More Applications of Derivatives 195
9.5 Rapid Review
- 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:y− 1 =−1(x−0) ory=−x+1.
- 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
.
- 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 - (See Figure 9.5-1.)
(a) When is the acceleration zero? (b) Is the particle moving to the right or left?
024
v
t
v(t)
Figure 9.5-1
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.
- 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. Thusa(t)=8. The maximum acceleration is 8. - Find the slope of the tangent to the curve defined byx= 3 t−5, y=t^2 −9 whent=3.
Answer:
dx
dt
=3 and
dy
dt
= 2 t
∣∣
t= 3 =6, so
dy
dx
=
dy
dt
÷
dx
dt