Derivatives & Applications: Solving Problems with Trig, Parametric, & Implicit Functions -, Exams of Analytical Geometry and Calculus

Download Derivatives & Applications: Solving Problems with Trig, Parametric, & Implicit Functions - and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Name: Math 170-03 Test 2B For problems 1 through 6, find the derivative. Do not simplify. (5 points each) 1. y = cscx For this, we simply need to remember the derivatives of the trigonometric functions. Recall that d dx (cscx) = − cscx cotx 2. y = tanx Recall that d dx (tanx) = sec2 x 3. y = secx tanx Here we use the product rule in addition to the trigonometric derivatives: y′ = secx · d dx (tanx) + tanx · d dx (secx) y′ = secx · sec2 x+ tanx · secx tanx 4. y = − csc2 x Here we use the chain rule in addition to the trigonometric derivatives: y′ = −2 cscx · d dx (cscx) y′ = 2 cscx · (− cscx cotx) 5. y = (x2 + 5x− 3)−4 Here we use the chain rule: y′ = −4(x2 + 5x− 3)−5 · d dx (x2 + 5x− 3) y′ = −4(x2 + 5x− 3)−5 · (2x+ 5) 6. y = ( x+ 1 x+ 2 )4 1 We again use the chain rule. Note that we also use the quotient rule when taking the derivative of the “inside”: y′ = 4 ( x+ 1 x+ 2 )3 · d dx ( x+ 1 x+ 2 ) y′ = 4 ( x+ 1 x+ 2 )3 · ( (x+ 2) ddx (x+ 1)− (x+ 1) d dx (x+ 2) (x+ 2)2 ) y′ = 4 ( x+ 1 x+ 2 )3 · ( (x+ 2) · 1− (x+ 1) · 1 (x+ 2)2 ) 7. Find the linearization of f(x) = 3 √ x at x = 8. (5 points) Recall that the linearization of f(x) at x = a is the function L(x) = f(a) + f ′(a)(x− a) This really just a form of the tangent line to f(x) at x = a. In this case, we choose a = 8. Then f(a) = 3 √ 8 = 2. We also need to find f ′(x). Let us rewrite f(x): f(x) = x1/3 Applying the power rule, we get: f ′(x) = 1 3 x−2/3 Evaluating f ′(8): f ′(27) = 1 3 · 8−2/3 f ′(27) = 1 3 · (81/3)−2 f ′(27) = 1 3 · 2−2 f ′(27) = 1 3 · 1 4 f ′(27) = 1 12 Plugging into L(x), we get: L(x) = 2 + 1 12 (x− 8) 8. Approximate 3 √ 8.12. (5 points) Recall that f(a+ ∆x) ≈ f(a) + f ′(a)∆x Here we choose f(x) = 3 √ x, a = 2, and ∆x = .12. We did most of the work already in #7, and so we can just plug values into our formula: 3 √ 8.12 ≈ 2 + 1 12 · .12 2 y′′ = − 2y2 y + 4x2 y y2 y′′ = −2y 2 + 4x2 y · 1 y2 y′′ = −2y 2 + 4x2 y3 For problems 13 through 16, consider the function f(x) = x2 − 6x + 5 defined on the interval [0, 4]. (5 points each) 13. Find all critical points of f(x). Recall that a critical point occurs when either the function is not differentiable or its derivative is zero. Since f(x) is a polynomial function, it is always differentiable, and thus we need only find where its derivative is zero. f ′(x) = 2x− 6 0 = 2x− 6 2x = 6 x = 3 So the only critical point occurs when x = 3. 14. Find the absolute maximum value. To find absolute extrema, we evaluate the function at all its critical points and endpoints. Doing this: f(0) = 5 f(3) = 9− 18 + 5 = −4 f(4) = 16− 24 + 5 = −3 The largest y-value occuring on this interval is 5, and therefore it is the absolute maximum value. 15. Find the absolute minimum value. From #14, we notice that the least y-value on the interval is −4, and therefore it is the absolute minimum value. 16. Find the value or values of c that satisfy the conclusion of the Mean Value Theorem. The Mean Value Theorem tells us that, for a function which is continuous on [a, b] and differentiable on (a, b), there exists some c ∈ (a, b) such that f ′(c) = f(b)− f(a) b− a 5 We have already done most of the calculations in the above problems. Evaluating the right hand side of the equality, we get: f(4)− f(0) 4− 0 = −3− 5 4 = −8 4 = −2 So we want to find c such that f ′(c) = 2, or 2c− 6 = −2 Solving for c, 2c = 4 c = 2 This value of c certainly lies inside the interval (0, 4), so we are done. 17. A 5-yd ladder is leaning against a house when its base starts to slide away. By the time the base is 4 yd from the house, the base is moving at the rate of 6 yd/sec. How fast is the top of the ladder sliding down the wall then? (20 points) First, we draw a picture of the situation. The ladder leaning against the house makes a right triangle. Let’s call the distance between the house and the base of the ladder x and the distance between the ground and the top of the ladder y. Note that the length of the ladder is constant, so the hypotenuse of this triangle will always have length 5. x y 5 Our goal is to find the rate at which the top of the ladder is moving, or dy dt . Using the Pythagorean theorem, we can write an equation which relates x and y: x2 + y2 = 52 Now we differentiate both sides of this equation with respect to t: 2x dx dt + 2y dy dt = 0 Dividing by 2: x dx dt + y dy dt = 0 In order to solve for dy dt , we need to plug in values for x, y, and dx dt . We are given that x = 4 and dx dt = 6. We apply the Pythogorean theorem again to determine y: 42 + y2 = 52 6 y2 = 25− 16 y2 = 9 y = ±3 Since we are dealing with lengths, we throw out the negative solution and determine that y = 3. Now we can plug all the appropriate values into our equation: 4 · 6 + 3dy dt = 0 3 dy dt = −24 dy dt = −8 So the top ladder is moving at 8 yd/sec. Note that the derivative is negative since the distance between the top of the ladder and the ground is getting smaller. Extra credit. The parametric equations in problems 9 and 10 and the implicitly defined function in problems 11 and 12 both describe the graph of the same geometric shape. What is this shape called? This shape is an ellipse. Just like a circle, it can be described parametrically using trigonometric functions or implicitly by a polynomial in x and y. Ellipses should be familiar if you read section 3.5 carefully or did the extra credit assignment in Math 143. The one in problems 9 through 12 in particular has a semimajor axis of length √ 2 and a semiminor axis of length 1. Here is a graph: x y 7