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,so
the 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 vector
is
〈
dx
dt
,
dy
dt
〉
=
〈
30,
25
3
cos
t
3
〉
, and
the 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−^1
18
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