst point, its maximum is at the second, and its period is a year, i.e., 360 days. We also draw a horizontal dotted line half-way between the coldest and warmest temperature, i.e., at y = 51. 30 120 210 300 37 51 65 Now we read the four constants A, B, C, D off the graph. The amplitude A is 14, the period B is 360, and the vertical shift (the distance up to the dotted line) is 51. To Þnd C we want to know when the curve crosses the dotted line on its way from a minimum to a maximum, i.e., when the curve looks like sinx looks when it crosses the origin. This can be found by taking the t-value half-way between the coldest day (which is t = 30) and the warmest day (which is t = 210). This t-value is (30 + 210)/2 = 120 (i.e., April 30). Thus, the phase shift is C = 120, and our equation is y(t) = 14 sin µ 2π 360 (t− 120) ¶ + 51. Homework 8. 1. Convert to radians: (a) 15◦, (b) 120◦, (c) a quarter revolution, (d) 10 revolutions. 2. A bicycle wheel has radius 40 cm. Find the rate in rad/sec at which the wheel is turning if the bicycle is traveling at (a) 10 m/sec, (b) 10 km/hr. 3. A drawbridge whose two spans are each 30 meters long is opened as shown below. Express in terms of θ: (a) the height y of the end of a span, (b) the slope of the left span, (c) the distance x between the ends of the spans. x y θ 60 8. TRIG FUNCTIONS AND SINUSOIDAL FUNCTIONS 4. Using the diagram at the right, show that sin(α+ β) = sinα cosβ + cosα sinβ . 5. Sketch the graph of each of the following functions: (a) y = 1 + sin(πx), (b) y = 2 sin ¡ π 3 t ¢ , (c) y = 2 + 2 sin ¡ π 2 (t− 1) ¢ , (d) y = 3 + 2 sin(2t+ 1). 1 β α α A F E D C B 6. The mean daily temperature in Fairbanks, Alaska was tabulated over a 30-year period, the average was taken over the 30 values for each day, and the result was found to be very close to the sinusoidal function 37 sin ¡ 2π 365(t− 101) ¢ + 25, where t is the day of the year.1 (Here we are NOT assuming 30 days per month.) Sketch the graph of this temperature function, Þnd the maximum and minimum mean daily temperatures, and Þnd the warmest and coolest days of the year. 7. Studies are made of the variation in population of foxes and rabbits in a certain large forest. The number changes from year to year. Roughly speaking, when there are a lot of rabbits, the foxes will start to increase in number, because they can eat rabbits plentifully; but then this decreases the number of rabbits, causing the foxes to be short of food, and this causes a decrease in the fox population; and, Þnally, when there are fewer foxes the rabbits will be able to increase in number without being eaten, so we are back to where we started. Thus, we might expect a sinusoidal oscillation in the number of foxes and in the number of rabbits. (This is the so-called “predator—prey” problem in mathematical biology.) Suppose that t stands for the number of years since the study began, F stands for the number of foxes, and R stands for the number of rabbits. Suppose that F and R are each a sinusoidal function of t. Use the following table of data (which gives the maximum and minimum population of each) to Þnd a formula for F and for R in terms of t: t 0 1 2 3 4 F 150 200 150 100 150 R 10000 7500 5000 7500 10000 8.Write a formula for each of the following functions of t, assuming that the function is sinusoidal (also suppose that a year consists of 12 months of 30 days each): (a) the mean low daily temperature, if the coldest is 20◦F on January 19 and the highest low temperature reading is 60◦F on July 19; (b) the time of sunset each day, if the sun sets at 9 p.m. on June 21 and at 5 p.m. on December 21; (c) the temperature in a well-insulated building, which varies between 25◦C at 5 p.m. (which is 17:00 on the 24-hour clock) and 15◦C at 5 a.m; (d) the velocity of a weight attached to the end of a spring, which is extended and released at time t = 0, if the weight reaches its maximum velocity of 10 cm/sec after 0.1 sec; (e) the height above the ground of a pebble that is picked up in the tread of a bicycle tire at time t = 0, if the tire has radius 40 cm and the bicycle is traveling at 10 m/sec. 1This example is from an article by B. M. Lando and C. A. Lando in The Mathematics Teacher, Sept. 1977. CHAPTER 9 Derivative of Trig Functions 1. The Derivative of sin x To Þnd the slope of the tangent line to the curve y = f(x) = sin x, we must go back to the deÞnition of the derivative as the limit of the difference quotient: dy dx = lim ∆x−→0 f(x+∆x)− f(x) ∆x = lim ∆x−→0 sin(x+∆x)− sinx ∆x . Recall the procedure used to Þnd the derivative of the much simpler function x2: expand (x+∆x)2, simplify algebraically, and then take the limit as ∆x gets closer and closer to 0. We do essentially the same thing for sin x, but we need the following three basic