204 STEP 4. Review the Knowledge You Need to Score High
Find
dr
dθ
=12 cos 4θand substitute.
dy
dx=
(3 sin 4θ) cosθ+sinθ(12 cos 4θ)
−(3 sin 4θ) sinθ+cosθ(12 cos 4θ).
Whenθ=
5 π
6
,4θ=
10 π
3=
4 π
3
+ 2 π,sothe functions of
10 π
3
are equal to those of
4 π
3. Evaluate
dy
dx
=
(3 sin(4π/3)) cos(5π/6)+sin(5π/6)(12 cos(4π/3))
−(3 sin(4π/3)) sin(5π/6)+cos(5π/6)(12 cos(4π/3))
at=(
3(
−√
3 / 2)) (
−√
3 / 2)
+(1/2)(12(− 1 /2))
−(
3(
−√
3 / 2))
(1/2)+(
−√
3 / 2)
(12(− 1 /2))=(9/4)− 3
(3
√
3 /4)+ 3√
3=−
√
3
15. The slope of
the tangent line is−
√
3
15.
- If the position of the object is given by〈
30 t, 25 sin
t
3〉
, then the velocity vectoris〈
dx
dt,
dy
dt〉
=〈
30,25
3
cos
t
3〉
, andthe acceleration vector is
〈
d^2 x
dt^2,
d^2 y
dt^2〉
=〈
0,− 25
9
sin
t
3〉. The
magnitude of the acceleration is equal to
∣∣
∣∣−^25
9
sin
t
3∣∣
∣∣=2 when sint
3=
18
25
. Solve to
findt=3 sin−^118
25
≈ 2 .411.
- Ifr=〈lnt,ln(t+4)〉, then
dr
dt
=
〈
1
t,
1
t+ 4〉. Evaluate att=4 for
dr
dt
=
〈
1
4,
1
8
〉
, then find
∣∣
∣∣∣∣
∣∣dr
dt∣∣
∣∣∣∣
∣∣=√(
1
t) 2
+(
1
t+ 4) 2
, which, at
t=4, is equal to
∣∣
∣∣∣∣
∣∣dr
dt∣∣
∣∣∣∣
∣∣=√
1
16+
1
64
=
√
5
8. The tangent
vector isT=
〈 1 /t,1/(t+4)〉
√
5 / 8=〈
8
t√
5,
8
(t+4)√
5〉. Whent=4,
T=
〈
2√
5
5,
√
5
5〉
.9.9 Solutions to Cumulative Review Problems
- Using product rule, letu=x;v=sin−^1 (2x).
dy
dx
=(1) sin−^1 (2x)+1
√
1 −(2x)^2(2)(x)=sin−^1 (2x)+
√^2 x
1 − 4 x^2- Lety=f(x)⇒y=x^3 − 3 x^2 + 3 x−1. To
findf−^1 (x), switchxand
y:x=y^3 − 3 y^2 + 3 y−1.
dx
dy
= 3 y^2 − 6 y+ 3
dy
dx=
1
dx/dy=
1
3 y^2 − 6 y+ 3
dy
dx∣∣
∣∣
y= 2=
1
3(2)^2 −6(2)+ 3
=
1
3
- Substitutingx=100 into the expression
x− 100
√
x− 10
would lead to
0
0
. Apply
L’Hoˆpital’sRule and you have limx→ 100
1
1
2
x−(^12)
or