r Parametric, Vector, and Polar Functions (BC Only) J

Chapter 8. r Parametric, Vector, and Polar Functions (BC Only) J When a particle moves in the xy-plane, the parametric equations x = f (t) and y = g (t) can be used to represent the location of the particle and the path that it traces in the plane. The distinctive characteristic of all parametric curves is that both the x- and y-coordinates of a point on the curve are defined in terms of another variable, the parameter t. Unlike Cartesian curves where the independent variable is commonly named x and the dependent variable y, on a parametric curve, the parameter t is the independent variable, while both x and yare dependent variables. When a curve is defined parametrically, equations for x(t) and y(t) are given. In addition, a domain of values for t is specified. Parametric equations can be used to describe the motion of a particle in the coordinate plane. In this context, we think of the parameter t as representing time. Often the motion starts at t = 0 (or at some point t> 0) and proceeds in the direction of increasing t. Before we get into the calculus of parametrically defined curves, let us look at a few examples that illustrate some important skills you need to have. rt There are many ways to solve this problem. One is to first find a Cartesian equation for the line containing the segment. 1-(-3) The slope is --- = 2, and an equation for the line is 4-2 y - (-3) = 2(x - 2) or y = 2x -7. With this equation, creating the parametric equations is easy. Just define x(t) = t and y(t) = 2t -7. Since the smaller value of x is 2 and the larger is 4, choose values of t between 2 and 4, inclusive. calculator in parametric mode. Try graphing this with your Another possible solution is to pretend that a particle is moving with a constant speed from the point (2, -3) straight to the point (4, 1). Suppose the particle starts at t = 0 and reaches the destination point at t = 1. Then the equations should be x(t) = 2 + 2t and y(t) = -3 + 4t. Ii

IA graph of any function y = f(x) can be parametrized by setting x(t) = t and yet) = f(t). For example, a graph of the function f(x) = Inx can be parametrized with x(t) = t and y(t) = In t, for t> O. Of course, there are other ways to create parametric curves. One of the most basic parametrized curves is the unit circle, which we can represent as x = cost, y = sint, 0 ~ t < 21r. Ifwe follow the movement of two particles simultaneously, their paths can cross any number of times. Such points are called intersection points. It is further possible that the particles will occupy the same point in the plane at the same value of t. Such points are called collision points. The motion of Particle 1 in the plane for 0 ~ t ~ 10 is described by the equations x] (t) = t + 1 and YI (t) = t 3-5t Z + 7t + 3. The motion of Particle 2 over the same time interval is described by the equations X z (t) = 7e' - 6 and Yz (t) = 4ge' - 46. Find all points where the paths of the particles intersect and all points where the particles collide, if any. ri For the paths to intersect, we need x] (t]) = Xz (tz) and Y] (t 1 ) = Yz (tj. Here we get t] + 1= 7e'2-6 and t l 3-5t]z + 7t[ + 3 = 4ge'2-46. Multiplying the first equation by 7 and subtracting it from the second to eliminate e t2, we get t]3-5tlz+7t]+3-7(t]+i)=-4 => t I 3-5t1 Z =0 => t I Z (t l -5)=0. Therefore,the intersection points are at t l = 0 and t] = 5. These points are (1, 3) and (6,38). At these. d!n 12. 1 I h.. ( pomts, t z = 0 an t z = -, respective y. As we can see, on y t e startmg pomt 1,3) at 7 t = t] = t z = 0 is a collision point. At t l = 5 and t z = In ~, respectively, the particles pass 7 through the same intersection point (6,38), but they pass through it at different times and do not collide. PJ

Tangent Lines A parametrized curve x = x (t), Y = Y (t) is differentiable at t = to if x and y are differentiable at t = to. The curve is differentiable on an open interval t 1 < t < t 2 if x' (t) and y' (t) exist for all values of t in the interval. If both x' (t) and y' (t) exist and are not simultaneously zero, then the curve has a tangent line at t. IThe slope of that tangent line is dy = dy/dx. dx dt dt This follows from the Chain Rule: if y = y (x (t)) then dy = dy. dx. Note that the dt dx dt tangent line is vertical where dx = 0 and dy *- O. dt dt In our earlier example of the unit circle x = cost, y = sint; 0 ~ t < 2Jr, dx = -sint and dt dy = cas t. The slope ofthe curve is dy = - c~s t, which is defined for all points at dt dx smt which sin t*-o. If sin t = 0, then the y-coordinate of the point on the circle is 0, and the circle has a vertical tangent. A curve is defined by the parametric equations x = 5t 2-7t, y =.!..t 3 + 6t. What is the 3 (A) 10 13 ri dy dy/dx t 2 + 6 dx = dt dt = 1Ot- 7. JJl (B).!2 10 10 13 The answer is A.

