294 STEP 4. Review the Knowledge You Need to Score High
- Find the length of the arc defined byx=t^2 andy= 3 t^2 −1 fromt=2tot=5.
Answer:
dx
dt
= 2 tand
dy
dt
= 6 t.L=
∫ 5
2
√
( 2 t)^2 +( 6 t)^2 dt=
∫ 5
2
√
40 t^2 dt
=
∫ 5
2
2 t
√
10 dt=
[
t^2
√
10
] 5
2
= 25
√
10 − 4
√
10 = 21
√
10.
- Find the area bounded by ther= 3 +cosθ.
Answer:To trace out the graph completely, without retracing, we need 0≤θ≤ 2 π.
Then,
A=
1
2
∫ 2 π
0
( 3 +cosθ)^2 dθ=
1
2
∫ 2 π
0
(
9 +6 cosθ+cos^2 θ
)
dθ
=
1
2
[
9 θ+6 sinθ+
1
2
θ+
1
4
sin 2θ
] 2 π
0
=
1
2
[( 18 π+π)− 0 ]=
19 π
2
.
- Find the area of the surface formed when the curve defined byx=sinθ, and
y=3 sinθon the interval
π
3
≤θ≤
π
6
is revolved about thex-axis.
Answer:
dx
dθ
=cosθand
dy
dθ
=3 cosθ,so
S=
∫π/ 3
π/ 6
2 π(3 sinθ)
√
cos^2 θ+9 cos^2 θdθ= 6 π
∫π/ 3
π/ 6
sinθ
√
10 cos^2 θdθ
= 3 π
√
10
∫π/ 3
π/ 6
2 sinθcosθdθ= 3 π
√
10
∫π/ 3
π/ 6
sin 2θdθ
=−
3
2
π
√
10 cos 2θ
]π/ 3
π/ 6
=−
3
2
π
√
10
[(
cos
2 π
3
)
−
(
cos
π
3
)]
=−
3
2
π
√
10
[(
−
1
2
)
−
(
−
1
2
)]
=
3
2
π
√
10.
- If
〈
dx
dt
,
dy
dt
〉
=
〈
5 −t^2 ,4t− 3
〉
and
〈
x 0 ,y 0
〉
=
〈
0, 0
〉
, find〈x,y〉.
Answer: x=
∫ (
5 −t^2
)
dt= 5 t−
t^3
3
+C 1 andy=
∫
(4t−3)dt= 2 t^2 − 3 t+C 2.
Since
〈
x 0 ,y 0
〉
=
〈
0, 0
〉
,C 1 =C 2 =0, so〈x,y〉=
〈
5 t−
1
3
t^3 ,2t^2 − 3 t
〉
.