Chapter 5, Problem 71RE

a.

To determine

To calculate: The rate of change of the area of the cone base.

Answer to Problem 71RE

The rate of change of the area of the cone base

  $\frac{dA}{dt}=\frac{4\pi }{3}$ .

Explanation of Solution

Given information: The volume V of a cone is increasing at the rate of $4\pi$ cubic inches per second. At the instant when the radius of the cone is $2$ inches, its volume is $8\pi$ cubic inches and the radius is increasing at $\frac{1}{3}$ inch per second.

Formula used: Area of the base of a cone

  $A=\pi r^2$

Calculation:

radius is increasing at $\frac{1}{3}$ inch per second

  $\frac{dr}{dt}=\frac{1}{3}$

Area of the base of a cone

  $A=\pi r^2$

Derivative of the area with respect to time

  $\begin{array}{l}A=\pi r^2\\ \frac{dA}{dt}=2\pi r\frac{dr}{dt}\\ \because \frac{dr}{dt}=\frac{1}{3},r=2\\ \frac{dA}{dt}=2\pi \times 2\times \frac{1}{3}\\ \frac{dA}{dt}=\frac{4\pi }{3}\end{array}$

b.

To determine

To calculate: The rate of change of the cone height h.

Answer to Problem 71RE

The rate of change of the cone height h

  $\frac{dh}{dt}=1$ .

Explanation of Solution

Given information: The volume V of a cone is increasing at the rate of $4\pi$ cubic inches per second. At the instant when the radius of the cone is $2$ inches, its volume is $8\pi$ cubic inches and the radius is increasing at $\frac{1}{3}$ inch per second.

Formula used: Volume of a cone

  $V=\frac{1}{3}\pi r^{2}h$

Calculation:

The volume V of a cone is $8\pi$ cubic inches

And the radius of the cone is $2$ inches

Volume of a cone

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ 8\pi =\frac{1}{3}\pi \times {\left(2\right)}^2\times h\\ h=6\end{array}$

The volume V of a cone is increasing at the rate of $4\pi$ cubic inches per second

  $\frac{dv}{dt}=4\pi$

Volume of a cone

  $V=\frac{1}{3}\pi r^{2}h$

Derivative of the volume with respect to time

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ \frac{dV}{dt}=\frac{2}{3}.\pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^2\frac{dh}{dt}\\ \because \frac{dv}{dt}=4\pi ,\frac{dr}{dt}=\frac{1}{3},h=6,r=2\\ 4\pi =\frac{2}{3}\pi \times 2\times 6\times \frac{1}{3}+\frac{1}{3}\pi \times 4\frac{dh}{dt}\\ \frac{4}{3}\pi =\frac{4}{3}\pi \frac{dh}{dt}\\ \frac{dh}{dt}=1\end{array}$

c.

To determine

To calculate: The rate of change of the area of cone base with respect to the cone height h.

Answer to Problem 71RE

The rate of change of the area of cone base with respect to the cone height h

  $\frac{dA}{dh}=\frac{4\pi }{3}$ .

Explanation of Solution

Given information: The rate of change of the area of the cone base

  $\frac{dA}{dt}=\frac{4\pi }{3}$

And The rate of change of the cone height h

  $\frac{dh}{dt}=1$

Calculation:

The rate of change of the area of the cone base

  $\frac{dA}{dt}=\frac{4\pi }{3}$

And The rate of change of the cone height h

  $\frac{dh}{dt}=1$

The rate of change of the area of cone base with respect to the cone height h

  $\begin{array}{l}\frac{dA}{dh}=\frac{\frac{dA}{dt}}{\frac{dh}{dt}}\\ \because \frac{dA}{dt}=\frac{4\pi }{3},\frac{dh}{dt}=1\\ \frac{dA}{dh}=\frac{\frac{4\pi }{3}}{1}\\ \frac{dA}{dh}=\frac{4\pi }{3}\end{array}$