What is an equation for the line tangent to the curve with parametric equations x =~, y = Jt+1 at the point where t = 3? t (A) Y-2=-~(X-±) (B) Y-2=±(X-±) (C) y-2 = - :( x-±) (D) y- ~ =-~(x+i) (E) y-±=-~(x+i) I The point of tangency is (x (3), y (3)) = ( ±' 2). The slope of the tangent is I I dy dy/dx 2Jt+1 dyi "4 9.. -.- = - - = 1 => - = -1 = --. Puttmg these two results together mto a dx dt dt =- dx 1=3 =- 4 t 2 9 linear equation, we get Y - 2 = - ~ ( x - ±), answer C. I A particle moves along a curve in the xy-plane according to the equations 3 2 I 3 X = t - 3t, Y = - t - 4t. particle's I 3 path is vertical? (A) 0 only (D) -2 and 2 only What are all the values of t for which the tangent line to the (B) 2 only (E) -2, 0, and 2 We must find the points where dx = 0 and dy *- O. dx = 3t 2-6t = 0 => t = 0,2 and dt dt dt dy = t 2 _ 4 *- 0 => t *- -2,2. The value t = 2 would make the slope indeterminate, but dt the curve does not have a vertical tangent at that point because lim t: - 4 = t + 21 = ~, and thus is finite. Therefore, the only value for which the 1-'> 2 3t -6t 3t 1=2 3 tangent line is vertical is t = O. The answer is A. I

A point where the slope is indeterminate, that is, where both dy and dx are zero, may be dt dt at a cusp on the curve. If both dy and dx are continuous, the curve still has a tangent dt dt line at the cusp point and we can find its slope by taking the limit of the slope when t approaches the value that produces the cusp. In the above example, the slope of the tangent line at t = 2 can be found as lim dy = lim t: - 4 = ~. The point on the curve H2 dx H2 3r - 6t 3 where t = 2 is (x (2), y (2)) = ( -4, - 1 3 6), so the tangent line at that point has an equation ( y +~ = ~ (x + 4) (Figure 8-1). The tangent at t = T can be vertical, too, if lim dy = 00 3 3 HT dx 5 3 1 dy 1" 3t 2 ) e.g.,x=t, y=t, Im-= Im- 4 =00. HO dx HO 5t Tangent line \'/ 2 To find d ~ dx for a parametrically defined curve, you need to carefully use the Chain Rule. Try the following example. 2 If x(t) = t 2 - t, yet) = e 21, determine d ~ dx at the point where t = 2.

I dx dy 2/ dy 2e 2 / -=2t-l and -=2e => -=-- dt dt dx 2t-l Length of a Parametric Curve In Chapter 6 we reviewed the methods for finding the arc length of a curve when y is a function of x. The arc length for a parametric curve can be found with a similar method. If a smooth curve x = x (t ), y = y (t ), a ~ t ~ b, is traced exactly once as t increases from a to b, then the arc length of the curve is The formula in Chapter 6 can be obtained from the this one by parametrizing the curve as x(t)=t,y=y(x(t)). Then dx=l, dy=dy,andl= r 1+(d y )2dt. dt dt dx a dx IBe careful: particle's distance from its initial position - ~( x(t) - X(O))2+ (y(t) - y(o))2 - is usually not the same as the length of particle's path when the particle travels from 0 to t. What is the length of the path described by the parametric equations x = cos(3t), y = sint; 0 ~ t ~ 21l"?

