5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 305

The acceleration of the object is known to be

a=〈−et,et〉=−eti+etj. Integrate to

find the velocity.v=−eti+etj+C, and

since the initial velocity isv 0 =

〈

3, 1

〉

,

v 0 =−i+j+C= 3 i+jand C= 4 i.

The velocity vector isv=−eti+etj+

4 i=(4−et)i+etj. Integrate again to find

the position vectors=(4t−et)i+etj+C.

The initial position at the origin means

thats 0 =(4. 0 −e^0 )i+e^0 j+C=−i+

j+C=0, and therefore,C=i−j. The

position vectors=(4t−et+1)i+

(et−1)jcan be evaluated att=1 to find

the position as (5〈 −e)i+(e−1)j=

5 −e,e− 1

〉

.

12.10 Solutions to Cumulative Review Problems

26. (See Figure 12.10-1.)

[−3,3] by [−1,7]

Figure 12.10-1

∫a

−a

ex^2 dx=

∫ 0

−a

ex^2 dx+

∫a

0

ex^2 dx

Sinceex^2 is an even function, thus

∫ 0

−a

ex^2 dx=

∫a

0

ex^2 dx.

k= 2

∫a

0

ex^2 dxand

∫a

0

ex^2 dx=

k

2

.

27. (See Figure 12.10-2.)

9

x

y

θ

Figure 12.10-2

sinθ=

9

x

Differentiate both sides:

cosθ

dθ

dt

=(9)(−x−^2 )

dx

dt

.

Whenx=15, 9^2 +y^2 = 152 ⇒ y= 12.

Thus, cosθ=

12

15

=

4

5

;

dx

dt

=−2 ft/sec.

4

5

dθ

dt

= 9

(

−

1

152

)

(−2)

=

dθ

dt

=

18

152

5

4

=

1

10

radian/sec.

28. (See Figure 12.10-3.)

[−2,5] by [−2,6]

Figure 12.10-3