5 Steps to a 5 AP Calculus BC 2019

196 STEP 4. Review the Knowledge You Need to Score High

7. 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

8. 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.

1. Find the linear approximation off(x)=
   (1+x)^1 /^4 atx=0 and use the equation to
   approximatef(0.1).


2. Find the approximate value of^3

√

28 using

linear approximation.

3. Find the approximate value of cos 46◦using
   linear approximation.


4. Find the point on the graph ofy=

∣∣

x^3

∣∣

such that the tangent at the point is parallel

to the liney− 12 x=3.

5. Write an equation of the normal to the
   graph ofy=exatx=ln 2.


6. If the liney− 2 x=bis tangent to the graph
   y=−x^2 +4, find the value ofb.


7. 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.


8. 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