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

MA 3972-MA-Book April 11, 2018 16:32

370 STEP 5. Build Your Test-Taking Confidence

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[0, 10] by [−1.5, 1]

Note that for 0≤t≤10,a(t)>0, and

for 4. 31729 <t≤10,v(t)>0.

Therefore, for 0≤t≤10,a(t)andv(t)

have the same sign, both positive on the

interval 4. 317 <t≤10. Thus, the speed

of the particle is increasing on the interval

4. 31729 <t≤10.

(C) The average value of a functionfon a

closed interval [a,b]is

∫b

a f(x)dx

b−a

.In

this case, the average velocity of the

particle on the interval [0, 10] is∫

10

0 v(t)dt

(10−0)

=

1

10

∫ 10

0

(−cos(ln(t+1))− 0 .1)dt=

1

10

(− 0 .175738)=− 0 .0175738, or

− 0 .018.

(D) The total distance traveled by the particle

for 0≤t≤10 is

∫ 10

0

|v(t)|dt=

∫ 10

0

|−cos(ln(t+1))− 0. 1 |dt= 4 .44663,

or 4.447. Another approach to find the

total distance traveled by the particle is to

add the distance traveled to the left with

the distance traveled to the right. From

the graph ofv(t), the particle changes

direction att= 4 .317. Therefore, the total

distance

=

∣

∣∣

∣

∫ 4. 31729

0

v(t)dt

∣

∣∣

∣+

∫ 10

4. 31729

v(t)dt, which

yields the same result, 4.447.

Section II Part B

3. (A)
   
   
   • 1 0
   
   
   
   

1

2

1

y

x

Rewrite

dy

dx

− 2 x=xyas

dy

dx

= 2 x+xy.

Below is a summary of

dy

dx

at the six given

points.

x=− 1 x= 0 x= 1

y= 1 − 3 0 3

y= 2 − 4 0 4

(B) At the point(0, 1),

dy

dx

=0. The equation

of the tangent line isy−y 1 =m

(

x−x 1

)

.

In this case,y− 1 = 0 (x− 0 ),ory=1.

(C) Since the tangenty=1 is a constant

function, the values ofystay at 1 for all

values ofx.Thus, using the tangent line at

(0, 1) to approximatef(0.02), you have

f(0.02)≈1.

(D) Rewrite

dy

dx

− 2 x=xyas

dy

dx

= 2 x+xy=x(2+y), and obtain

dy

2 +y

=xdx. Integrating both sides of the

equation, you have

∫

dy

2 +y

=

∫

xdx.

Integrate

∫

dy

2 +y

by lettingu= 2 +yand

therefore∫ du=dx. You obtain

du

u

=ln|u|+c=ln| 2 +y|+c. Thus,