196 STEP 4. Review the Knowledge You Need to Score High
- Find the slope of the tangent line to the graph ofr=−3 cosθ.
Answer:
dr
dθ
=3 sinθ. Since x=rcosθ,
dx
dθ
=−rsinθ+cosθ
dr
dθ
=3 cosθsinθ
+3 cosθsinθ =6 sinθcosθ =3 sin 2θ. Since y =rsinθ,
dy
dθ
=rcosθ
+ sinθ
dr
dθ
=−3 cos^2 θ + 3 sin^2 θ =−3(cos^2 θ − sin^2 θ) =−3 cos 2θ.
dy
dx
=
dy
dθ
·
dθ
dx
=
−3 cos 2θ
3 sin 2θ
=−cot 2θ
- Find
dr
dt
for the vector functionr(t)= 3 ti− 2 tj=〈 3 t,− 2 t〉.
Answer:
dx
dt
=3 and
dy
dt
=−2, so
dr
dt
=〈3,− 2 〉.
9.6 Practice Problems
Part A The use of a calculator is not
allowed.
- Find the linear approximation off(x)=
(1+x)^1 /^4 atx=0 and use the equation to
approximatef(0.1). - Find the approximate value of^3
√
28 using
linear approximation.
- Find the approximate value of cos 46◦using
linear approximation. - Find the point on the graph ofy=
∣∣
x^3
∣∣
such that the tangent at the point is parallel
to the liney− 12 x=3.
- Write an equation of the normal to the
graph ofy=exatx=ln 2. - If the liney− 2 x=bis tangent to the graph
y=−x^2 +4, find the value ofb. - If the position function of a par-
ticle iss(t)=
t^3
3
− 3 t^2 +4, find the velocity and
position of particle when its acceleration is 0. - The graph in Figure 9.6-1 represents the
distance in feet covered by a moving particle
intseconds. Draw a sketch of the
corresponding velocity function.
5
4
3
2
1
0 1 2345
s(t)
t
Seconds
Feet s
Figure 9.6-1
9.The position function of a moving particle
is shown in Figure 9.6-2.
t 1 t 2
t 3
s(t)
s
t
Figure 9.6-2