5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:14

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

10.4 Rapid Review

1. Write an equation of the normal line to the graphy=exatx=0.
   Answer:
   dy
   dx

∣∣

∣∣

x= 0

ex=ex|x= 0 =e^0 = 1 ⇒mnormal=− 1

Atx=0,y=e^0 = 1 ⇒you have the point (0, 1).

Equation of normal line:y− 1 =−1(x−0) ory=−x+1.

2. Using your calculator, find the values ofxat which the functiony=−x^2 + 3 xandy=lnx
   have parallel tangents.
   Answer:y=−x^2 + 3 x⇒
   dy
   dx

=− 2 x+ 3

y=lnx⇒

dy

dx

=

1

x

Set− 2 x+ 3 =

1

x

. Using the [Solve] function on your calculator, enter

[Solve]

(

− 2 x+ 3 =

1

x

,x

)

and obtainx=1orx=

1

2

.

3. Find the linear approximation off(x)=x^3 atx=1 and use the equation to findf(1.1).
   Answer:f(1)= 1 ⇒(1, 1) is on the tangent line and f′(x)= 3 x^2 ⇒ f′(1)=3.
   y− 1 =3(x−1) ory= 3 x−2.
   f(1.1)≈3(1.1)− 2 ≈ 1. 3


4. (See Figure 10.4-1.)
   (a) When is the acceleration zero? (b) Is the particle moving to the right or left?

024

v

t

v(t)

Figure 10.4-1

Answer: (a) a(t)=v′(t) andv′(t) is the slope of the tangent. Thus,a(t)=0att=2.

(b) Sincev(t)≥0, the particle is moving to the right.

5. Find the maximum acceleration of the particle whose velocity function isv(t)=t^2 + 3
   on the interval 0≤t≤4.
   Answer:a(t)=v′(t)=2(t) on the interval 0≤t≤4,a(t) has its maximum
   value att=4. Thus,a(t)=8. The maximum acceleration is 8.