Chapter 4, Problem 70EP

$\text{(a)}$

To determine

To find: The rate of change of the area of it base.

Answer to Problem 70EP

The answer: $\frac{dA}{dr}=\frac{4\pi }{3}$

Explanation of Solution

Given information:

The volume $V$ of a cone $\left(V=\frac{1}{3}\pi r^{2}h\right)$ increasing at the rate of $4\pi$ cubic inches per second.

At the instant when the radius of the cone is $2\text{ inches}$ .

Calculation:

Area of the base of a cone is the area of a circle:

  $A=\pi r^2$

Then,

  $\begin{array}{c}A=\pi r^2\\ \frac{dA}{dt}=2\pi r\frac{dr}{dt}\\ \frac{dA}{dt}=2\pi (2)\left(\frac{1}{3}\right)\\ =\frac{4\pi }{3}\end{array}$

Therefore, the required rate of change of the area of it base is:

  $\frac{dA}{dr}=\frac{4\pi }{3}$ .

$\text{(b)}$

To determine

To find: The rate of change of the area of its height $\text{h}$ .

Answer to Problem 70EP

The answer: $\frac{dh}{dt}=1$

Explanation of Solution

Given information:

The volume $V$ of a cone $\left(V=\frac{1}{3}\pi r^{2}h\right)$ increasing at the rate of $4\pi$ cubic inches per second.

At the instant when the radius of the cone is $2\text{ inches}$ .

Calculation:

Must use the volume equation to solve for the missing value of the height

  $V=\frac{1}{3}\pi r^{2}h$

Then,

  $\begin{array}{c}V=\frac{1}{3}\pi r^{2}h\\ \to 8\pi =\frac{1}{3}\cdot \pi \cdot 2^2\cdot h\\ \to h=6\\ \frac{dV}{dt}=\frac{2}{3}\cdot \pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^2\frac{dh}{dt}\\ \to 4\pi =\frac{2}{3}\pi \cdot 2\cdot 6\cdot \frac{1}{3}+\frac{1}{3}\cdot \pi \cdot 2^2\frac{dh}{dt}\\ \to \frac{dh}{dt}=1\end{array}$

Therefore, the required rate of change of the area of its height $\text{h}$ is:

  $\frac{dh}{dt}=1$

$(c)$

To determine

To find: The instantaneous rate of change of the area of its base with respect to height $\text{h}$ .

Answer to Problem 70EP

The answer: $\frac{dA}{dh}=\frac{4\pi }{3}$

Explanation of Solution

Given information:

The volume $V$ of a cone $\left(V=\frac{1}{3}\pi r^{2}h\right)$ increasing at the rate of $4\pi$ cubic inches per second.

At the instant when the radius of the cone is $2\text{ inches}$ .

Calculation:

  $\begin{array}{c}\frac{dA}{dh}=\frac{\frac{dA}{dt}}{\frac{dh}{dt}}\\ \to \frac{dA}{dh}=\frac{\frac{4\pi }{3}}{1}\\ \to \frac{dA}{dh}=\frac{4\pi }{3}\end{array}$

Therefore, the required instantaneous rate of change of the area of its base with respect to height $\text{h}$ is:

  $\frac{dA}{dh}=\frac{4\pi }{3}$