Chapter 3.8, Problem 16E

(a) The volume of a growing spherical cell is $V=\frac{4}{3}\pi r^3$, where the radius r is measured in micrometers (1 μm = 10^–6 m). Find the average rate of change of V with respect to r when r changes from

(i) 5 to 8μm

(ii) 5 to 6μm

(iii) 5 to 5.1μm

(b) Find the instantaneous rate of change of V with respect to r when r = 5μm.

(c) Show that the rate of change of the volume of a sphere with respect to its radius is equal to its surface area. Explain geometrically why this result is true. Argue by analogy with Exercise 13(c).

To determine

To find: The average rate of change of volume of sphere $V$ with respect to when $r$, when $r$ changes from

1. i. $5\text{to}8\text{μm}$
2. ii. $5\text{to}6\text{μm}$
3. iii. $5\text{to}5.1\text{μm}$

Answer to Problem 16E

1. i. Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{172\pi {\text{μm}}^{\text{3}}/\text{μm}}$.
2. ii. Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{121.333\pi {\text{μm}}^{\text{3}}/\text{μm}}$.
3. iii. Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{102.013\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

Explanation of Solution

Calculation:

Volume of a sphere is as below.

$V\left(r\right)=\frac{4}{3}\pi r^3$

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from $5\text{μm}$ to $8\text{μm}$.

$V\left(r\right)=\frac{V\left(8\right)-V\left(5\right)}{8-5}$

$\begin{array}{l}V\left(r\right)=\frac{\frac{4}{3}\pi {\left(8\right)}^3-\frac{4}{3}\pi {\left(5\right)}^3}{3}\\ V\left(r\right)=\frac{\frac{4}{3}\pi \left(512\right)-\frac{4}{3}\pi \left(125\right)}{3}\\ V\left(r\right)=172\pi {\text{μm}}^{\text{3}}/\text{μm}\end{array}$

Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{172\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from $5\text{μm}$ to $6\text{μm}$.

$\begin{array}{l}V\left(r\right)=\frac{V\left(6\right)-V\left(5\right)}{6-5}\\ V\left(r\right)=\frac{\frac{4}{3}\pi {\left(6\right)}^3-\frac{4}{3}\pi {\left(5\right)}^3}{1}\\ V\left(r\right)=\frac{\frac{4}{3}\pi \left(216\right)-\frac{4}{3}\pi \left(125\right)}{1}\\ V\left(r\right)=121.333\pi {\text{μm}}^{\text{3}}/\text{μm}\end{array}$

Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{121.333\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from $5\text{μm}$ to $5.1\text{μm}$.

$\begin{array}{l}V\left(r\right)=\frac{V\left(5.1\right)-V\left(5\right)}{5.1-5}\\ V\left(r\right)=\frac{\frac{4}{3}\pi {\left(5.1\right)}^3-\frac{4}{3}\pi {\left(5\right)}^3}{0.1}\\ V\left(r\right)=\frac{\frac{4}{3}\pi \left(132.651\right)-\frac{4}{3}\pi \left(125\right)}{0.1}\\ V\left(r\right)=102.013\pi {\text{μm}}^{\text{3}}/\text{μm}\end{array}$

Therefore, the average rate of change of the volume of a sphere is $\underset{\_}{102.013\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

To determine

To find: The instantaneous rate of change when radius $r$ is $5\text{μm}$.

Answer to Problem 16E

The instantaneous rate of change is $\underset{\_}{100\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

Explanation of Solution

Calculate the instantaneous rate of change $V$ when $r$ is $5\text{μm}$

$V\left(r\right)=\frac{4}{3}\pi r^3$

Differentiate the above volume equation.

$\begin{array}{l}V\text{'}\left(r\right)=\frac{4}{3}\times 3\times \pi r^2\\ V\text{'}\left(r\right)=4\pi r^2\end{array}$

Substitute the value $5$ for r.

$\begin{array}{l}V\text{'}\left(5\right)=4\pi {\left(5\right)}^2\\ V\text{'}\left(5\right)=100\pi {\text{μm}}^{\text{3}}/\text{μm}\end{array}$

Therefore, when the value of $r$ is $5$ the instantaneous rate of change is $\underset{\_}{100\pi {\text{μm}}^{\text{3}}/\text{μm}}$.

To determine

To find: The rate of change of the volume of a sphere with respect to its radius is equal to the surface area of the volume and why it is geometrically true.

Answer to Problem 16E

It is geometrically true. Hence, the resulting change in volume $\Delta V$ if the value of change in radius $\Delta r$ is small is $4\pi r^2$.

Explanation of Solution

Surface area of the sphere is given as below.

$S\left(r\right)=4\pi r^2$ (1)

First derivative of the volume expression is as given below.

$V\text{'}\left(r\right)=4\pi r^2$ (2)

Compare the equations (1) and (2).

$S\left(r\right)=V\text{'}\left(r\right)=4\pi r^2$

Show that if the change in radius $\Delta r$ is small, then the change in the circle of area is approximately equal to its circumference. Prove the statement through geometry.

Approximate the resulting change in volume $\Delta V$ of the sphere of radius r, if change in radius $\Delta r$ is small.

Write the $\Delta A$ expression as below.

$\begin{array}{l}\Delta V=\text{Volume}\text{of}\text{sphere}\text{with}\text{increasing}\text{radius -Volume}\text{of}\text{sphere }\\ \Delta V=\frac{4\pi }{3}{\left(r+\Delta r\right)}^3-r^3\\ \Delta V=\frac{4\pi }{3}\left(3r^2\left(\Delta r\right)+3r{\left(\Delta r\right)}^2+{\left(\Delta r\right)}^3\right)\end{array}$

The value of change in radius $\Delta r$ is small and negligible.

The above equation for change in volume $\Delta V$ is transformed as below

$\begin{array}{l}\Delta V=4\pi \left(r^2\left(\Delta r\right)\right)\\ \frac{\Delta V}{\Delta r}=4\pi r^2\end{array}$

Hence, it is geometrically true. Hence, the resulting change in volume $\Delta V$ if the value of change in radius $\Delta r$ is small is $4\pi r^2$.