304 STEP 4. Review the Knowledge You Need to Score High
- The area enclosed by the curve is the
upper half of an ellipse.
Find
dx
dt
=−2 sint.
A=∫π0(3 sint)(−2 sint)dt=− 6∫π0sin^2 tdt=− 6
[
1
2
t−1
4
sin 2t]π0=− 3 πThe negative simply indicates that the
area has been swept from right to left,
rather than left to right, and so may be
ignored. The area enclosed by the curve is
3 π.- Differentiate to find
dr
dθ
=sin
(
θ
2)
cos(
θ
2)
, and calculater^2 =sin^4(
θ
2)
and(
dr
dθ) 2
=sin^2(
θ
2)cos^2(
θ
2). Then the length of the arc is
L=
∫π0√
sin^4(
θ
2)
+sin^2(
θ
2)
cos^2(
θ
2)
dθ=
∫π0√
sin^2(
θ
2)(
sin^2(
θ
2)
+cos^2(
θ
2))
dθ=
∫
π
0∣∣
∣∣sin(
θ
2)∣∣
∣∣dθ=−2 cos(
θ
2)∣∣
∣∣π0=−2 cos
π
2
+2 cos 0=−2(0)+2(1)= 2.- Find
dx
dt
=etcost+etsintand
dy
dt
=etcost−etsint. Square
each derivative.
(
dx
dt) 2
=(etcost+etsint)^2=e^2 t(cos^2 t+2 sintcost+sin^2 t)
=e^2 t(1+2 sintcost) and
(
dy
dt) 2
=(etcost−etsint)^2=e^2 t(cos^2 t−2 costsint+sin^2 t)
=e^2 t(1−2 costsint).ThenS= 2 π∫ π 20etcost×
√
e^2 t(1+2 sintcost)+e^2 t(1−2 sintcost)dt= 2 π∫ π 20e^2 tcost×
√
1 +2 sintcost+ 1 −2 sintcostdt= 2 π∫ π 20(e^2 tcost)√
2 dt= 2
√
2 π
e^2 t
5
(sint+2 cost)∣∣
∣∣π 20= 2√
2 π
eπ− 2
5≈ 37. 5702
- Squarer^2 = 4 +8 sinθ+4 sin^2 θ. The area
A=
1
2
∫ 2 π0(
4 +8 sinθ+4 sin^2 θ)
dθ=
1
2
[
4 θ−8 cosθ+ 4(
1
2
θ−1
4
sin 2θ)] 2 π0=[
3 θ−4 cosθ−1
2
sin 2θ] 2 π0
=(6π−4)−(−4)= 6 π.- The acceleration vector for an object
moving in the plane is〈−et,et〉. Find the
position of the object att=1, if the initial
velocity isv 0 =
〈
3, 1〉
and the initial
position of the object is at the origin.