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Chapter 3: Problem 22

Changing dimensions in a rectangular box Suppose that the edge lengths \(x, y,\) and \(z\) of a closed rectangular box are changing at the following rates: $$\frac{d x}{d t}=1 \mathrm{m} / \mathrm{sec}, \quad \frac{d y}{d t}=-2 \mathrm{m} / \mathrm{sec}, \quad \frac{d z}{d t}=1 \mathrm{m} / \mathrm{sec}.$$ Find the rates at which the box's (a) volume, (b) surface area, and (c) diagonal length \(s=\sqrt{x^{2}+y^{2}+z^{2}}\) are changing at the instant when \(x=4, y=3,\) and \(z=2\).

Short Answer

Expert verified
(a) 2 m鲁/s, (b) 0 m虏/s, (c) 0 m/s.

Step by step solution

01

Define Volume and Differentiate

The volume of the box is given by the formula \( V = x y z \). To find the rate of change of the volume, differentiate with respect to time \( t \):\[ \frac{dV}{dt} = \frac{d}{dt}(xyz) = x \frac{dy}{dt} z + y \frac{dz}{dt} x + z \frac{dx}{dt} y \].Substitute the known values: \( x = 4, y = 3, z = 2\) and their rates \( \frac{dx}{dt} = 1, \frac{dy}{dt} = -2, \frac{dz}{dt} = 1 \).\[ \frac{dV}{dt} = (4)(-2)(2) + (3)(1)(4) + (2)(1)(3) = -16 + 12 + 6 = 2 \, \text{cubic meters per second}. \]
02

Define Surface Area and Differentiate

The surface area \( A \) of the box is given by the formula \( A = 2(xy + xz + yz) \). To find the rate of change of the surface area, differentiate:\[ \frac{dA}{dt} = 2 \left( y \frac{dx}{dt} + x \frac{dy}{dt} + z \frac{dx}{dt} + x \frac{dz}{dt} + z \frac{dy}{dt} + y \frac{dz}{dt} \right). \]Substitute the known values: \( x = 4, y = 3, z = 2 \), and their rates:\[ \frac{dA}{dt} = 2 \left( 3(1) + 4(-2) + 2(1) + 4(1) + 2(-2) + 3(1) \right) = 2 \left( 3 - 8 + 2 + 4 - 4 + 3 \right) = 2 \times 0 = 0 \, \text{square meters per second}. \]
03

Define Diagonal Length and Differentiate

The diagonal length \( s \) is given by \( s = \sqrt{x^2 + y^2 + z^2} \). Differentiate with respect to time:\[ \frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} (2x \frac{dx}{dt} + 2y \frac{dy}{dt} + 2z \frac{dz}{dt}). \]Substitute the known values: \( x = 4, y = 3, z = 2 \), and their rates:\[ s = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{16 + 9 + 4} = \sqrt{29}, \]\[ \frac{ds}{dt} = \frac{1}{\sqrt{29}} (4(1) + 3(-2) + 2(1)) = \frac{1}{\sqrt{29}} (4 - 6 + 2) = \frac{0}{\sqrt{29}} = 0 \, \text{meters per second}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume of Rectangular Box
The volume of a rectangular box measures the total space contained within. It is calculated by the formula \( V = x y z \), where \( x \), \( y \), and \( z \) are the box's length, width, and height, respectively. When these dimensions change with time, the volume also changes, and we represent this rate of change with \( \frac{dV}{dt} \).

This rate is critical for understanding how substances fill or empty in real-time scenarios. To derive \( \frac{dV}{dt} \), we differentiate the volume formula with respect to time:\[ \frac{dV}{dt} = x \frac{dy}{dt} z + y \frac{dz}{dt} x + z \frac{dx}{dt} y \].

After substituting the values for each dimension and their rates, you can find the numerical rate at which the volume changes.
Surface Area of Rectangular Box
The surface area of a rectangular box is the total area of all its six faces. It follows the formula \( A = 2(xy + xz + yz) \). If the dimensions \( x, y, \) and \( z \) vary with time, the surface area changes as well. To find the rate of this change, we use \( \frac{dA}{dt} \).

First, differentiate the formula for surface area with respect to time: \[ \frac{dA}{dt} = 2 \left( y \frac{dx}{dt} + x \frac{dy}{dt} + z \frac{dx}{dt} + x \frac{dz}{dt} + z \frac{dy}{dt} + y \frac{dz}{dt} \right) \].

Substitute the given values and their rates to compute how the surface area is changing at a specific instant. This is particularly important in applications such as painting or wrapping where the surface coverage needs to be monitored.
Diagonal Length of Rectangular Box
The diagonal length of a rectangular box gives the longest straight line that can be drawn from one corner to its opposite. Calculated by the formula \( s = \sqrt{x^2 + y^2 + z^2} \), it reflects a key spatial property of the box. When the box dimensions shift over time, the diagonal length can also change, and this is depicted as \( \frac{ds}{dt} \).

This is derived by first differentiating the formula for diagonal length with respect to time: \[ \frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} (2x \frac{dx}{dt} + 2y \frac{dy}{dt} + 2z \frac{dz}{dt}) \].

By plugging in the known values and their rates, you assess how the diagonal shifts as the box reshapes. Knowing \( \frac{ds}{dt} \) can be crucial for operations that depend on maximizing or minimizing space, like storage or shipping logistics.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=x^{3}+x^{2}-x, x_{0}=1$$

Changing dimensions in a rectangle The length \(l\) of a rectangle is decreasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec}\) while the width \(w\) is increasing at the rate of 2 \(\mathrm{cm} / \mathrm{sec} .\) When \(l=12 \mathrm{cm}\) and \(w=5 \mathrm{cm},\) find the rates of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of the rectangle. Which of these quantities are decreasing, and which are increasing?

The diameter of a tree was 10 in. During the following year, the circumference increased 2 in. About how much did the tree's diameter increase? The tree's cross-sectional area?

Find \(d y / d x\) if \(y=x^{3 / 2}\) by using the Chain Rule with \(y\) as a composite of $$\begin{array}{l}{\text { a. } y=u^{3} \text { and } u=\sqrt{x}} \\ {\text { b. } y=\sqrt{u} \text { and } u=x^{3}}\end{array}$$

The derivative of \(\cos \left(x^{2}\right) \quad\) Graph \(y=-2 x \sin \left(x^{2}\right)\) for \(-2 \leq\) \(x \leq 3 .\) Then, on the same screen, graph $$y=\frac{\cos \left((x+h)^{2}\right)-\cos \left(x^{2}\right)}{h}$$ for \(h=1.0,0.7,\) and \(0.3 .\) Experiment with other values of \(h .\) What do you see happening as \(h \rightarrow 0 ?\) Explain this behavior.
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