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

MA 3972-MA-Book April 11, 2018 16:20

398 STEP 5. Build Your Test-Taking Confidence

(B) g′(t) =−4 sec^2

(

t

20

)(

1

20

)

g′(10)=−

1

5

sec^2

(

10

20

)

≈− 0. 26

(C) Set the temperature of the liquid equal to

86 ◦F. Using your calculator, let

y 1 = 90 −4 tan

(

x

20

)

; andy 2 = 86.

To find the intersection point ofy 1 andy 2 ,

lety 3 =y 1 −y2 and find the zeros ofy 3.

Using the [Zero] function of your

calculator, you obtainx= 15 .708. Since

y 1 <y 2 on the interval 15. 708 <x≤20,

the temperature of the liquid is below

86 ◦F when 15. 708 <t≤20.

(D) Average temperature below 86◦F

=

1

20 − 15. 708

∫ 20

15. 708

(

90 −4 tan

(

x

20

))

dx.

Using your calculator, you obtain:

Average temperature =

1

4. 292

(364.756)

≈ 84. 9851

≈ 84. 99 ◦F.

Section II Part B

3. (A) The midpoints of three subintervals of
   equal length are:
   t=4, 12, and 20.
   The length of each interval is

24 − 0

3

=8.

Thus,

∫ 24

0

v(t)dt≈ 8 [v(4)+v(12)+v(20)]

≈8[25+ 15 +20]

=8(60)= 480.

(B) Average velocity =

1

24

∫ 24

0

v(t)dt

≈

1

24

(480)=20 ft/s.

(C) Average acceleration=

v(24)−v(0)

24 − 0

=

30

24

= 1 .25 ft/s^2.

(D) a(t)=0att=6 andt=14, since the

slopes of tangents att=6 andt=14 are 0.

(E) a(20)≈

v(22)−v(18)

22 − 18

≈

25 − 15

4

≈

10

4

≈ 2 .5ft/s^2

4. (A) f′(x)= 3

(

e−^2 x^2

)

(− 4 x)=− 12 xe−^2 x^2

Setting f′(x)=0,− 12 xe−^2 x^2 = 0

⇒x=0.

(B) f′′(x)=(−12)

(

e−^2 x^2

)

+(− 12 x)

(

e−^2 x^2

)

(− 4 x)

=− 12 e−^2 x

2

+ 48 x^2 e−^2 x

2

Setting f′′(x)=0, 12e−^2 x^2 + 48 x^2 e−^2 x^2 = 0

⇒ 48 x^2 e−^2 x^2 = 12 e−^2 x^2

⇒ 48 x^2 = 12 ⇒x^2 =

1

4

or

x=±

1

2

.

(C) limx→∞ 3 e−^2 x^2 =xlim→∞

3

e^2 x^2

= 0

x→lim−∞^3 e−^2 x^2 =xlim→−∞

3

e^2 x^2

= 0

(D)

incr. decr.

rel. max.

f′ +

f′

0 –

x

0