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=∫ 52√
( 2 t)^2 +( 6 t)^2 dt=∫ 52√
40 t^2 dt=∫ 522 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 π0( 3 +cosθ)^2 dθ=1
2
∫ 2 π0(
9 +6 cosθ+cos^2 θ)
dθ=
1
2
[
9 θ+6 sinθ+1
2
θ+1
4
sin 2θ] 2 π0=
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,soS=
∫π/ 3π/ 62 π(3 sinθ)√
cos^2 θ+9 cos^2 θdθ= 6 π∫π/ 3π/ 6sinθ√
10 cos^2 θdθ= 3 π√
10∫π/ 3π/ 62 sinθcosθdθ= 3 π√
10∫π/ 3π/ 6sin 2θdθ=−
3
2
π√
10 cos 2θ]π/ 3π/ 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〉
.