AP Calculus BC Review Chapter (Parametric Equations and Polar Coordinates) Things to Know and Be Able to Do Understand the meaning of equations given in parametric and polar forms, and develop a sketch of the appropriate graph Find the slope of a tangent line to a parametrically-given curve at a particular point using the equation =, and understand how this is derived using the Chain Rule Find the second derivative of a parametrically-given curve at a particular point using the equation d ( ) =, and understand how this is derived using the Chain Rule Find the area enclosed by a parametrically-given curve Find the arc length of a parametrically-given curve and the area of the surface generated by revolving a parametrically-given curve Find the area enclosed by a polar curve using the equation A= r dθ (understanding that r= f ( θ )) and understand how this is derived. Be very careful with finding the endpoints for evaluating the integral! Find the arc length of a polar curve Practice Problems These problems should be done without a calculator, with the exception of, as explained below. The original test, of course, required that you show relevant work for free-response problems. x = cos t For t, an object travels along an elliptical path given parametrically by. At the point at which y = sint t =, the object leaves the path and travels along the line tangent to the path at that point. What is the slope of the line on which the object travels? tan a b c d e tan tan x= t t The position of a particle moving in the xy-plane is given by the parametric equations. y = t t t values of t is the particle at rest? a only b only c only d and only e,, and For what A curve C is defined by the parametric equations tangent to the graph of C at the point (,8 )? x= t t+ y = t a x = b x = c 8. Which of the following is an equation of the line y = 7 d y= ( x+ ) + e y ( x ) 8 = + + 8
x= t A particle moves so that its position at time t is given by. y = sin( t) cos a 8sin b 6 5 Which of the following integrals represents the area shaded in the graph shown at right? The curve is given by r = sin θ. a sin ( θ b 8sin ( θ c sin θ dθ d sin ( θ e sin θ dθ 6 Which of the following integrals represents the arc length of the polar function r = + cosθ from θ? a b + cosθ + sinθ dθ + sin θ dθ c ( + cosθ d + θ What is the speed of the particle when t =? c ( cos) + 6 d ( sin) + 8 e ( cos) + 6 cos dθ e ( + cos θ ) sin θ ( + cos θ ) + ( sin θ ) d θ 7 Consider the graph of the vector function r () t = + t,+ t. What is the value of where x =? a b 8 A particle moves so that at time c 8 d ln t t, t. t > its position vector is 8 e 9 at the point on the graph 8 + At time t =, its velocity vector is a,8 b, c,8 d, e 5 8 8 9 Consider the curves r = cosθ and r =. a Sketch the curves on the axes provided at right. b Show use of calculus to find the area of the region common to both graphs. x = t t Consider the curve given parametrically by. y = t t a In terms of t, find. b Write an equation for the line tangent to the curve at the point at which t =. c Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent. 6,
A projectile is launched from the edge of a cliff 9 feet above the ground below. The projectile has an initial velocity of 8 feet per second and is launched at an angle of to the horizontal. A horizontal wind is also blowing against the projectile at a velocity of 8 feet per second. The acceleration of gravity is feet per second per second. a Sketch the situation and the path of the projectile. b Write a vector equation that represents the path of the project tile, using t in seconds as the parameter. c When does the projectile reach its maximum height, and what is this maximum height? Justify your answer. d How far, horizontally, does the projectile travel before landing on the ground? When this test was originally administered, the following question was to be taken home and completed with the use of a calculator but with no other resources permitted to be used, such as other people, books, computers, or anything else. Students had one weekend to complete this. An object moving along a curve in the xy-plane has position ( x( t), y( t )) at time t with t t = and = ln( + ( t ) ). At time t =, the object is at position ( ),5. At time t =, the object is at point P with x- coordinate. a Find the acceleration vector and the speed at time t =. b Find the y-coordinate of point P. c Write an equation for the line tangent to the curve at point P. d For what value(s) of t, if any, is the object at rest? Justify your answer.
Answers d c a c 5 b 6 a 7 d 8 a 9a see solutions 9b 7 6 t t a = y = x+ 5 6t 6t t t 8,6, 6 ; b c horizontal at ( ) and vertical at (, ) and (, ) a see solutions b r () t t t t = 6 8,6 6 + 9 c 56 ft at t = s d 8 8 ft a a =, ; v = + ( ln7). 7 ln7 b.67 c y.67 = ( x ) d t = only Solutions We find = sint and cos t. = Then (by the Chain Rule) cost = = =, so at t =, sint tant. = tan This is choice d. We want a time t at which both = and ; = that is, both t 6t= and 6t 6t =. The first equation is t = 6t or t =, and the second equation is the same as t t =, which factors to ( t )( t+ ) =. Therefore t, ; the intersection of the two solution sets is t = only, choice c. = for { } Firstly, find that ( xy, ) = (,8) occurs at. t = Now find t = and t, = so t = =. Then t = =. This is undefined, so the tangent line must be vertical. Since the point has an x-value of t =, the relevant tangent line is x =, choice a. The particle s velocity is given by v () t =, = t,cos t. We are interested in v = 6 + cos = cos + 6, choice c. which is 6,cos, 5 Recall that the area enclosed by a polar graph is given by r dθ. Therefore the integrand must be ( sinθ ) = ( 6) sin θ = 8sin θ. Without even having to consider the limits, we alrea know that the correct choice must be b. dr r + dθ. Since 6 Recall that the arc length of a curve represented in polar coordinates is given by dθ dr = sin θ, the integral must take the form ( cos ) ( sin ) d. dθ + θ + θ θ We are given the limits, and the correct answer is a.