facts in order to do this: Fact 1. sin(α+ β) = sinα cosβ + cosα sin β; Fact 2. lim θ−→0 sin θ θ = 1; Fact 3. lim θ−→0 1− cos θ θ = 0. Fact (1) was proved in the last section (Problem 4 of the homework). Facts (2) and (3) can be seen from the diagram below, where θ is a small angle in radians. By the deÞnition of the radian measurement of an angle, θ is the distance along the circumference from A to C. Meanwhile, sin θ = AB and 1− cos θ = BC. θ 1 A B C O sin(θ) θ Thus, Fact (2) says that although the line AB is smaller than the arc AC, the ratio of the two lengths gets closer and closer to 1 as θ gets smaller. Fact (3) says that the ratio of the line BC to the arc AC gets closer and closer to 0 as θ gets smaller. We are now ready to simplify the above difference quotient. We have: sin(x+∆x)− sin x ∆x = sin x cos(∆x) + cosx sin(∆x)− sin x ∆x (by Fact (1) with α = x and β = ∆x) = cosx sin(∆x)− sin x(1− cos∆x) ∆x 61 CHAPTER 10 Curves Given Parametrically So far we have had two different ways of giving a curve. The Þrst was explicitly, i.e., in the form y = f(x). For example, the equation y = √ r2 − x2 gives us the upper semicircle of radius r centered at the origin. The second way was implicitly, usually in the form F (x, y) =constant. For example, the equation x2+y2 = r2 gives us the circle (both upper and lower semicircles) of radius r centered at the origin. There is a third fundamental way a curve can be given: parametrically. This means that we have a third variable, called the parameter, which is used to de- termine where we are on the curve in the xy-plane. To give a curve parametrically means to give two formulas, one for x and one for y, each in terms of the parameter. This parameter might have a geometrical meaning – for example, an angle denoted θ – or it may be the time t. In the case of an angle parameter, our two formulas take the form x = g(θ), y = h(θ); in the case of a time parameter, the formulas take the form x = g(t), y = h(t). (Often we don’t bother to use separate letters like g and h, instead writing simply x(t) for the formula for x in terms of t and y(t) for the formula for y in terms of t.) The best way to understand parameterized curves is through examples. Example 10.1. (Circles) Suppose we want to describe the points on the circum- ference of the circle of radius r centered at the origin. A common way to do this is in terms of the angle θ which the line from the point to the origin makes with the positive x-axis (measured counterclockwise). That is, as θ goes from 0 to 2π radians, we sweep once around the circle: θ r−r (x(θ), y(θ)) What we want to do is express the x- and y-coordinates of our points on the cir- cumference in terms of θ. To do this, we drop the perpendicular from the point to the x-axis, obtaining a triangle with hypotenuse r and with legs x and y. So we can use the trig functions sin and cos to express x and y in terms of θ: x = x(θ) = r cos θ, y = y(θ) = r sin θ, 0 ≤ θ ≤ 2π. 65 66 10. CURVES GIVEN PARAMETRICALLY This says that, as θ goes from 0 to 2π, if we determine x and y by these two equations, then the corresponding point (x, y) goes exactly once around the circumference of the circle. A special case of this is the unit circle – this is simply the case r = 1. Thus, the points on the unit circle can be described parametrically by the equations x = x(θ) = cos θ, y = y(θ) = sin θ, 0 ≤ θ ≤ 2π. The fact that the x-coordinate is simply the cosine and the y-coordinate is simply the sine is not surprising, since we deÞned the cosine and sine of an angle to be the x- and y-coordinates of the point that you get by going around the unit circle through the given angle. If we want to shift the circle of radius r so that its center is at the point (h, k), we can do this by adding h to the x-formula and k to the y-formula: x = x(θ) = h+ r cos θ, y = y(θ) = k + r sin θ, 0 ≤ θ ≤ 2π. This is the circle which is given implicitly by the equation (x− h)2 + (y− k)2 = r2. If instead of 0 ≤ θ ≤ 2π we had written, say, 0 ≤ θ ≤ π, then we would have described only half of the circle (the upper semicircle). If we had written 0 ≤ θ ≤ 4π, then we would have described going around the circle twice. Example 10.2. (Ellipses) Let us suppose that we have put our ellipse on the axes in the following convenient way: its center is at the origin, its longer axis of symmetry – of length 2a – is along the x-axis, and its shorter axis of symmetry – of length 2b – is along the y-axis. Such an ellipse can be obtained by taking the unit circle and dilating it by a factor of a in the x-direction and by a factor of b in the y- direction. (If a or b is less than 1, the “dilation” is actually a contraction.) In terms of the parametric equations, we can accomplish this dilation simply by multiplying the