Chapter 5.6, Problem 10E

a.

To determine

To find: the rates of the volume of a rectangular box.

Answer to Problem 10E

The value of $\frac{dv}{dt}$ is $2\text{ m}{\text{.}}^2/\sec .$

Explanation of Solution

Given information:

  $\frac{dy}{dt}=-2\text{ m}\text{./sec}\text{.}$ .

  $\frac{dz}{dt}=1\text{ m}\text{./sec}\text{.}$

  $\frac{dx}{dt}=1\text{ m}\text{./sec}\text{.}$

  $x=4$

  $y=3$

  $z=2$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rates of the volume of a rectangular box.

Therefore,

Volume of a rectangular box-

  $\text{V}=xyz$

Differentiate with respect to t using product rule.

  $\frac{dv}{dt}=xy\frac{dz}{dt}+xz\frac{dy}{dt}+yz\frac{dx}{dt}$

Putting the value of $x,y,z,\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$ in above equation,

  $\begin{array}{l}\frac{dv}{dt}=\left(4\right)\left(3\right)\left(1\right)+\left(4\right)\left(2\right)\left(-2\right)+\left(3\right)\left(2\right)\left(1\right)\\ \frac{dv}{dt}=12+\left(-16\right)+6\\ \frac{dv}{dt}=12+6-16\\ \frac{dv}{dt}=18-16\\ \frac{dv}{dt}=2\text{ m}{\text{.}}^2\text{/sec}\text{.}\end{array}$

Hence, the value of $\frac{dv}{dt}$ is $2\text{m}{\text{.}}^2/\sec .$

b.

To determine

To find: the rates of the surface area of a rectangular box.

Answer to Problem 10E

The value of $\frac{ds}{dt}$ is $\text{0m}{\text{.}}^2/\sec .$

Explanation of Solution

Given information:

  $\frac{dy}{dt}=-2\text{ m}\text{./sec}\text{.}$

  $\frac{dz}{dt}=1\text{ m}\text{./sec}\text{.}$

  $\frac{dx}{dt}=1\text{ m}\text{./sec}\text{.}$

  $x=4$

  $y=3$

  $z=2$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rates of the surface area of a rectangular box.

Therefore,

Surface area of a rectangular box-

  $\text{S}=2xy+2yz+2xz$

Differentiate with respect to t using product rule.

  $\frac{ds}{dt}=2\left[x\frac{dy}{dt}+y\frac{dx}{dt}+y\frac{dz}{dt}+z\frac{dy}{dt}+x\frac{dz}{dt}+z\frac{dx}{dt}\right]$

Putting the value of $x,y,z,\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$ in above equation,

  $\begin{array}{l}\frac{dv}{dt}=2\left[\left(4\right)\left(-2\right)+\left(3\right)\left(1\right)+\left(3\right)\left(1\right)+\left(2\right)\left(-2\right)+\left(4\right)\left(1\right)+\left(2\right)\left(1\right)\right]\\ \frac{dv}{dt}=2\left[-8+3+3-4+4+2\right]\\ \frac{dv}{dt}=2\left[12-12\right]\\ \frac{dv}{dt}=0\text{ m}{\text{.}}^2\text{/sec}\text{.}\end{array}$

Hence, the value of $\frac{ds}{dt}$ is $\text{0m}{\text{.}}^2/\sec .$

c.

To determine

To find: the rates of the diagonal length of a rectangular box.

Answer to Problem 10E

The value of $\frac{ds}{dt}$ is

  $0$

Explanation of Solution

Given information:

  $\frac{dy}{dt}=-2\text{ m}\text{./sec}\text{.}$

  $\frac{dz}{dt}=1\text{ m}\text{./sec}\text{.}$

  $\frac{dx}{dt}=1\text{ m}\text{./sec}\text{.}$

  $x=4$

  $y=3$

  $z=2$

All variables are differentiable functions of t .

Calculation :

We have to calculate the rates of the diagonal length of a rectangular box.

Therefore,

Diagonal length of a rectangular box-

  $\begin{array}{l}\text{S}=\sqrt{x^2+y^2+z^2}\\ \text{S}={\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\end{array}$

Differentiate with respect to t using product rule.

  $\begin{array}{l}\frac{ds}{dt}=\frac{1}{2}{\left[x^2+y^2+z^2\right]}^{\frac{1}{2}-1}\cdot \left(2x\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}\right)\\ \frac{ds}{dt}=\frac{1}{2}{\left[x^2+y^2+z^2\right]}^{-\frac{1}{2}}\cdot 2\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)\\ \frac{ds}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}{\sqrt{x^2+y^2+z^2}}\end{array}$

Putting the value of $x,y,z,\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}$ in above equation,

  $\begin{array}{l}\frac{ds}{dt}=\frac{\left(4\right)\left(1\right)+\left(3\right)\left(-2\right)+\left(2\right)\left(1\right)}{\sqrt{{\left(4\right)}^2+{\left(3\right)}^2+{\left(2\right)}^2}}\\ \frac{ds}{dt}=\frac{4+\left(-6\right)+2}{\sqrt{16+9+4}}\\ \frac{ds}{dt}=\frac{6-6}{\sqrt{29}}\\ \frac{ds}{dt}=\frac{0}{\sqrt{29}}\\ \frac{ds}{dt}=0\end{array}$

Hence, the value of $\frac{ds}{dt}$ is $0$ .