5 Steps to a 5 AP Calculus BC 2019

Applications of Derivatives 151

Letr, hbe the radius and height of the cone respectively.

Sinceh= 3 r, the volume of the coneV=

1

3

πr^2 h=

1

3

πr^2 ( 3 r)=πr^3.

V=πr^3 ;

dV

dt

= 3 πr^2

dr

dt

.

When

dV

dt

= 12

dr

dt

,12

dr

dt

= 3 πr^2

dr

dt

⇒ 4 =πr^2 ⇒r=

2

√

π

.

Thus,V=πr^3 =π

(

2

√

π

) 3

=π

(

8

π

√

π

)

=

8

√

π

.

TIP • Go with your first instinct if you are unsure. Usually that is the correct one.

Inverted Cone (Water Tank) Problem

A water tank is in the shape of an inverted cone. The height of the cone is 10 meters and the

diameter of the base is 8 meters as shown in Figure 8.1-1. Water is being pumped into the

tank at the rate of 2 m^3 /min. How fast is the water level rising when the water is 5 meters

deep? (See Figure 8.1-1.)

10m

5m

8m

Figure 8.1-1

Solution:

Step 1: Define the variables. LetVbe the volume of water in the tank;hbe the height of

the water level attminutes;rbe the radius of surface of the water attminutes;

andtbe the time in minutes.

Step 2: Given:

dV

dt

=2m^3 /min. Height=10 m, diameter=8m.

Find:

dh

dt

ath=5.

Step 3: Set up an equation:V=

1

3

πr^2 h.

Using similar triangles, you have

4

10

=

r

h

⇒ 4 h = 10 r;orr =

2 h

5

. (See
Figure 8.1-2.)