x-equation for the unit circle by a and multiplying the y-equation for the unit circle by b: x = x(θ) = a cos θ, y = y(θ) = b sin θ, 0 ≤ θ ≤ 2π. This is the parametric form for our ellipse – the ellipse which can be given implicitly in the form x2 a2 + y2 b2 = 1. Note: A word of warning, though, about the parameter θ for an ellipse. It no longer has the meaning of the angle formed by the line from the point (x, y) to the origin. For example, suppose a = 2, b = 1. Then the point on the ellipse corresponding to the parameter value θ = π/4 = 45◦ is the point x = 2cos(π/4) =√ 2, y = sin(π/4) = √ 2/2; and it is easy to see that the line joining the point (1.414, 0.707) to the origin does not make a 45◦ angle with the x-axis. 1. Trajectories It is very common to want to describe a curve in terms of the time parameter t. In that case we sometimes call the curve the “path” or “trajectory” of our point (which we think of as a small object or particle). As t goes from an initial time a to a Þnal time b, the equations x(t), y(t) tell us where the particle is at that particular time t. 2. THE VELOCITY VECTOR 69 Þgure). This triangle has hypotenuse 10 m/sec, mak- ing an angle of 60◦ above the horizontal. Then the horizontal leg of the triangle – the initial x-velocity v0,horiz = úx(0) – can be found using the cosine, and the vertical leg of the triangle – the initial y-velocity v0,vert = úy(0) – can be found using the sine. We thus have: 60◦ 10 m/sec 10 sin 60◦ m/sec 10 cos 60◦ m/sec v0,horiz = úx(0) = 10 cos 60 ◦ = 10 · 0.5 = 5m/sec, v0,vert = úy(0) = 10 sin 60 ◦ = 10 · 0.866 = 8.66m/sec. We are now ready to write the equations for x(t) and y(t), as we did in Exam- ple 10.5. The horizontal component of velocity (which remains constant) is not 10 m/sec, but rather only the horizontal part, namely, 5 m/sec. So x = 5t. In the falling body formula y = −12gt2 + v0t+y0, we now have a nonzero initial vertical component of velocity v0 = v0,vert = 8.66 m/sec. Thus, we obtain: x = x(t) = 5t, y = y(t) = −4.9t2 + 8.66t+ 19.6. The trajectory of the ball is shown in the Þgure at the right. To Þnd the time interval until the ball hits the ground, we again set y = 0, and use the quadratic for- mula to Þnd the time t when we’re on the ground. There will be 2 roots, of which we want the positive one: 5 10 15 20 5 10 15 20 x y (−8.66− p 8.662 − 4 · (−4.9) · 19.6)/(2 · (−4.9)) = 3.07 sec. So the time interval is 0 ≤ t ≤ 3.07. Example 10.7. The same as Example 10.6, except that the ball is thrown at an angle of 60◦ below the horizontal. Here the only change is that the tip of the velocity triangle points downward, i.e., its vertical leg úy(0) points in the negative direction. Thus, in the equations of motion the only thing that changes is the sign of v0,vert in the falling body formula: x = x(t) = 5t, y = y(t) = −4.9t2 − 8.66t+ 19.6. Of course, the ball hits the ground much sooner (see Þgure on the following page). In all three cases in Examples 10.5—10.7, the trajectory is part of a parabola, as we can see by eliminating t and writing y in terms of x rather than in terms of t. Suppose that we have a trajectory x(t), y(t), as in Examples 10.3—10.7. There are three types of derivatives we might be interested in. One is the horizontal component of velocity dx/dt = úx(t). The second is the vertical component of 70 10. CURVES GIVEN PARAMETRICALLY velocity dy/dt = úy(t). The third is not a velocity at all, but rather the slope of the curve in the xy-plane: dy/dx. If we Þnd dy/dx at some instant, we can determine the direction in which the object is headed at that instant. How do we Þnd dy/dx if we do not have y written in terms of x, i.e., if we do not have a formula for the function y = f(x)? Well, according to the chain rule, at any instant our three derivatives are connected by the relation dy dt = dy dx · dx dt . In the past we have used the chain rule to Þnd the left hand side. Now we want to Þnd dy/dx. So the form of the chain rule we need is: dy dx = dy/dt dx/dt = úy úx . 2 4 6 8 10 5 10 15 x y In words, this says that the slope of the path is equal to the vertical component of velocity divided by the horizontal component of velocity. This makes sense: if the horizontal component of velocity is small and the vertical component of velocity is large, we are traveling at a steep angle; while if the horizontal component of velocity is large and the vertical component of velocity is small, we are traveling at a shallow angle. In Þnding dy/dx it is important always to remember that you take the ratio of velocities úy over úx, NOT simply the ratio of y to x. In other words, you must take the derivatives (with respect to t) of the y(t) and x(t) formulas before taking their ratio. Example 10.8. Suppose we want to know the instant when the ball in