Topics for the First Exam - Analytical Geometry and Calculus III | MATH 208, Exams of Analytical Geometry and Calculus

Download Topics for the First Exam - Analytical Geometry and Calculus III | MATH 208 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Math 208H Topics for the first exam Chapter 9: Parametric curves The motivation: think of the graph of y = f(x) as a path that we are walking along. The ’right’ way to think of this is that we are visiting each point of the graph at various times t, e.g., x = t , y = f(x) = f(t) But we need not be limited to having x = t ; we can more generally describe our path as x = x(t) , y = y(t) This is a parametric curve; it describes a curve in the plane, and how we traverse it through time. The advantage is that the curve we describe need not be the graph of a function. t = the parameter = the independent variable ; x and y = dependent variables A circle of radius 1 centered at (0, 0) : x2 + y2 = 1 x(t) = cos t , y(t) = sin t 0 ≤ t ≤ 2π Twice as fast around: x(t) = cos 2t , y(t) = sin 2t 0 ≤ t ≤ π A circle of radius r centered at (a, b) : (x− a)2 + (y − b)2 = r2 Think: x− a = r cos t , y − b = r sin t x(t) = a+ r cos t , y(t) = b+ r sin t 0 ≤ t ≤ 2π An ellipse: (x/a)2 + (y/b)2 = 1 x(t) = a cos t , y(t) = b sin t 0 ≤ t ≤ 2π A line through (a, b) and (c, d) x(t) = a+ t(c− a) , y(t) = b+ t(d− b) Finding an (x, y) equation from a parametric equation: (if possible) solve for x = x(t) or y = y(t) as t = expression in x or y, then plug into the other equation. Ex: x = t2 − 1 , y = t3 + t − 1 , then x + 1 = t2 so t = ±√x+ 1, so y = (±√x+ 1)3 + (±√x+ 1) − 1 Calculus of curves Thinking of a parametric curve as a path that we are traversing, we are at each instant aware of (at least) two things: how fast we are going and what direction we are going. Each can be computed essentially as we would for a graph. Speed = the limit of (distance)/(time interval) as the time interval shrinks to 0. average speed = √ (∆x)2 + (∆y)2/∆t = √ (∆x/∆t)2 + (∆y/∆t)2 instantaneous speed = √ (dx/dt)2 + (dy/dt)2 = √ (x′(t))2 + (y′(t))2 direction = slope of tangent line - limit of slopes of secant lines secant lines: slope = ∆y/∆x = (∆y/∆t)/(∆x/∆t) tangent lines: slope = (dy/dt)/(dx/dt) = y′(t)/x′(t) We can encode both of these in the velocity vector (x′(t), y′(t)) A parametric curve x = x(t) , y = y(t) , a ≤ t ≤ b with x(a) = x(b) , y(a) = y(b) ends where it begins; it is a closed curve. Such a curve surrounds and encloses a region R in the plane. If the curve goes around the region counterclockwise, then the area of the region can be computed as Area = ∫ b a x(t)y′(t) dt = − ∫ b a y(t)x′(t) dt We will see why this formula is true later in this class.... 1 Arclength and surface area Just as with graphs of functions, we can compute the length of a paramentric curve and the surface area when a curve is rotated around an axis: Length: we approximate it the same way, as a sum of lengths of line seqments that ap- proximate the curve. Each segment has length√ (∆x)2 + (∆y/)2 = √ (∆x/∆t)2 + (∆y/∆t)2∆t ≈ √(x′(t))2 + (y′(t))2 dt so the length of the curve is ∫ b a √ (x′(t))2 + (y′(t))2 dt Surface area: if we spin the curve x = x(t) , y = y(t) , a ≤ t ≤ b around the line y = c then, just like before, we can approximate the surface by frustra of cones, each having area approximately 2π|y(t)− c|√(x′(t))2 + (y′(t))2 dt = (2π)(radius)(length) and so the area of the surface of revolution is 2π ∫ b a |y(t) − c| √ (x′(t))2 + (y′(t))2 dt Ex: for the ellipse x = 3 cos t , y = 5 sin t , 0 ≤ t ≤ 2π, spun around y = 7, we have Area = 2π ∫ 2π 0 (7 − 3 sin t) √ 9 sin2 t+ 25 cos2 t dt = 2π ∫ 2π 0 (7 − 3 sin t) √ 9 + 16 cos2 t dt Polar coordinates Idea: describe points in the plane in terms of (distance,direction). r = (x2 + y2)1/2 = distance , θ = arctan(y/x) = angle with the positive x-axis. x = r cos θ , y = r sin θ The same point in the plane can have many representations in polar coordinates: (1, 0)rect = (1, 0)pol = (1, 2π)pol = (1, 16π)pol = . . . A negative distance is interpreted as a positive distance in the opposite direction (add π to the angle): (−2, π/2)pol = (2, π/2 + π)pol = (0,−2)rect An equation in polar coordinates can (in principal) be converted to rectangular coords, and vice versa: E.g., r = sin(2θ) = 2 sin θ cos θ can be expressed as r3 = (x2 + y2)3/2 = 2(r sin θ)(r cos θ) = 2yx, i.e., (x2 + y2)3 = 4x2y2 Given an equation in polar coordinates r = f(θ) , i.e., the curve (f(θ), θ)pol, θ1 ≤ θ ≤ θ2 we can compute the slope of its tangent line, by thinking in rectangular coords: x = f(θ) cos θ, y = f(θ) sin θ , so dy dx = dy/dθ dx/dθ = f ′(θ) sin θ + f(θ) cos θ f ′(θ) cos θ − f(θ) sin θ Arclength: the polar curve r = f(θ) is really the (rectangular) parametrized curve x = f(θ) cos θ, y = f(θ) sin θ, and (x′(θ))2 + (y′(θ))2)1/2 = (f ′(θ))2 + (f(θ))2)1/2, so the arclength for a ≤ θ ≤ b is displaystyle ∫ b a (f ′(θ))2 + (f(θ))2)1/2 dθ Area: if r = f(θ) , a ≤ θ ≤ b describes a closed curve (f(a) = f(b) = 0), then we can compute the area inside the curve as a sum of areas of sectors of a circle, each with area approximately 2 in 3-space, there is a third vector which is