294 STEP 4. Review the Knowledge You Need to Score High
7. Find the length of the arc defined by x = t2 and y = 3t2 − 1 from t = 2 to t = 5.
Answer: dx = 2t and dy = 6t. L = 5 5 40t2d t
dt dt 2
(2t)2 + (6t)2d t =
2
55
= 2t 10d t = t2 10 = 25 10 − 4 10 = 21 10.
22
8. Find the area bounded by the r = 3 + cos θ.
Answer: To trace out the graph completely, without retracing, we need 0 ≤ θ ≤ 2π .
Then,
A = 1 2π (3 + cos θ )2d θ = 1 2π 9 + 6 cos θ + cos2 θ dθ
2 0 2 0
1 9θ + 6 sin θ + 1 + 1 sin 2θ 2π = 1 [(18π + π) − 0] = 19π .
=2 2θ 4 0 2 2
9. Find the area of the surface formed when the curve defined by x = sin θ, and
π π
y = 3 sin θ on the interval 3 ≤ θ ≤ 6 is revolved about the x-axis.
Answer: dx = cos θ and dy = 3 cos θ, so
dθ dθ
π/3 π/3
S = 2π (3 sin θ ) cos2 θ + 9 cos2 θ d θ = 6π sin θ 10 cos2 θ d θ
π/6 π/6
π/3 π/3
= 3π 10 2 sin θ cos θd θ = 3π 10 sin 2θd θ
π/6 π/6
3 10 cos 2θ π/3 3 10 cos 2π − cos π
= −2π π/6 = − 2 π 3 3
= − 3 π 10 − 1 − − 1 = 3 π 10.
2 2 2 2
10. If dx , dy = 5 − t2, 4t − 3 and x0, y0 = 0, 0 , find x , y .
dt dt
Answer: x = 5 − t2 d t = 5t − t3 + C1 and y = (4t − 3)d t = 2t2 − 3t + C2.
3
Since x0, y0 = 0, 0 , C1 = C2 = 0, so x , y = 5t − 1 t 3, 2t2 − 3t .
3
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AP calculus BC 2019
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AP calculus BC 2019
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