Example 10.6 reaches the peak of its trajectory. At that instant the path has a horizontal tangent line, i.e., dy/dx = 0. Since dy/dx = úy/ úx, this slope is zero if and only if the numerator úy is 0. Thus, to Þnd the instant when the ball reaches its highest point we set the vertical velocity úy equal to 0 and solve for t. This is the same principle that we saw in our earlier one-dimensional falling body problems: the ball is at the peak when its vertical velocity is zero. (Warning: Be sure not to confuse this procedure, where we set úy equal to zero, with setting y itself equal to 0, which we do to Þnd the time when the ball hits the ground.) Since úy = ddt(−4.9t2 + 8.66t+ 19.6) = −9.8t+ 8.66, we have −9.8t+ 8.66 = 0, i.e., t = 8.66/9.8 = 0.88 sec. If we also want to know the maximum height reached by the ball, we take this value of t and substitute it in the formula for the height y, i.e., ymax = y(0.88) = −4.9 · (0.88)2 + 8.66 · 0.88 + 19.6 = 23.4 m. Example 10.9. Suppose that we want to know the direction in which the ball in Example 10.6 is headed after 0.5 sec, and also at the instant when it hits the ground. We Þrst take the time derivatives of y and x: dy dt = d dt (−4.9t2 + 8.66t+ 19.6) = −9.8t+ 8.66 and dx dt = d dt (5t) = 5 . Then we have: dy dx = úy(t) úx(t) = (−9.8t+ 8.66)/5 . 2. THE VELOCITY VECTOR 71 Thus, when t = 0.5 we have dy/dx = (−9.8 · 0.5 + 8.66)/5 = 0.752. At that instant the ball’s location is (x(0.5), y(0.5)) = (2.5, 22.7), so that the tangent line to the path at the point (2.5, 22.7) has slope 0.752. We can also determine the angle at which the ball is traveling at time t = 0.5. By this we mean the angle above the horizontal of the tangent line to the path. To do that we observe that the slope of a line ∆y/∆x is opposite over adjacent, i.e., the tan of the angle that the line makes with the horizontal. So to determine the angle we use the inverse-tan on our calculator1: Arctan(0.752) = 37◦. At the same time t = 0.5 we could also construct the velocity triangle, having horizontal leg úx(0.5) = 5 and vertical leg úy(0.5) = 3.76, and compute the hypotenuse (which is the speed): speed at time 0.5 = p 52 + 3.762 = 6.26 m/sec. 37◦ t = 0.5 sec 5 m/sec 3.76 m/sec −77◦ t = 3.07 sec 5 m/sec −21.4 m/sec Next, we make the same computations at the instant when the ball hits the ground, i.e., at t = 3.07 sec. We have úx(3.07) = 5, úy(3.07) = −9.8 · 3.07 + 8.66 = −21.4, and so speed=p52 + (−21.4)2 = 22.0. The angle at which the ball hits the ground is Arctan(−21.4/5) = −77◦. Thus, the ball hits the ground traveling at 22 m/sec at an angle of 77◦ below the horizontal. 1Arctan is usually denoted tan−1 on calculators. CHAPTER 11 Combined Parametric Motions Sometimes we are interested in the motion of a point P which has a simple motion relative to some other point C, which, in turn, also is moving in a particular way. It is not hard to write parametric formulas x(t), y(t) for the point P , if we proceed carefully in a step-by-step manner. Example 11.1. (The baton) Suppose we want to describe the motion of a tip of a baton as it rotates after being thrown into the air. More precisely, suppose that at time t = 0 a 2-foot long baton is situated vertically with the tip we are interested in at height 6 ft above ground level and the other tip at height 4 ft. At that moment the baton is tossed at 20 ft/sec at an angle of 45◦ above the horizontal, and it is also set spinning at 3 rev/sec clockwise in the plane of its trajectory. (It could have been set spinning in a different plane, but that would have made this into a three-dimensional problem, and that would be too complicated for our present purposes.) To set up the parametric equations for the motion of the tip of the baton, the basic procedure is to divide the motion into two parts: the motion of the center of the baton and the rotation of the tip around the center. The Þrst of these is a falling body problem, and the second is a circular motion problem. We shall take the y-axis to be the vertical line along which the baton is situated at time t = 0, and we shall take the x-axis to be the line on the ground under the baton’s trajectory. According to a principle of physics, the center of the baton follows the same path as a small object (say, a ball) that is thrown with the same initial velocity. In other words, the Þrst half of our procedure – the motion of the center of the baton – is like Example 10.6 of the last chapter. Let (xcent(t), ycent(t)) denote the x- and y-coordinates of the center of the baton at time t. Repeating the procedure used in Example 10.6 – with g = 32 ft/sec2, v0,horiz = 20 cos 45 ◦ = 14 ft/sec (rounded to the nearest ft/sec), v0,vert = 20 sin 45 ◦ = 14 ft/sec – we obtain: xcent(t) = 14t, ycent = −16t2 + 14t+ 5. (Notice that y0 = 5 ft, because the initial location of the center is half-way between 4 and 6.) Now let (x(t), y(t)) denote the x- and y-coordinates of the tip of the baton at time t. The second half of our procedure consists in determining how much x(t) differs from xcent(t) and how much y(t) differs from ycent(t), i.e., how far the tip is from the center in the x- and y-directions. Let θ be the angle in radians through which the baton has spun at time t. We have the following diagram: 75 76 11. COMBINED PARAMETRIC MOTIONS θ (x, y) (xcent, ycent) P Pq 1 ft 1 ft Since the tip is 1 ft from the center, the hypotenuse of this triangle is 1. Hence x− xcent = sin θ and y − ycent = cos θ, i.e., x(t) = 14t+ sin θ, y(t) = −16t2 + 14t+ 5 + cos θ. It remains to write θ in terms of t. We do this as in Example 10.3 of the last chapter, obtaining θ = 6πt. We conclude that x(t) = 14t+ sin(6πt), y(t) = −16t2 + 14t+ 5 + cos(6πt). (Notice that here it is the x-coordinate that involves the sine and the y-coordinate that involves the cosine, the reverse of what we got in our earlier examples of circular motion. The reason is that in this problem it was convenient to measure the angle θ clockwise from the positive y-direction rather than counterclockwise from the positive x-direction. This different way of deÞning the angle accounts for the reversal of the sine and cosine.) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0 1 23 4 5 6 7 8 9 10 11 12 13 14 15 16 178 19 20 The dotted line is the path of the center of the baton: xcent(t) = 14t ycent(t) = −16t2 + 14t+ 5 The solid line is the path of the top of the baton: x(t) = 14t+ sin(6πt) y(t) = −16t2 + 14t + 5 + cos(6πt) Numbers indicate how many twientieths of a second have elapsed. The graph above shows the trajectory of the center of the baton (dotted line) and the trajectory of the tip of the baton (solid line) between t = 0 and t = 1 sec. 11. COMBINED PARAMETRIC MOTIONS 79 The Þgure below shows the point on the curve for several values of θ. θ = 0 θ = 60◦ θ = 120◦ θ = 180◦ θ = 240◦ θ = 300◦ θ = 360◦ θ = 420◦ θ = 480◦ θ = 540◦ θ = 600◦ θ = 660◦ θ = 720◦ Now suppose that we know how the bicycle is moving with time. For example, suppose that it is traveling at constant speed v, and it passes the origin at time t = 0. That is, ds/dt = v. To express the position of the pebble in terms of t, all we have to do is express θ in terms of t, and then substitute in the above equations for (x, y). Since θ has a simple relation to s – namely, θ = 1r s – it follows that dθ/dt has a simple relation to ds/dt, namely (taking d/dt of both sides): dθ dt = 1 r ds dt = 1 r v. Because θ is 0 at time t = 0, and it increases as the constant rate v/r, it follows that we have: θ = (v/r)t. Thus, when the bicycle is traveling at constant velocity v, we obtain the following equations for the motion of the pebble: x(t) = r(v/r)t− r sin((v/r)t) = vt− r sin(vt/r), y(t) = r − r cos(vt/r). 80 11. COMBINED PARAMETRIC MOTIONS Example 11.3. (The trochoid) The same as in Example 11.2, except that when the wheel rolls over the origin, instead of getting stuck in the tread on the outside of the tire, the pebble ßies up and gets stuck in a spoke at a distance a directly below the center (where a < r). Thus, the pebble again rotates around the center, but at a distance a rather than r from the center. The path of this pebble is called a trochoid. In this problem everything is just like in Example 11.2, except that in the circular motion part the hypotenuse of the triangle (which is the line from the center to the pebble after the wheel has rotated through an angle of θ radians) is now a rather than r. This leads to the formulas x(θ) = xcent(θ)− a sin θ = rθ − a sin θ, y(θ) = ycent(θ)− a cos θ = r − a cos θ. Here is a picture of the trochoid path in the case when a = r/2: πr 2πr 3πr r 2r 11. COMBINED PARAMETRIC MOTIONS 81 Homework 11. 1. Suppose that the wheel in Example 11.2 has radius 1. (a) Graph the cycloid by plotting the points where the pebble is located after the wheel rotates through all multiples of 30◦ from 0 to 360◦ (i.e., θ goes through all multiples of π/6 from 0 to 2π). (b) Use the chain rule in the form dy dx = dy/dθ dx/dθ to Þnd a formula for dy/dx in terms of θ. (c) Draw the tangent lines to the curve in part (a) at all of the plotted points from θ = π/6 to θ = 11π/6, and use part (b) to Þnd the slope of each tangent line. 2. A spring is causing an object to oscillate up and down sinusoidally between y = 3 and y = −3. At t = 0 the object is at maximum height y = 3, and it makes one full motion (from 3 to −3 and back to 3) twice a second. Meanwhile, the whole apparatus is moving at constant velocity 2 units/sec down the x-axis, starting at 0 at time t = 0. (a) Find x(t). (b) Find y(t). Namely, use the information given to (i) graph y as a sinusoidal function of time in the ty-plane, and (ii) Þnd A, B, C and D in the formula y = A sin ¡ 2π B (t− C) ¢ +D. (c) Graph the path of the object in the xy-plane. Find the formula for y = f(x) in two ways: (i) Use the graph of the path to Þnd a, b, c and d in y = a sin ¡ 2π b (x− c) ¢ + d (where we’re using small letters for the constants, so as to avoid confusion with the sinusoidal function in part (b)). (ii) Eliminate t, and use part (b) to write y directly in terms of x. Use both methods, and check that your answers agree. 