perpendicular to both of them. Given the two vectors v = 〈a1, a2, a3〉 , w = 〈b1, b2, b3〉 we can solve the pair of equations a1x+ a2y + a3z = 0, b1x+ b2y + b3z = 0 to find that〈a2b3 − a3b2,−(a1b3 − a3b1), a1b2 − a2b1〉 is a solution. We call this vector the cross product v × w for v and w . How do you remember this formula? Most people remember it using the notation v × w = ∣∣∣∣∣∣ i j k a1 a2 a3 b1 b2 b3 ∣∣∣∣∣∣ = ∣∣∣∣ a2 a3b2 b3 ∣∣∣∣i− ∣∣∣∣ a1 a3b1 b3 ∣∣∣∣j + ∣∣∣∣ a1 a2b1 b2 ∣∣∣∣k where ∣∣∣∣ a bc d ∣∣∣∣ = ad− bc is the determinant of the 2 × 2 matrix The cross product satisfies several useful equalities: v • (v × w) = 0, w • (v × w) = 0 v × w = −w × v (kv) × w = k(v × w) v × (w + u) = v × w + v × u u • (v × w) = v • (w × u) = w • (u× v) (the triple product) u× (v × w) = (u • w)v − (u • v)w For our standard vectors in 3-space we have i×j = k , j × k =i , k ×i = j Our formula for the cross product was worked out just by solving a pair of equations; any other multiple of our vector would have been perpendicular to v and w, too. But in a precise sense, the formula we came up with is the right one, because the length of our vector has geometric significance: ||v × w|| = ||v||||w|| sin θ , where θ = angle between v and w . But! The area of a parallelogram with sides equal to the vectors v and w is Area = (base)×(height) = ||w|| · h = ||w|| · ||v|| · sin(θ) (from trigonometry). So: ||v × w|| = ||v|| · ||w|| · sin(θ) = the area of that parallelogram! The cross product can be used to carry out many calculations which we will find useful. For example, to compute the distance d from a point P to the line through the points Q and R, we find that (setting v = −−→QP , w = −−→QR) using right triangles we have d = ||v|| sin θ = (||v||||w|| sin θ)/||w|| = ||v × w||/||w|| Also, to compute the volume of a parallelopiped with sides u,v, w, we can compute volume = (area of base)·(height) = ||u× v|| · (||w||| cosψ|) where ψ = angle between u× v and w, so volume = |(u× v) • w| = absolute value of the triple product! Lines and planes in 3-space Just as with lines in the plane, we can parametrize lines in space, given a point on the line, P , and the direction v that the line is travelling: L(t) = (x(t), y(t), z(t)) = P + vt = (x0 + at, y0 + bt, z0 + ct) This involves a (somewhat arbitrary) parameter t to describe; we can find a more symmetric description of the line by determining, for each coordinate, what t is and setting them all equal to one another: x− x0 a = y − y0 b = z − z0 c 5 To determine if and where two lines in space intersect, if we use the parametrized forms, we need to remember that the two lines might pass through that same point at different times, and so we really need to use different names for the parameters: P + vt = Q+ ws This gives us three equations (each of the three coordinates) with two variables; it therefore usually does not have a solutions. Two lines in 3-space that do not meet are called skew. If two lines do meet, then we can treat them much like in the plane; we can, for example, determine the angle at which they meet by computing the angle between their direction vectors v, w . For planes, three points P , Q and R that do not lie on a single line will have exactly one plane through them. To describe that plane, we can think of it as all points X so that−−→ PX can be expressed as a combination of −−→PQ and −→PR. This in turn means that −−→PX is perpendicular to anything that is simultaneously perpendicular to both −−→PQ and −→PR. But the cross product is such a vector; and so we can describe the plane by insisting that−−→ PX • (−−→PQ×−→PR) = 0 If we write −−→PQ × −→PR = 〈a, b, c〉 and −−→PX = 〈x − x0, y − y0, z − z0〉, then this equation becomes a(x− x0) + b(y − y0) + c(z − z0) = 0 What is really needed to describe this plane, in some sense, is the point P = (x0, y0, z0) and the vector (N) = 〈a, b, c〉 = the normal vector to the plane. In other words, to completely describe a plane we can also use knowledge of a single point that the plane passes through, P , and what direction “up” is, namely the vector (N) perpendicular to the plane (i.e, the vector perpendicular to every vector lying in the plane). We can then write the equation for the plane as 〈x, y, z〉 • N = P • N Note that if we are given the equation for the plane, we can quickly read off its normal vector; it is the coefficients of x, y, and z. Intersecting planes: typically, two planes will intersect in a line (unless they are parallel, i.e., their normals are multiples of one another). We can find the parametric equation for the line by solving each equation of the plane for x, say, as an expression in y and z. Setting these two expressions equal, we can express y, say, as a function of z. Plugging back into our original expression for x, we get x as a function of z. So x, y, and z have all been expressed in terms of a single variable, z, which is exactly what a parametric equation does! The direction vector for this line, it is worth pointing out, is the cross product of the normals to the two planes; this direction vector points in a direction lying in both planes, and so much be perpendicular to both normals. 6