5 Steps to a 5 AP Calculus BC 2019

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

24. (Calculator) At what value(s) ofxdoes the
    tangent to the curvex^2 +y^2 =36 have a
    slope of−1.
    25. (Calculator) Find the shortest distance
    between the point (1, 0) and the curve
    y=x^3.

8.6 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1. Volume:V=

4

3

πr^3 ;

Surface Area: S= 4 πr^2

dS

dt

= 8 πr

dr

dt

.

Since

dS

dt

= 8

dr

dt

,

8

dr

dt

= 8 πr

dr

dt

⇒ 8 = 8 πr

orr=

1

π

.

Atr=

1

π

,V=

4

3

π

(

1

π

) 3

=

4

3 π^2

cubic

units.

2. Pythagorean Theorem yields
   x^2 +y^2 =(13)^2.
   Differentiate: 2x
   dx
   dt
   
   
   • 2 y
     dy
     dt
   
   
   
   

= 0 ⇒

dy

dt

=

−x

y

dx

dt

.

Atx=5, (5)^2 +y^2 = 132 ⇒y=±12, since

y>0,y=12.

Therefore,

dy

dt

=−

5

12

(−2) ft/sec=

5

6

ft/sec.

The ladder is moving away from the wall

at

5

6

ft/sec when the top of the ladder is

5 feet from the ground.

3. Volume of a sphere isV=

4

3

πr^3.

Differentiate:

dV

dt

=

(

4

3

)

(3)πr^2

dr

dt

= 4 πr^2

dr

dt

.

Substitute: 100= 4 π(5)^2

dr

dt

⇒

dr

dt

=

1

π

cm/sec.

Letxbe the diameter. Since

x= 2 r,

dx

dt

= 2

dr

dt

.

Thus,

dx

dt

∣∣

∣∣

r= 5

= 2

(

1

π

)

cm/sec

=

2

π

cm/sec. The diameter is increasing at

2

π

cm/sec when the radius is 5 cm.

4.(See Figure 8.6-1.) Using similar triangles,

withythe length of the shadow you have:

5

20

=

y

y+x

⇒ 20 y= 5 y+ 5 x⇒

15 y= 5 xory=

x

3

.

Differentiate:

dy

dt

=

1

3

dx

dt

⇒

dy

dt

=

1

3

(6)

=2ft/sec.

5 ft

20 ft

yx

Light

Figure 8.6-1

5.(See Figure 8.6-2.) Volume of a cone

V=

1

3

πr^2 h.

Using similar triangles, you have

12

18

=

r

h

⇒ 2 h= 3 rorr=

2

3

h, thus

reducing the equation to

V=

1

3

π

(

2

3

h

) 2

(h)=

4 π

27

h^3.