3. The same as Problem 2, except suppose that the whole apparatus is moving down the x-axis according to the formula x(t) = t2. Notice that the vertical motion y(t) as a function of time is unaffected by this change. However, in the horizontal direction the object is accelerating, i.e., its velocity, rather than being constant, is steadily increasing according to the formula v(t) = úx(t) = 2t; and the acceleration a(t) = ẍ(t) = 2 is constant. Here x(t) = t2 can be thought of as 12gt 2, where the constant acceleration g here is 2 units/sec2 (and of course, unlike gravity, which pulls in the negative y-direction, here the constant acceleration is pulling in the positive x-direction). Also notice that your graph of the path y = f (x) is no longer sinusoidal: the distance between peaks becomes greater as the object accelerates to the right. The formula for y = f(x) can be obtained by substituting t = √ x in place of t in the formula in part (b). 4. A wheel of radius 2 ft is centered at the point (0, 100 ft) and is rotating clockwise at the constant rate of one revolution every 2 seconds. At time t = 0 a pebble on the perimeter of the wheel is located at the point (0, 102). At time t = 0 the wheel is dropped. Write down formulas for the x- and y-coordinates of the pebble as functions of time, and graph its trajectory for t between 0 and 2.5 sec (plot points at intervals of 1/4 sec). 5. A wheel rolls down the x-axis from left to right without slipping at constant speed 6 m/sec. The diameter of the wheel is 1.5 m. As it passes over the origin at 84 12. RELATED RATES Note: A crucial point to notice in this and similar problems is that, even though a numerical value is given for r, r must be regarded as a variable, not a constant. The value of the variable at the instant in question cannot be plugged into the equation until after taking d/dt of the equation. Example 12.2. You are inßating a spherical balloon at the rate of 7 cm3/sec. How fast is its radius increasing at the instant when r = 4? Here the variables are the radius r and the volume V . We know dV/dt, and we want dr/dt. The two variables are related by means of the equation V = 43πr 3. Taking d/dt of both sides gives dV/dt = 4πr2 · dr/dt. We now substitute the values we know at the instant in question: 7 = 4π42 · dr/dt. Solving for dr/dt, we obtain dr/dt = 7/64π = 0.0348 cm/sec. Example 12.3. A plane is ßying at 500 mph at an altitude of 3 miles in a direction away from where you’re standing (i.e., the point on the ground directly beneath the plane is moving away from you). How fast is the plane’s distance from you increasing at the moment when the plane is ßying over a point on the ground 4 miles from you? To see what’s going on, we Þrst draw a triangle whose hypotenuse is the line from you to the plane (see Þgure at right). The vertical leg of the triangle is a constant – 3 mi – which does not change in the course of the problem. The horizontal leg x is the variable whose rate we know – its rate úx is the speed of the plane. The hypotenuse y is the variable whose rate we want. The equation relating x and y is the Pythagorean theorem: x2 + 32 = y2. We next take d/dt of both sides, getting 2x · úx = 2y · úy. At the instant in question x y 3 we’re told that x = 4 and úx = 500. What about y? From the Pythagorean theorem, we have: y = √ 42 + 32 = 5 at the instant in question. Thus, we obtain: 2 · 4 · 500 = 2 · 5 · úy, and solving for úy gives the value 400 mph. That is, the plane is receding from you at 400 mph. (Until you think about this for a while, it might seem counterintuitive that, although the side y is greater than the side x, its rate of increase is less.) Example 12.4. Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down (see side view at right), and it has a height of 30 cm and a base radius of 10 cm. How fast is the water level rising when the water is 4 cm deep (at its deepest point)? In this problem, the water forms a conical shape within the big cone whose height and base radius – and hence also its volume – are all increasing as water is poured into the container. This means that we ac- tually have three things varying with time: the water level h (the height of the cone of water), the radius r of the circular top surface of water (the base radius of the cone of water), and the volume of water V . And we have the relation V = 13πr 2h. We know dV/dt, and we want dh/dt. At Þrst something seems to be wrong: we have a third variable r whose rate we don’t know. h r 10 30 But the dimensions of the cone of water must have the same proportions as those of the container. In other words, r and h Þt into the above diagram of similar 12. RELATED RATES 85 triangles, from which it follows that rh = 10 30 , i.e., r = h/3. So we can eliminate r from the problem entirely: V = 13π(h/3) 2h = π27h 3. We now take d/dt of both sides and then plug in h = 4 and dV/dt = 10, obtaining 10 = π273 · 42 · dhdt . Thus, dh/dt = 90/16π = 1.79 cm/sec. Example 12.5. A swing consists of a board at the end of a 10 ft long rope. Think of the board as a point P at the end of the rope, and let Q be the point of attachment at the other end. Suppose that the swing is directly below Q at time t = 0, and is being pushed by someone who walks at the speed of 6 ft/sec from left to right. Find (a) how fast the swing is rising after 1 sec; (b) the angular speed of the rope in deg/sec after 1 sec. In doing this problem, we must start out by ask- ing: What is the geometric quantity whose rate of change we know, and what is the geometric quantity whose rate of change we’re being asked about? Note that the person pushing the swing is moving hor- izontally at a rate we know. In other words, the horizontal coordinate of P is increasing at 6 ft/sec. In the xy-plane let us make the convenient choice of putting the origin at the location of P at time t = 0, i.e., a distance 10 directly below the point of attach- ment. Then the rate we know is dx/dt, and in part Q θ P x y 10 (a) the rate we want is dy/dt (the rate at which P is rising). In part (b) the rate we want is úθ = dθ/dt, where θ stands for the angle in radians through which the swing has swung from the vertical. (Actually, since we want our answer in deg/sec, at the end we must convert dθ/dt from rad/sec by multiplying by 180/π.) (a) From the diagram above we see that we have a right triangle whose legs are x and 10 − y, and whose hypotenuse is 10. Hence x2 + (10 − y)2 = 100. Taking d/dt of both sides – and recalling that the derivative of the square of a function of t is twice the function being squared times the time derivative of that function – we obtain: 2x úx+ 2(10 − y)(0 − úy) = 0. We now look at what we know after 1 sec, namely x = 6 (because x started at 0 and has been increasing at the rate of 6 ft/sec for 1 sec), y = 2 (because we get 10− y = 8 from the Pythagorean theorem applied to the triangle with hypotenuse 10 and leg 6), and úx = 6. Putting in these values leaves us with 2 · 6 · 6 − 2 · 8 úy = 0, from which we can easily solve for úy, getting úy = 4.5 ft/sec. (b) Here our two variables are x and θ, so we want to use the same right triangle as in part (a), but this time to relate θ to x. Since the hypotenuse is constant (equal to 10), the best way to do this is to use the sine: sin θ = x/10. Taking d/dt of both sides, we obtain (cos θ) úθ = 0.1 úx. At the instant in question (t = 1 sec), when we have a right triangle with sides 6-8-10, the cos θ in this equation is 8/10. We also substitute úx = 6 on the right. We get 0.8 úθ = 0.6, i.e., úθ = 0.6/0.8 = 0.75 rad/sec = 43 deg/sec. Sometimes there are several variables that change with time, you know the rates of all but one of them, and you want to know the remaining rate. As in the case when there are just two variables, take d/dt of both sides of the equation relating all of the variables, and then plug in all of the known values and solve for the unknown rate. 86 12. RELATED RATES Example 12.6. A road going South—North crosses a road going West—East at the point O. Car A is driving North along the Þrst road, and car B is driving East along the second road. At time t car A is a(t) to the North of O, and car B is b(t) to the East of O. Suppose that at some instant you know the values of a(t), b(t), the velocity v1(t) = úa(t) of car A, and the velocity v2(t) = úb(t) of car B. Find a formula in terms of these known values for the rate at which the distance between the two cars is increasing. Let c be the distance from car A to car B. By the Pythagorean Theorem, c2 = a2 + b2. We now take d/dt of both sides of the Pythagorean Theorem, obtaining 2c úc = 2a úa+ 2búb. Dividing by 2c gives the following formula for the unknown rate: úc = a úa+ búb c = av1 + bv2√ a2 + b2 . B = (b(t), 0) A = (0, a(t)) O North South West East Notice how this problem differs from Example 12.3. In both cases we took d/dt of the Pythagorean Theorem. However, in Example 12.3 one of the sides was a constant (the altitude of the plane), and so d/dt of the square of that side was simply zero. In Example 12.6, on the other hand, all three sides of the right triangle are variables. As always in a story problem, it’s important to read the problem carefully enough to determine at the start what are the variables and what are the constants. 