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Can you help me with this AP Calculus problem?

The graphs of the polar curves r=3 and r=4-2sin(theta) are shown in the figure above. The curves intersect when theta=pi/6 and theta=5pi/6.
 
a) Let S be the shaded region that is inside the graph of r=3 and also inside the graph of r=4-2sin(theta). Find the area of S. 
 
b) A particle moves along the polar curve r=4-2sinθ so that at time t seconds, theta=t^2. Find the time t in the interval 1≤t≤2 for which the x-coordinate of the particle's position is -1. 
 
c) For the particle described in part (b), find the position vector in terms of t. Find the velocity vector at time t=1.5. 
 
This is AP Calculus BC free response question in 2013. Please show all your work step by step. 
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2 Answers

All right,
 
For part a)
 
We are asked to find the area of S. I assume that you have the figure of this curve plotted out. It should look like a wide squashed circle (shifted down) and smaller circle about the origin.
 
One way to think about the problem is to cut up the shaded area into two parts.
 
We know the area of a circle (constant radius) is
 
A=πr2
 
We can also think of this equation as an integral of the radius
 
A = 0 r(θ)2/2 
 
If we work out the integral for a constant radius (r(θ) = R) we can prove this
 
A = 0 R2/2 dθ (notice there are no θ terms in the integral, or θ0)
 
A = R2/2θ |0 (raise the power (0 → 1), and divide by power (/1))
 
A =  R2(0)/2 + R2(2π)/2 (evaluate from 0 to 2π)
 
A = πR(We are left with the equation we wanted)
 
 
Looking at our graph we see that the plots of r=3 and r = 4-2sin(θ) intersect twice. Let's find out where they intersect by setting them equal.
 
3 = 4-2sin(θ)
 
-1 = -2sin(θ)
 
1/2 = sin(θ)
 
θ = sin-1(1/2)  (this actually has two solutions)
 
θ = π/4, θ=3π/4
 
This solution will tell us what range for which to evaluate the area of the curves.
 
Looking at the graph and starting at θ = 0 and moving counter-clockwise, the shaded area is bounded  by the curve r = 3 up to the point where θ = π/4.
After that, the shaded area is bounded by the curve r = 4-2sin(θ) from θ = π/4 to θ = 3π/4
After that, the shaded area is again bounded by the curve r = 3 from  θ = 3π/4 all the way back around to  θ = 2π.
 
So we can find the area of the shaded region by adding up the following areas
 
A10π/4 32/2 dθ
 
A2π/43π/4 [4-2sin(θ)]2/2 dθ
 
A3 = 3π/4 32/2 dθ
 
Let's solve them one at a time
 
A1 = 9θ/2 |0π/4
 
A1 = 0 + 9/2*π/4 = 9π/8
 
A2 =   π/43π/4 (4-2sin(θ)) (4-2sin(θ))/2 dθ
 
 
A2 = π/43π/4 (16-16sin(θ)+4sin2(θ))/2 dθ
 
A2 = π/43π/4 8-8sin(θ)+2sin2(θ) dθ
 
 
A2 = π/43π/4 8 - π/43π/4 8sin(θ) + π/43π/4 2sin2(θ)
 
A2 = 8θ + 8cos(θ) + 2/2 (θ-sin(θ)cos(θ))  |π/43π/4 (evaluate this at π/4 to 3π/4)
 
A2 =  8π/4 + 8cos(π/4) + π/4-sin(π/4)cos(π/4) + 24π/4 + 8cos(3π/4) + 3π/4-sin(3π/4)cos(3π/4)
 
A2 = 8π/4 + 8√2/2 + π/4-√2/2*√2/2 + 24π/4 - 8√2/2 + 3π/4+√2/2*√2/2
 
A2 = (8+1+24+3)π/4 =
 
A3 = 3π/42π 32/2 dθ
 
A3 = 9θ/2  |3π/4
 
A3 = 9*3π/(2*4) + 9*2π/2
 
A3 = 27π/8 + 9π
 
Add them all together and we get
 
A1 + A2 + A3 = 9π/8 + 6π + 27π/8 + 9π = 9π/2+15π = 39π/2
 
 
 
This is not correct. I made an error right at the beginning. The two curves intersect at π/6 and 5π/6
 
 
a) The shaded region is symmetric about the y-axis, so it suffices to find the area in the first and fourth quadrant, i.e., from θ=-(π/2) to (π/2), and multiply the result by 2. Between θ=-(π/2) and (π/6) the boundary curve is the circle r=3. Between θ=(π/6) and (π/2) the boundary curve is r=4-2sin(θ).
The area element in polar coordinates is dA=rdrdθ.
Since we are given r as a function of θ, we can integrate r from 0 to r(θ) and write the area element as

dA=(1/2)(r(θ))² dθ,

so that the total area is

A=2*(1/2) ∫-π/2π/2 (r(θ))² dθ

We need to split this integral into two parts:

A = ∫-π/2π/6(3)² dθ+∫π/6π/2(4-2sin(θ))² dθ
= 12π -(15/2)√3 ≈ 24.7 (using a graphing calculator).

b) You set
x= r cos(θ) = (4-2sin(θ))cos(θ) = -1.
This transcendental equation can only be solved numerically/graphically. Again using a graphing calculator, you get
θ ≈ 2.039 ⇒ t=√θ ≈ 1.428
 
c) The position vector can be written as
[x,y] = r [cos(θ), sin(θ)] = (4-2sin(θ)) [cos(θ), sin(θ)]
= [(4-2sin(t²))cos(t²), (4-2sin(t²))sin(t²)]
 
You can find the velocity vector [vx,vy] by taking the t-derivative of [x,y]  and evaluate it at t=1.5.
 

Comments

Andre,

For part b, you say set (4-2sin(θ))cos(θ) = -1 where θ = t2 and t is in the interval [1,2].  However, I don't understand exactly how you are solving this either graphically or on a calculator.
 
Can you please elaborate on how you determined the solution of part b?
Kenneth,
 
I graphed (4-2sin(θ))cos(θ)+1 on the interval θ =1²...2², and found a zero between 2.0 and 2.1. I then plugged in a few values between 2.0 and 2.1 into my calculator to get the approximation 2.039. I could have used something like Newton's method for a better approximation.
 
It later occurred to me that the equation is not transcendental, but can be turned into a 4-th order polynomial equation using the substitution x=sin(θ) (so that cos(θ)=sqrt(1-x²)). There exists a (complicated) solution formula for 4-th order equations, which should give the exact answer, though the computational search engines like WolframAlpha only return numerical approximations, like x=.892324..., which gives θ=2.039124...