dx = -3 sin (3t) and dy = cos t. The arc length formula gives dt dt L= J:Jr ~(-3sin(3t)r +(cost)2 dt lil~l3.065,answerd. Integrals for arc length usually cannot be evaluated by finding an antiderivative. Even when they can, it is often a very tedious task. If the problem is on the calculator portion of the exam, be prepared to use the numerical integrator on your calculator. Since the formula is complicated, double-check your input as to the number and placement of parentheses, exponents, and so on. Your calculator may take some time to complete the integral; be patient and start the next problem while waiting for your calculator to finish. Ii The length of the curve described by the parametric equations x = (In t)2, Y = ~ (3t + 1)3, 1 ::::;t ::::; 2 is given by 3 (A) r JI l 2 (3t + 1)2 J2 1+ dt [ 2Int 12 2In t 2 (B) -+3(3t+l) dt I t (C) r~(2int)2 +(3t+lr dt (D) f UJ +(31+1)' dl (E) f el;i)' +9(31+1)' dl ri dx = 2ln t and dy = ~. 3(3t + 1)2.3 = 3(3t + 1)2. dt t dt 3 The arc length formula gives L ~r e~i)' + (3(31+ 1)')' dl, which is equivaientto answer E. Ii

II dx = (1+ cos t)( - sin t) + (cos t)( - sin t) = (- sin t) (1+ 2 cos t) and dt dy = (1+ cos t)( cos t) + (sin t)( - sin t) = cos t + cos 2 t - sin 2 t. dt starts at t = 0 and stops at t = 2Jr. The arc length is One pass around the curve L= f:it Ii ((-sint)(1+2cost)f +(cost+cos 2 t-sin 2 tfdt=8. Algebraically, an n-dimensional vector is an ordered set of n numbers. A two-dimensional vector is a pair of numbers (X, y) called components of the vector. We can add two vectors and multiply a vector by a real number. These operations are performed on respective components: (Xl' y\) + (x2, Y2) = (XI + x2' YI + Y2) e (x, Y) = (ex, ey) Geometrically, a two-dimensional vector is a directed segment between two points on the plane. Two vectors are deemed equal if they have the same length and direction (Figure 8-2).

A vector (x, y) can be drawn as a directed segment that goes from the origin to the point on the plane with coordinates x and y. Thus, there is a one-to-one correspondence between two-dimensional vectors and points on the plane. But a vector can be drawn between any two points on the plane as needed for better visualization. For example, when a particle is moving on the plane, its position can be described as a vector (x, y). This vector connects the origin to the point at which the particle is located. The velocity of the particle is also a vector. It is often convenient to draw the velocity vector from the point of the current location. The acceleration of the particle is also a vector, but its geometric representation is rarely useful. I.... The length (or magnitude) of a vector (x, y) is given by ~ x 2 +l " direction f) is described by tan f) = Y. x The length of a vector v is represented by Ivl. (Some textbooks use the vector notations (x, y) or xi + yj. You are allowed to use both of these standard notations in your answers on the AP exam, but AP questions use parentheses, and in this book we do the same.) When a particle moves on the xy-plane, the coordinates of its position can be given as parametric functions x = x (t) and y = y (t) for some time interval a::::; t ::::; b. The particle's position vector s(t) = (x(t), y(t)) is a vector function of t. The coordinates of the position at time t are the same as the components of s at time t. Therefore, II a vector function is essentially a different notation for a parametric function.

A vector function s(t) = (x (t), y (t)) is differentiable at t if x and y have derivatives at t. The derivative of s(t),!!-s(t), is defined as the vector (x' (t), y' (t)). A vector function dt is said to be differentiable if it is differentiable at every point in its domain. If (x (t ), y (t )) is the position vector of a particle moving along a smooth curve in the xy-plane, then, at any time t, 1. The particle's velocity vector v (t) is (x' (t), y' (t)) ; if drawn from the position point, it is tangent to the curve. 2. The particle's speed along the curve is the length of the velocity vector, Iv (t)1 = ) (x' (t )r + (y' (t )r. 3. The particle's acceleration vector a(t)is (x"(t), y"(t)), is the derivative ofthe velocity vector and the second derivative of the position vector. If the position vector of a particle moving along a curve is defined by s (t ) = (e-i, - sin t), then the velocity vector of the particle at t = 0 is (A) (-1,0) (B) (1,0) (C) (-1,1) (D) (-1, -1) (E) (1, -1) g v(t)=(-e- I, -cost) => v(o) = (_e- O, -coso)=(-1, -1). TheanswerisD. JJ A particle moves along a curve so that x (t ) = 6t 2 + In t and dy = sin t + 2. What is the dt speed of the particle when t = 2?