To evaluate this, recall the identity 6 6 d 7 Recall that when a curve is given parametrically, =. Therefore we must find.. t d 8 ( ) t t 8 8 Therefore = = = 5. The point at which x = is the point at which t =, so =, t t 9t 9 t = choice d. 8 The velocity vector is v () t = t +, =, t. At t =, we have v ( ) =,8, choice a. t + t 9a The first equation is a circle of diameter tangent to the y-axis and shifted along the positive x-axis; the second is a circle of diameter centered at the θ = 6 origin. The circles along with the region referred to in part b are shown at right. 9b As shown in the diagram, the circles intersect at the points ( r, θ ) =, 6 and,. Note that the region is symmetric about the line θ =, so we 6 θ = 6 have the luxury of considering only the portion of it with θ, and then doubling its area. For θ,, the outer limit of the region is r = 6, so that part s area is 6 dθ =. For θ,, 6 the outer limit of the region is r = cos θ, so that part s area is cos d = cos d. cos θ = + cos θ. Therefore the inte- gral may also be written as ( + cosθ θ ] θ 6 6 = + sin = + =. The union of these 6 two regions is then + = 7, which is half of the area of the entire region. Therefore the area of the whole region is 7. 6 t a The Chain Rule gives =. This is. If you are so inclined, you may factor out a from the numerator and a 6 from the denominator, yielding 6t 6t t, but such simplification is entirely optional. t t ( ) 9 b We need = = =. We also must find x and y values: x( ) = ( ) ( ) t = 6( ) 6( ) = 5 and y( ) = ( ) ( ) =. Therefore an equation for the tangent line is given by, in point-slope form, y = ( x+ 5 ). c Critical points are those at which the curve s derivative is either undefined (vertical tangent line) or zero (horizontal tangent line). The derivative will be undefined iff (if and only if) its denominator is zero; that is, =, so we
solve t t ; = this gives { } t,. At, has vertical tangent lines at (, ) and so we solve t = to get t =±. At, the curve has horizontal tangent lines at ( 8,6) t = xy, =, ; at, t = xy, =,. Therefore the curve,. The derivative is zero iff its numerator is zero; that is,, = xy, = 8,6, and at, xy, =, 6. Therefore t = and (, 6 ). t = a An annotated hand-sketch is presented below. It is very much not to scale, and the path should be symmetric and parabolic. θ = wind = 8 ft s a g = ft s b The component of the initial velocity in the horizontal direction is 8cos, and the wind slows the projectile down by 8 feet every second. Therefore x() t = ( 8cos ) t 8t= ( 6 8 ) t. In the vertical direction, the initial velocity s component is 8sin, the particle has an initial height of 9 ft, and gravity s contribution is given by at = t = 6 t. The total is y t = 8sin t 6t + 9 = 6t 6t + 9. Therefore the g vector equation is r () t t t t = 6 8,6 6 + 9. c We want to maximize y(), t so we find = 6 t and set it equal to zero, finding. gives 6 + 9 = 56 ft. d To begin, find the time at which the particle lands by setting y( t= ) ; that is, t {,6 }, from which we can immediately discard t = Evaluating y 6t 6t 9. + = This gives t = because it is nonsensical to have negative time given the situation. Therefore the particle hits the ground at t = 6 s; we find its horizontal displacement by finding x ( 6) = ( 6 8)( 6) = 8 8 ft. a The acceleration vector () t h = 9 ft t a is given by a () t =, = 6 t,. Evaluating this at t = gives + t a =, 7. The speed at t = is given by the magnitude of the velocity vector at t = ; since the velocity vector is v () t, t t,ln( ( t ) ), we have v =,ln7, so v = + ( ln7).. b The change in the object s y-coordinate is given by integrating with respect to time over the time interval in question; this is ln ( + ( t ) ). The integrand cannot be antidifferentiated without the techniques of com-
plex analysis, but the calculator will (after rather a while) give the approximation 8.67. However, note that this is the change in the object s y-coordinate; to get the actual y-coordinate, we must add its initial y-coordinate of 5, giving.67. ( + ( t ) ) ( ) ln ln + ln7 c We find the slope at P by = = ; at t =, = =. Since we know t t the x-coordinate of P (it is given) and the y-coordinate (from part b), we can write the equation for the line in ln7 point-slope form as y.67 = ( x ). d The object is stationary in the x-direction when ; = that is, t t =, or t {, }. The object is stationary in the y-direction when = ; that is, ln( + ( t ) ) =, or t =. The object is wholly stationary at the in- tersection of the solution sets, which is t =.