344 STEP 4. Review the Knowledge You Need to Score High
Step 4. Verify result by differentiating
y=2(2x+1)^1 /^2
dy
dx= 2
(
1
2)
( 2 x+ 1 )−^1 /^2 ( 2 )=
2
√
2 x+ 1.
Compare this with:dy
dx=
y
2 x+ 1=
2 ( 2 x+ 1 )^1 /^2
2 x+ 1=
√^2
2 x+ 1.
Thus, the function is
y= f(x)=2(2x+1)^1 /^2.(d) f(x)=2(2x+1)^1 /^2
f(0.1)=2(2(0.1)+1)^1 /^2 =2(1.2)^1 /^2
≈ 2. 191- See Figure 13.12-1.
[−π,π] by [−4,4]
Figure 13.12-1(a)Intersection points: Using the
[Intersection] function on the
calculator, you havex=0 and
x= 1 .37131.Area ofR=∫ 1. 371310[3 sinx−(ex−1)]dx.Enter∫
(3 sin(x))−(e∧(x)−1),x,0,1.37131 and obtain 0.836303.
The area of regionRis approximately
0.836.(b) Using the Washer Method, volume of
R=π∫ 1. 371310[
(3 sinx)^2 −(ex−1)^2]
dx.Enterπ∫
((3 sin(x))∧ 2 −(e∧(x)−1)∧2,
x,0,1.37131) and obtain 2.54273π
or 7.98824.
The volume of the solid is 7.988.(c) Volume of Solid=π∫ 1. 371310(Area of
Cross Section)dx.
Area of Cross Section=1
2
πr^2=1
2
π(
1
2(3 sinx−(ex−1))) 2
.Enter(
π
2)
1
4∗
∫
((3 sin(x)−
(e∧(x)−1))∧2,x,0,1.37131)
and obtain 0.077184πor 0.24248.
The volume of the solid is
approximately 0.077184πor 0.242.- Integrate the acceleration vector to get the
velocity vector:
a ̄(θ)=〈sinθ,−cosθ〉=sinθ ̄ı−cosθj ̄
v ̄(θ)=−cosθ ̄ı−sinθj ̄+C
v ̄(
π
3)
=−cos(
π
3)
ı ̄−sin(
π
3)
j ̄+C=− ̄ı−
1
2
̄ı−√
3
2
j ̄+C=− ̄ıC=−
1
2
̄ı+√
3
2
j ̄v ̄(θ)=(
−cosθ−1
2
)
̄ı+(√
3
2
−sinθ)
j ̄.Integrate the velocity vector to get the
position vector:s ̄(θ)=(
−sinθ−1
2
θ)
ı ̄+
(√
3
2
θ+cosθ)
j ̄+C