II The particle's speed along the curve is IV(I)I = (:)' +(:)' ~ (12t+D' +( sinl+ 2)'. When 1~ 2, the speed is Ii ffi]] (12. 2+~)' +(sin2+ 2)' = 24.672, answer C. A particle moves along the graph of y = x 2 + X, with its x-coordinate changing at the rate dx 2 ) of - = 3"" for t> O. If x(1 = 1, find dt t (a) the particle's position at t = 2; (b) the speed of the particle at t = 2. fdx x(t)= -dt= f3""dt=-z+c. 2 1 Smce.) x(1 =1, C=2. Now x=-z+2 1 and dt t t t x(2) = 2. Substituting the value of x(2) = 2 into the given equation for y, we get 4 4 7 7 77.... 7 77 Y = - ( ) 2 +- = -. So the particle's position is ( -, - ). 4 4 16 4 16 The speed of the particle at t = 2 is Iv (2)1 = function of x, we use the Chain Rule to get dy, as follows: dy = dy. dx = (2x + 1)( ~ ). dt dt dx dt t When t = 2 and x = 2 dy = (~) 4' dt 2 4 8 IV(2)1~ (H +(H.=1152. (.!-) = 2.. So, the speed is

An alternative method in both Parts (a) and (b) is to find x(t), then a formula for y in 2 (1 )2 (1 ) 1 5. terms oft: y=x +x= -(2+2 + -(2+2 =1-(2+6. Then we can use ltto find.. 1 5 77 dyl (4 10) 9 the position and the speed: y(2)=---+6=- and - = -5+-3 = 16 4 16 dt 1=2 t t 1=2 8 JJ A particle moves along a curve with its position vector given by (3cos ( ±). 5 sin ( ~ ) : for t :::::O. Find its acceleration vector when the particle is at rest for the first time. II When the particle is at rest, the velocity vector, ( - %sin ( ±). ~cos( ~ ) ), must be equal to the zero vector (0,0). -%sin(±)=o => t=0,2n,4n,... and % cos ( ±) = 0 => t = 2n, 6n, 1On,..., so the first value for t when the particle is at rest is 2,.. The acceleration vector is ( -~cos( ±). - 1 5 6sin( ~)) and at t ~ 2,., the acce 1. eratlon vector IS. (3-,- - 5). 4 16 JJ A roller coaster track has an inverted loop as a portion of its course. The position of the car on the track (in feet) at time t seconds, 0 S; t S; 7, is given by the equations x = 5t -12 sin (t + 2) + 10 y = 15 + 12cos (t + 2)

g y reaches maximum when cos(t + 2) is equal to 1. This occurs when t + 2 = 2trn => t = 2trn - 2, where n is any integer. between 0 and 7. The speed at t = 2tr - 2 is ~( X'(t))2 + (y'(t))2 = ~( 5 -I2cos(2tr) r + (-I2sin(2tr) r = 7 ft/sec. Ii t = 2tr - 2 is the only such value Polar coordinates allow us to write some very complicated equations in a much simpler lorm..c: For examp Ie, thc' e artesian equation.42322222240' x + x + x y +.xy - y + y = IS equivalent to the polar equation r = 1- cos (). Some very simple polar equations have graphs that are very complicated (but beautiful) curves. The polar coordinates for a point Pare (r, ()), where r represents the distance from the origin to the point P and () is the measure of an angle from the positive x-axis (sometimes called the polar axis) to the ray joining the origin to point P. By convention, angles measured in the counterclockwise direction are positive, and angles measured in the clockwise direction are negative (Figure 8-3). In working with polar forms of equations, it is sometimes necessary to convert the coordinates to Cartesian form. The Cartesian coordinates (x, y) and polar coordinates (r, ()) are related by the following equations: (). () 2 2 2 () Y X = r cos, y = r SIn, x + y = r, tan = - x

Note that in Cartesian coordinates, every point in the plane has exactly one ordered pair that describes it. This is not true in a polar coordinate system, however. In fact, every point has an infinite number of pairs of coordinates, because (r, B + 2:rrn), for all integers n, correspond to the same point. Most graphing calculators have a polar graphing mode, which makes it very easy to draw polar curves. Be sure that you are familiar with the use of this mode in case you need to draw a polar curve. You should also be familiar with the graphs of the most common polar equations, namely circles, lines, cardioids, and three- and four-leaf roses (Figure 8-4), in case there are polar questions on the non-calculator portion of the exam. y y y a y y r=-- e=~ cose r=a 4 r = acose r = asine a x x Limayon Cardioid Three-leaf rose Four-leaf rose Lemniscate r = b+2acose r = a(l + cos e) r = asin3e r = acos2e r 2 = a 2 sin2e y y y y y IThe slope of a tangent line to a polar enrve is still fonnd by :' not by :~. We can use the Chain Rule to get dy in terms of rand B: dx dy ~ dy / dx ~ -!e(rsinlj) ~ r'sinlj+rcoslj dx db db ~ (r cos B) r' cos B - r sin B db