12. RELATED RATES 89 (a) Express y in terms of a, θ, and β. (b) Suppose you know the rate dθ/dt at which the angle between the blades is decreas- ing. Express dy/dt in terms of a, θ, β, and dθ/dt. (c) Suppose that the distance a is 20 cm, and the angle β is 5◦. Further suppose that you are closing the scissors at the rate of 50 deg/sec. At the instant when θ = 30◦, Þnd the rate (in cm/sec) at which the paper is being cut. A y B C Dθ β 16. In the previous problem, (a) express in terms of dθ/dt the speed of the point on the scissors that moves fastest (this point is B); (b) if we have a = 20 kilometers (i.e., we have scissors of grand proportions) and if β = 0.001 radians, Þnd a value of dθ/dt for which dy/dt is a little faster than the speed of light (c = 300000 km/sec) when the pair of scissors is almost closed; (c) would the situation in your answer to part (b) be possible, or would it violate the Theory of Relativity? Explain. 17. In the situation of Example 12.6, Þnd the rate at which the distance between the cars is increasing or decreasing if (a) car A is 300 meters north of O, car B is 400 meters east of O, both cars are going at constant speed toward O, and the two cars will collide in 10 seconds; (b) 8 seconds ago car A started from rest at O and has been picking up speed at the steady rate of 5 m/sec2, and 6 seconds after car A started car B passed O moving east at constant speed 60 m/sec; (c) the functions a(t) and b(t) are given by the graphs below and you want to Þnd the rate at which the cars are separating at time t = 2 sec. Hint: Estimate derivatives by drawing tangent lines. 1 2 3 4 5 time (seconds) -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 a ( t ) ( m e t e r s ) 1 2 3 4 5 time (seconds) -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 b ( t ) ( m e t e r s ) 90 12. RELATED RATES 18. In Example 12.6 suppose that instead of car B you have a helicopter ßying at speed v2 to the east of O. Let h be the altitude of the helicopter, and let c be the distance from car A to the helicopter. Then the three-dimensional Pythagorean Theorem states: c2 = a2 + b2 + h2. Find a formula for úc in terms of a, b, h and the various rates if: (a) the helicopter is ßying at speed úb at constant altitude; (b) the helicopter is ßying at horizontal speed úb (i.e., the point on the ground under the helicopter is moving at this speed), and at the same time is gaining altitude at rate úh. B A h c b a CHAPTER 13 Curve Sketching In this section we discuss how to sketch the graph of a function y = f(x) without plotting many points. Without having to make up a table of values for the function, we can obtain a good qualitative picture of the graph using certain crucial information – local maxima and local minima, inßection points, asymptotes, etc. Our aim is not to draw an exact graph, but rather to get an accurate overall picture of the graph and to pinpoint the points where something special happens. We start by describing the steps to take in curve sketching. For a particular f(x), not all of the steps below will necessarily lead to useful information. The various possibilities will be illustrated later in the examples. 1. Maxima/Minima If (x, f(x)) is a point where f(x) reaches a maximum (or a minimum), then the tangent line at that point is horizontal, i.e., f 0(x) = 0. Such points can thus be found by setting the derivative equal to zero, and solving for x. (There may be one x for which f 0(x) = 0, there may be many, or there may be none.) Once you Þnd the x for which f 0(x) = 0 (and the corresponding y-coordinate of each possible max/min point), you have to determine whether it really is a maximum or minimum. In simple examples it might be obvious. A systematic way to tell is by the second derivative test, which will be described below. x=a x=b First, we make a few remarks. (1) We will use the term local maximum point instead of “maximum point,” because it may not actually be a maximum value of the function. As in the above drawing, the local maximum might only be the maximum f(x) for all nearby x – the function might later turn again and go up to a higher value. Similarly, we will use the term local minimum instead of “minimum point”. Sometimes, local maxima and minima are referred to as “upper turning points” and “lower turning points.” (2) A maximum or minimum might occur at a point where there is no tangent line, i.e., where f 0(x) does not exist. For example, f (x) = |x| has its minimum point at (0, 0). 91