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

MA 3972-MA-Book April 11, 2018 14:46

196 STEP 4. Review the Knowledge You Need to Score High

12. (See Figure 9.6-7.)

xx

River

(200 – 2 x)

Figure 9.6-7

Step 1: Area:

A=x(200− 2 x)= 200 x− 2 x^2

with 0≤x≤100.

Step 2: A′(x)= 200 − 4 x

Step 3: SetA′(x)= 0 ⇒ 200 − 4 x=0;

x=50.

Step 4: Apply the Second Derivative Test:

A′′(x)=−4; thus atx=50, the

area is a relative maximum.

A(50)=5000 m^2.

Step 5: Check the endpoints.

A(0)=0 andA(100)=0;

therefore atx=50, the area is the

absolute maximum and 5000 m^2

is the maximum area.

Part B Calculators are allowed.

13. Step 1: Lethbe the height of the trough
    and 4 be a side of one of the two
    equilateral triangles. Thus, in a
    30---60 right triangle,h= 2

√

3.

Step 2: Volume:

V=(area of the triangle)· 10

=

[

1

2

(h)

(

2

√

3

h

)]

10 =

10

√

3

h^2.

Step 3: Differentiate with respect tot.

dV

dt

=

(

10

√

3

)

(2)h

dh

dt

Step 4: Substitute known values:

1 =

20

√

3

(2)

dh

dt

;

dh

dt

=

√

3

40

m/min.

The water level is rising√

3

40

m/min when the water level

is 2 m high.

14. (See Figure 9.6-8.)

3000 m

S

Z

θ

Camera

Figure 9.6-8

Step 1: tanθ=S/ 3000

Step 2: Differentiate with respect tot.

sec^2 θ

dθ

dt

=

1

3000

dS

dt

;

dθ

dt

=

1

3000

(

1

sec^2 θ

)

dS

dt

=

1

3000

(

1

sec^2 θ

)

(200t)

Step 3: Att=5;S=100(5)^2 =2500;

Thus,Z^2 =(3000)^2 +(2500)^2 =

15, 250, 000. Therefore,

Z=± 500

√

61, sinceZ>0,

Z= 500

√

61. Substitute known
    values into the equation:
    dθ
    dt

=

1

3000

⎛

⎜⎜

⎜⎝

1

500

√

61

3000

⎞

⎟⎟

⎟⎠

2

(1000),

since secθ=

Z

3000

.