It is not necessary to memorize the last form of this rule: just remember the Chain Rule and the equations relating polar and Cartesian coordinates. For the polar curve r = cos(2b), 0::; B::; 2n, find equations for all the tangent lines at the pole (origin). n 3n 5n 7n The curve is at the pole when r=o ~ cos(2b)=0 ~ B=- - --. 4' 4 ' 4 ' 4 dy = dy/dx = -!e(cos(2b)sinb) _ cos(2b)cosb+sinb(-2sin(2b)). dx db db ~(cos(2b)cosb) -cos(2b)sinb+cosb( -2sin(2B)) db I 0.J2+J2.(_2).1 dy = h h -1. When B = n, the tangent line has equation y = x. dx e _!!... 2 2 4-4 -0'2+2 (-2).1 Similarly, when B = 3n, the tangent line has equation y = -x. When B = 5n, the 4 4 tangent line has equation y = x and when B = 7n, the tangent line has equation y = -x. 4 Thus, equations for the tangent lines at the pole are: y = x and y = -x. Ii Area Enclosed by a Polar Curve Finding the area enclosed by a polar curve, or by a part of a polar curve, offers no new calculus concepts to explore. Rather, it is another setting in which you transform a Riemann sum into a definite integral, and then use the integral to answer questions. Whereas you use rectangles as a first approximation in finding the area under the graph of a function, here you use sectors of a circle (Figure 8-5).

A,,,,,,,., \ \ \ The area of a sector with a small angle tie in radians and radius r is ~ r 2 tie. 2 IThe area of the polar region from e = a to e = j3 A = s: ~[r(e)j de. is given by The area inside the circle with polar equation r = 2 sin e and above the lines with equations y = x and y = -x is given by ri Jr (A) L~2sin t 2 ede (B) 2sinede (C) 4 3Jr (D) f.:: 4 sin ede (E) f.:: 4 2sin 2 ede 4 4 The line y = x is equivalent to e = 1r, and the line y = -x is 4 equivalent to e = - 1r or e = 31r. Since the graph of 4 4 3Jr r = 2 sin e is a circle in the first and second quadrants, we want to use e = 31r 4 for the second line. So we need the portion of the circle between these two angles. The area is J:: ~(2sine)2 de. The answer is E. "4 2 t(2 sin 2 e -1) de

CHAPTER 8 - PARAMETRIC, VECTOR, AND POLAR FUNCTIONS 217 Just as you can find the area between two function graphs, you can find the area between two polar graphs. Since we are looking at sectors of a circle, we must think of the area between two graphs as an outside area minus an inside area. So the area formula becomes A = J: ~ [( 1] ( B)r-(r 2 ( B)r]dB, where 1] ( B) is the formula for the outside curve and r 2 ( B) is the formula for the inside curve. Sometimes it can be difficult to determine the limits of integration, though. 2 - r=-- cose Find the area of the region R enclosed by the graphs of r = _2_ cosb and r = 4sin(3B). rt Use the numeric solver on your calculator to find the values of B where the curves intersect. Solving gives a;;:; 0.1776 and j3;;:; 0.7854. Store these in variables A and B. The area is given by ~ r[( 4sin(3B))2 _(_2_)2]dB 1i!I;;:; 2.040. 2 A cose ~

Arc Length for Polar Curves The arc length for a polar curve r = r( B) between B = a and B = fj is given by This topic is not mentioned in the course description. In any event, you can always convert the polar curve to parametric form and use the parametric curve arc length formula. II x = rcos B = cosb+ 2cos 2 B = cosb + cos2b+ 1. y = r sin B = sin B + 2 sin B cos B = sin B + sin 2B. dx = _ sin B-2 sin 2B. dy = cos B + 2 cos 2B. db db {%)'do= L=r (~)' r 21T ~ 2 2 Jo (sinb+2sin2b) +(cosb+2cos2b) dt=ii~13.365. JJ