Related Rates with Areas and Volumes

Spheres

Example 1

A spherical snowball melts so that its volume decreases at

16

cubic centimeters per minute. When its diameter is

8

centimeters, find the rate at which (a) its radius and (b) its surface area are changing.

(a)

V = \frac{4}{3}\pi r^3

, so

\frac{dV}{dt} = 4\pi r^2\,\frac{dr}{dt}

. Melting means

\frac{dV}{dt} = -16

, and the radius is

4

-16 = 4\pi(16)\,\frac{dr}{dt}, \qquad \frac{dr}{dt} = -\frac{1}{4\pi}.

The radius is decreasing at

\frac{1}{4\pi} \approx 0.080

centimeter per minute.

(b)

S = 4\pi r^2

, so

\begin{aligned} \frac{dS}{dt} &= 8\pi r\,\frac{dr}{dt} \\[4pt] &= 8\pi(4)\left(-\frac{1}{4\pi}\right) = -8. \end{aligned}

The surface area is decreasing at

8

square centimeters per minute.

Cones

Example 2

A reservoir is a cone with its vertex down,

6

meters in radius at the top and

8

meters deep. Water flows in at

18

cubic meters per minute. When the water is

4

meters deep, find the rate at which (a) the depth and (b) the area of the water's surface are changing.

A cross-section of the reservoir. The water (blue) is a smaller cone, similar to the reservoir, with r/h = 6/8.

(a) The water is a cone with radius

r

and depth

h

, similar to the reservoir, so

\frac{r}{h} = \frac{6}{8}

and

r = \frac{3h}{4}

. Substituting that before differentiating leaves only

h

, and the rate of

r

isn't needed:

\begin{aligned} V &= \frac{1}{3}\pi\left(\frac{3h}{4}\right)^2 h = \frac{3\pi h^3}{16}, \\[4pt] \frac{dV}{dt} &= \frac{9\pi h^2}{16}\,\frac{dh}{dt}. \end{aligned}

With

\frac{dV}{dt} = 18

and

h = 4

18 = 9\pi\,\frac{dh}{dt}

, so

\frac{dh}{dt} = \frac{2}{\pi} \approx 0.637

meter per minute.

(b) The surface is a circle of radius

\frac{3h}{4}

, so

A = \frac{9\pi h^2}{16}

and

\begin{aligned} \frac{dA}{dt} &= \frac{9\pi h}{8}\,\frac{dh}{dt} \\[4pt] &= \frac{9\pi(4)}{8} \cdot \frac{2}{\pi} = 9. \end{aligned}

The surface area is increasing at

9

square meters per minute.

Troughs

Example 3

A trough is

8

feet long. Its ends are trapezoids

2

feet wide at the bottom,

4

feet wide at the top, and

2

feet high. Water drains out at

3

cubic feet per minute. Find the rate at which the water level is changing when the water is

1

foot deep.

One end of the trough. Each side slants out 1 foot over the 2-foot height, so water h feet deep is 2 + h feet wide.

When the water is

h

feet deep, its surface is

2 + h

feet wide, so the end of the water is a trapezoid with area

\frac{1}{2}(2 + 2 + h)h = 2h + \frac{1}{2}h^2

. The volume is

8

times that:

\begin{aligned} V &= 16h + 4h^2, \\[4pt] \frac{dV}{dt} &= (16 + 8h)\,\frac{dh}{dt}. \end{aligned}

Draining means

\frac{dV}{dt} = -3

. At

h = 1

-3 = 24\,\frac{dh}{dt}

, so

\frac{dh}{dt} = -\frac{1}{8}

. The water level is falling at

\frac{1}{8}

foot per minute, or

1.5

inches per minute.

An area with an angle

Example 4

Two sides of a triangle are

6

and

5

centimeters long, and the angle

\theta

between them increases at

0.2

radian per second. Find the rate at which the area of the triangle is changing when

\theta = \frac{\pi}{6}

Sides 6 and 5 with the angle θ between them. The height on the side of length 6 is 5 sin θ.

The height on the side of length

6

5\sin\theta

, so

A = \frac{1}{2}(6)(5\sin\theta) = 15\sin\theta

. Then

\begin{aligned} \frac{dA}{dt} &= 15\cos\theta\,\frac{d\theta}{dt} \\[4pt] &= 15\left(\frac{\sqrt{3}}{2}\right)(0.2) = \frac{3\sqrt{3}}{2}. \end{aligned}

The area is increasing at

\frac{3\sqrt{3}}{2} \approx 2.60

square centimeters per second.