5 Steps to a 5 AP Calculus BC 2019

Take a Diagnostic Exam 31

22. v(t)=

t^3

3

− 2 t^2 + 5

a(t)=v′(t)=t^2 − 4 t

See Figure DS-9.

The graph indicates that for 0≤t≤6, the

maximum acceleration occurs at the endpoint

t=6.a(t)=t^2 − 4 tanda(6)= 62 −4(6)= 12.

Figure DS-9

23. y=x

(^3) ,x≥0;dy
dx
= 3 x^2
f′(x)= 12 ⇒
dy
dx
= 3 x^2 = 12
⇒x^2 = 4 ⇒x= 2
Slope of normal=negative reciprocal of slope
of tangent=−

1

12

.

Atx=2,y=x^3 = 23 =8; (2, 8)

y− 8 =−

1

12

(x−2).

Equation of normal line:⇒y=−

1

12

(x−2)+ 8

ory=−

1

12

x+

49

6

.

24. f(x)=

lnx

x

; f′x=

(1/x)(x)−(1)lnx

x^2

=

1

x^2

−

lnx

x^2

y=−x^2 ;

dy

dx

=− 2 x

Perpendicular tangents

⇒(f′(x))

(

dy

dx

)

=− 1

⇒

((

1

x^2

)

−

lnx

x^2

)

(− 2 x)=− 1.

Using the [Solve] function on your calculator,

you obtainx≈ 1. 37015 ≈ 1 .370.

25. f

(

π

2

)

= 3 ⇒

(

π

2

,3

)

is on the graph.

f′

(

π

2

)

=− 1 ⇒slope of the tangent atx=

π

2

is− 1.

Equation of tangent line:y− 3 =

− 1

(

x−

π

2

)

ory=−x+

π

2

+ 3.

Thus, f

(

π

2

+

π

180

)

≈−

(

π

2

+

π

180

)

+

π

2

+ 3

≈ 3 −

π

180

≈ 2. 98255

≈ 2. 983.

26. Position is given byx= 4 t^2 andy=

√

t,

so velocity is

dx

dt

= 8 tand

dy

dt

=

1

2

√

t

. The

acceleration will be

d^2 x

dt^2

=8,

d^2 y

dt^2

=−

1

4

t−^3 /^2.

Evaluate−

1

4

t−^3 /^2

]

t= 4

=−

1

4

(

1

8

)

=−

1

32

to get

the acceleration vector

〈

8,−

1

32

〉

.

27. The slope of the tangent line is
    dy
    dx

∣∣

∣∣

t= 1

and

dx

dt

=2,

dy

dt

= 2 t+2, so

dy

dx

∣∣

∣∣

t= 1

=

2 t+ 2

2

∣

∣∣

∣t= 1 =t+^1

∣∣

t= 1 =2. Att=1,

x=5,y=3,⇒(5, 3),

So, the equation of the tangent line is

y− 3 =2(x−5)⇒y=2(x−5)+ 3 ⇒y= 2 x−7.