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Millersville University Name Answer Key
Department of Mathematics
MATH 211, Calculus II, Test 3
April 18, 2008
Please answer the following questions. Your answers will be evaluated on their correctness,
completeness, and use of mathematical concepts we have covered. Please show all work and
write out your work neatly. Answers without supporting work will receive no credit. The
point values of the problems are listed in parentheses.
- (14 points) Find the area of the region inside the graph of r = 3 + 2 sin θ and outside the graph of r = 4. - 4 - 2 2 4
x
2
4
y
The curves intersect when
3 + 2 sin θ = 4
2 sin θ = 1
sin θ =
θ =
π
6
or θ =
5 π
6
Thus the area between the curves is
A =
∫ (^5) π/ 6
π/ 6
(3 + 2 sin θ) 2 − 4 2 dθ
∫ (^5) π/ 6
π/ 6
( 9 + 12 sin θ + 4 sin 2 θ − 16
) dθ
∫ (^5) π/ 6
π/ 6
( −7 + 12 sin θ + 4 sin 2 θ
) dθ
∫ (^5) π/ 6
π/ 6
(−7 + 12 sin θ) dθ +
∫ (^5) π/ 6
π/ 6
4 sin 2 θ dθ
( −
θ − 6 cos θ
)∣ ∣ ∣ ∣
5 π/ 6
π/ 6
∫ (^5) π/ 6
π/ 6
(1 − cos 2θ) dθ
( 5 π
6
π
6
) − 6
( cos
5 π
6
− cos
π
6
)
( θ −
sin 2θ
)∣ ∣ ∣ ∣
5 π/ 6
π/ 6
7 π
3
( −
)
( 5 π
6
π
6
) −
( sin
5 π
3
− sin
π
3
)
7 π
3
2 π
3
( −
)
5 π
3
5 π
3 ≈ 6. 02234
- (10 points) Consider the parametric equations:
x = t − 2 sin t
y = 1 − 2 cos t
Find the slope of the tangent line to the graph of the parametric equations at t = 2π/3.
dy
dx
dy dt dx dt
=
2 sin t
1 − 2 cos t
Thus when t = 2π/3,
m =
2 sin 2π/ 3
1 − 2 cos 2π/ 3
- (14 points) Find a Taylor series for f (x) = ln(2 + x) about c = 0.
Let
f ′ (x) =
2 + x
1 + x/ 2
1 − (−x/2)
∑^ ∞
k=
( −
x
2
)k
∑^ ∞
k=
(−1)k
2 k+^
x k if |x| < 2.
Integrating term-by-term we have
f (x) =
∑^ ∞
k=
∫ (−1)k
2 k+^
x k dx =
∑^ ∞
k=
(−1)k
2 k+1(k + 1)
x k+ .
- (8 points each) Consider the function f (x) = ln(cos x).
(a) Find the Taylor polynomial of degree 2 about c = π/6 for f (x).
k f (k) (x) f (k) (π/6)
f (k)(π/6) k! 0 ln(cos x) ln(
3 /2) ln(
1 − tan x − 1 /
2 − sec 2 x − 4 / 3 − 2 / 3 3 −2 sec^2 x tan x
P 2 (x) =
∑^2
k=
f (k)(π/6)
k!
( x −
π
6
)k = ln
)
−
( x −
π
6
) −
( x −
π
6
) 2
(b) Find the second Taylor remainder for f (x) about c = π/6.
R 2 (x) =
f (3) (z)
3!
( x −
π
6
) 3
− sec 2 z tan z
3
( x −
π
6
) 3
where z is between x and π/6.
- (14 points) Consider the parametric equations:
x = cos 3 t
y = sin 3 t
Find the exact arclength (no approximations) of the graph of the parametric equations for 0 ≤ t ≤ π/2.
s =
∫ (^) b
a
√ √ √ √
( dx
dt
) 2
( dy
dt
) 2
dt
∫ (^) π/ 2
0
√ (3 cos^2 t(− sin t))^2 + (3 sin 2 t(cos t))^2 dt
∫ (^) π/ 2
0
√ 9 cos^4 t sin 2 t + 9 sin 4 t cos^2 t dt
∫ (^) π/ 2
0
sin t cos t
√ cos^2 t + sin 2 t dt
∫ (^) π/ 2
0
sin t cos t dt
∫ (^1)
0
u du
