MATH 152 - Recitation Quiz 6 - Spring 2015
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1. Find the exact length of the curve given by parametric equations x (t) = 3t − t3 and y (t) = 3t2 from
the point (0, 0) to (−2, 12) .
The arclength of a curve is given by the following formula:
ˆb q
L=
2
2
[x0 (t)] + [y 0 (t)] dt,
a
where a ≤ t ≤ b. Therefore, we need to nd three pieces of information: the values that t ranges over, x0 (t) ,
and y 0 (t) . Note that when t = a, y = 3t2 = 0 and when t = b, y = 3t2 = 12. Therefore, a = 0 and b = 2.
Also, x0 (t) = 3 − 3t2 and y 0 (t) = 6t. Therefore,
L=
ˆ2 p
ˆ2 p
ˆ2 p
ˆ2 q
2
2
(3 − 3t2 ) + (6t) dt =
9 − 18t2 + 9t4 + 36t2 dt =
9 + 18t2 + 9t4 dt =
9 (1 + 2t2 + t4 )dt.
0
0
ˆ2
3·
L=
0
q
0
ˆ2
2
(1 + t2 ) dt =
0
2
3 + 3t2 dt = 3t + t3 0 = (6 + 8) − (0 + 0) = 14 units.
0
1
2. The curve C is given by the equation x = 31 y 2 + 2
solid obtained when rotating C about the x-axis.
23
for 0 ≤ y ≤ 1. Compute the surface area of the
The surface area generated by revolving a curve around the axis is given by:
ˆb
2πr ·
A=
q
2
2
[x0 (t)] + [y 0 (t)] dt,
a
where a ≤ t ≤ b, and where r is the distance from the axis to the curve. Therefore, since we are rotating
C about the x-axis, r = y. Here, we can let y = t, so a = 0 and b = 1. Also, y 0 (t) = 1. Then, x (t) =
√
32
1
1
2
=⇒ x0 (t) = 32 · 13 · 2t t2 + 2 2 = t t2 + 2. Therefore,
3 t +2
ˆ1
t·
A = 2π ·
r
ˆ
ˆ p
2
p
p
2
2
2
t t + 2 + 1dt = 2π · t · t (t + 2) + 1dt = 2π t t4 + 2t2 + 1dt.
1
0
1
0
0
ˆ1
ˆ1
ˆ1 q
2
2
2
t3 + t dt.
A = 2π t (t + 1) dt = 2π t t + 1 dt = 2π
0
0
A = 2π
1 4 1 2
t + t
4
2
1
= 2π
0
2
1 1
+
4 2
0
− (0 + 0) = 2π ·
3
3π
=
units2 .
4
2