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

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

AP Calculus AB Practice Exam 1 369

Solutions for AP Calculus AB Practice Exam 1 Section II

Section II Part A

1. (A) Begin by finding the intersection points of
   the two curves, and obtainx=0 and
   x= 2 .60308.
   Area of a region enclosed by two curves∫ =
   b
   a

(Upper Curve−Lower Curve)dx.

In this case, the area of region

S=

∫ 2. 60308

0

((4x−x^2 )−(2x^4 − 5 x^3 ))dx

= 17 .258.

0

(2, 4)

y

y = 4

x

R

r

(B) Volume of a solid of revolution with hole

=π

∫b

a

(R^2 −r^2 )dx, whereRis the outer

radius andris the inner radius.

In this case,R= 4 −(2x^4 − 5 x^3 ), and

r= 4 −(4x−x^2 ). (Note that the highest

point in regionSis (2, 4).) Therefore, the

volume of the solid revolving about the

liney=4 is:

π

∫ 2. 60308

0

[(4−(2x^4 − 5 x^3 ))^2

−(4−(4x−x^2 ))^2 ]dx= 172. 896 πor 543. 17

(C) Volume of a solid with a given base

=

∫b

a

(area of cross section)dx. In this case,

the area of the cross section

=(0. 5 + 2 x)((4x−x^2 )−(2x^4 − 5 x^3 )).

Thus, the volume of the solid with baseS

is

∫ 2. 60308

0

(0. 5 + 2 x)((4x−x^2 )

−(2x^4 − 5 x^3 ))dx= 64. 334.

2. (A) Remember thatv(t)=

∫

a(t)dt, and in

this case,v(t)=

∫

sin(ln(t+1))

t+ 1

dt.

Using your TI-89 graphing calculator,

you havev(t)=−cos(ln(t+1))+c.

Sincev(0)=− 1 .1, you have

− 1. 1 =−cos(ln(t+1))+c,or

− 1. 1 =−cos(0)+c, which yieldsc=− 0 .1.

Therefore,v(t)=−cos(ln(t+1))− 0 .1,

andv(4)=−cos(ln(4+1))− 0. 1 =

− 0 .061368, or− 0 .061.

Another approach to findv(4)isas

follows. Note that

∫ 4

0

a(t)dt=v(4)−v(0)

=v(4)−(− 1 .1)=v(4)+ 1 .1. Therefore,

v(4)=

∫ 4

0

a(t)dt− 1 .1. Also note that

∫ 4

0

a(t)dt=

∫ 4

0

sin(ln(t+1))

t+ 1

dt=

1 .03863. Therefore,

v(4)= 1. 03863 − 1. 1 =− 0 .061368, or

− 0 .061.

(B) The speed of the particle is increasing

whena(t) andv(t) are both positive or

both negative. Examine the graphs ofa(t)

andv(t).