Chapter 4.6, Problem 10E

(a)

To determine

To find : The rates at that the box’s volume.

Answer to Problem 10E

The answer: $2{\text{m}}^3/\text{sec}$

Explanation of Solution

Given information:

The edge length $x,y,z$ of a closed rectangular box:

  $\begin{array}{c}\frac{dx}{dt}=1\text{m}/\text{sec},\\ \frac{dy}{dt}=-2\text{m}/\text{sec},\\ \frac{dz}{dt}=1\text{m}/\sec \end{array}$

Calculation:

  $\begin{array}{c}\frac{dx}{dt}=1\\ \frac{dy}{dt}=-2\\ \frac{dz}{dt}=1\\ x=4\\ y=3\\ z=2\end{array}$

Differentiate subtly. Apply the Product Rule to the inner $yz$ first, then to $x(yz)$

To determine the solution, enter each of the values given.

  $\begin{array}{c}V=xyz\\ \Rightarrow \frac{dV}{dt}=x\left(\frac{d}{dt}(yz)\right)+yz\frac{dx}{dt}\\ =x\left(y\frac{dz}{dt}+z\frac{dy}{dt}\right)+yz\frac{dx}{dt}\\ =xy\frac{dz}{dt}+xz\frac{dy}{dt}+yz\frac{dx}{dt}\\ =(4)(3)(1)+(4)(2)(-2)+(3)(2)(1)\\ =2\end{array}$

Therefore the required rates at that the box’s volume is $2{\text{m}}^3/\text{sec}$ .

(b)

To determine

To find: The rates at that the box’s surface area.

Answer to Problem 10E

The answer: $0{\text{m}}^2/\text{sec}$

Explanation of Solution

Given information:

The edge length $x,y,z$ of a closed rectangular box:

  $\begin{array}{c}\frac{dx}{dt}=1\text{m}/\text{sec},\\ \frac{dy}{dt}=-2\text{m}/\text{sec},\\ \frac{dz}{dt}=1\text{m}/\sec \end{array}$

Calculation:

Apply the Product Rule three times while implicitly differentiating the surface area equation $S$ .

To determine the solution, enter each of the values given.

  $\begin{array}{c}\frac{dx}{dt}=1\\ \frac{dy}{dt}=-2\\ \frac{dz}{dt}=1\\ x=4\\ y=3\\ z=2\end{array}$

So,

  $\begin{array}{c}S=2xy+2xz+2yz\\ =2(xy+xz+yz)\\ \frac{dS}{dt}=2\left(x\frac{dy}{dt}+y\frac{dx}{dt}+x\frac{dz}{dt}+z\frac{dx}{dt}+y\frac{dz}{dt}+z\frac{dy}{dt}\right)\\ =2(4(-2)+3(1)+4(1)+2(1)+3(1)+2(-2))\\ =0\end{array}$

Therefore the required rates at that the box’s surface area is $0{\text{m}}^2/\text{sec}$

(c)

To determine

To find: The rates at that the box’s diagonal length.

Answer to Problem 10E

The answer: $0\text{m}/\text{sec}$

Explanation of Solution

Given information:

The edge length $x,y,z$ of a closed rectangular box:

  $\begin{array}{c}\frac{dx}{dt}=1\text{m}/\text{sec},\\ \frac{dy}{dt}=-2\text{m}/\text{sec},\\ \frac{dz}{dt}=1\text{m}/\sec \end{array}$

Calculation:

Apply the Power and Chain Rules to implicitly differentiate the diagonal length equation $s$

To determine the solution, enter each of the values given.

  $\begin{array}{c}\frac{dx}{dt}=1\\ \frac{dy}{dt}=-2\\ \frac{dz}{dt}=1\\ x=4\\ y=3\\ z=2\end{array}$

So,

  $\begin{array}{c}s=\sqrt{x^2+y^2+z^2}\\ \Rightarrow \frac{ds}{dt}=\frac{1}{2}{\left(x^2+y^2+z^2\right)}^{-1/2}\left(2x\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}\right)\\ ={\left(x^2+y^2+z^2\right)}^{-1/2}\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)\\ ={\left(4^2+3^2+2^2\right)}^{-1/2}(4(1)+3(-2)+2(1))\\ =\frac{0}{\sqrt{29}}\\ =0\end{array}$

Therefore the required rates at that the box’s diagonal length $0\text{m}/\text{sec}$ .