Another AP Calculus BC free response question (will defiantly give 10 points)?

b. r = θ + sin 2θ

x = r cos θ = 2

(θ + sin 2θ) cos θ = -2

θ cos θ + sin 2θ cos θ + 2 = 0

θ ≈ 2.786

c. This means that r is decreasing on the interval from π/3 to 2π/3. The curve is getting closer to the x-axis.

d. d = √(x^2 + y^2) = r √(cos^2 θ + sin^2 θ) = r

To find the minimum distance, dr/dθ = 1 + 2 cos 2θ = 0

2 cos 2θ = -1

cos 2θ = -1/2

2θ = 2π/3

θ = π/3

r = π/3 + sin 2π/3 = π/3 - 1/2

c. It spirals counterclockwise inward in that region. Or put another way, r gets shorter. Be sure to remember that we are talking about polar coordinates here.

d. Take dr/dθ , = 1 + 2cos(2θ) . Set to zero in the usual way to find local maximum or minimum. In fact, they pretty much gave you the points at which this is 0 in part c = pi/3 and 2pi/3. You would take the second derivative to show whether this is a maximum or minimum. But in part c, they already said that the first derivative is negative over that region, so the curve must be spiraling inward. The max is therefore at pi/3. This being an essay question though, you'd better take the second derivative, then impress them with the depth of your reasoning later.

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