Problem 22 - University Calculus: Early Transcendentals

Question

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 \(z=2\)

Step-by-Step Solution

Answer

(a) 2 m³/sec, (b) 0 m²/sec, (c) 0 m/sec.

Step 1: Understand the Problem

We are given a rectangular box with edges changing at certain rates and need to find the rates of change for volume, surface area, and diagonal length. We need to derive the formulas for these, then differentiate them with respect to time.

Step 2: Volume Rate of Change

The volume of the box is given by the formula \( V = x imes y imes z \). To find how fast the volume is changing over time, we use the product rule for differentiation:\[\frac{dV}{dt} = \frac{dx}{dt}yz + x\frac{dy}{dt}z + xy\frac{dz}{dt}\]Substitute the given rates and dimensions: \[\frac{dV}{dt} = 1(3)(2) + 4(-2)(2) + 4(3)(1)\]

Step 3: Compute Change in Volume

Substituting the values from Step 2, we calculate:\[\frac{dV}{dt} = 6 - 16 + 12 = 2 \, \text{cubic meters per second}\]

Step 4: Surface Area Rate of Change

The surface area \( A \) of the box is \( 2(xy + xz + yz) \). Differentiate:\[\frac{dA}{dt} = 2 \left(\frac{dx}{dt}(y + z) + \frac{dy}{dt}(x + z) + \frac{dz}{dt}(x + y)\right)\]Substitute the dimensions and their rate of changes:\[\frac{dA}{dt} = 2 \left(1(3 + 2) + (-2)(4 + 2) + 1(4 + 3)\right)\]

Step 5: Compute Change in Surface Area

Substituting the values from Step 4, we calculate:\[\frac{dA}{dt} = 2 \left(5 - 12 + 7\right) = 0 \, \text{square meters per second}\]

Step 6: Diagonal Length Rate of Change

The diagonal length \( s \) is given by \( s = \sqrt{x^2 + y^2 + z^2} \). Differentiate using the chain rule:\[\frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} (2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt})\]Substitute the dimensions and their rate of changes:\[s = \sqrt{4^2 + 3^2 + 2^2} = \sqrt{29},\]\[\frac{ds}{dt} = \frac{1}{\sqrt{29}} \left(8 + (-12) + 4\right)\]

Step 7: Compute Change in Diagonal Length

Calculate the result from Step 6:\[\frac{ds}{dt} = \frac{1}{\sqrt{29}} (0) = 0 \, \text{meters per second}\]

Step 8: Summarize Results

We found that at the given moment:(a) Volume changes at a rate of \(2\, \text{m}^3/\text{sec}\).(b) Surface area changes at a rate of \(0\, \text{m}^2/\text{sec}\).(c) Diagonal length changes at a rate of \(0\, \text{m/sec}\).

Key Concepts

Rectangular Box

A rectangular box is a 3-dimensional geometric shape with six rectangular faces. It consists of three pairs of parallel faces, commonly characterized by its length, width, and height, denoted here as \( x \), \( y \), and \( z \). These dimensions can change over time in real-world scenarios, such as in packing or thermal expansion processes.

In the problem we're tackling, the dimensions of our rectangular box are not constant. They vary at specific rates given by \( \frac{d x}{dt} = 1 \ \text{m/s} \), \( \frac{d y}{dt} = -2 \ \text{m/s} \), and \( \frac{d z}{dt} = 1 \ \text{m/s} \). This rate of change is essential as it helps determine how other properties of the box—like its volume, surface area, and diagonal length—are affected over time.

Volume Change

The volume of a rectangular box is derived by multiplying its length, width, and height. For our problem, this can be expressed as \( V = x \times y \times z \). To find how fast the volume changes as the dimensions of the box change, we apply the product rule of differentiation.

• Using the differentiation formula, we find \( \frac{dV}{dt} = \frac{dx}{dt}yz + x\frac{dy}{dt}z + xy\frac{dz}{dt} \).


• Substitute the values: \( x = 4, \ y = 3, \ z = 2 \) into the derivative gives us \( \frac{dV}{dt} = 6 - 16 + 12 = 2 \, \text{m}^3/\text{sec} \).

The positive rate of \( 2 \, \text{cubic meters per second}\) indicates that the volume of the box is increasing as time progresses, despite the decrease in \( y \). The increase in the other two dimensions outweighs the shrinkage in \( y \).

Surface Area Change

To compute how surface area changes with time, we start with the formula for the surface area of a box: \( A = 2(xy + xz + yz) \). Differentiation is applied to find the rate of change of the surface area.

• We employ: \( \frac{dA}{dt} = 2 \left( \frac{dx}{dt}(y + z) + \frac{dy}{dt}(x + z) + \frac{dz}{dt}(x + y) \right) \).


• Plug in the specific values \( x = 4, \ y = 3, \ z = 2 \), and corresponding rates to get \( \frac{dA}{dt} = 0 \ \text{m}^2/\text{sec} \).

This zero rate means the surface area does not change at that moment, reflecting a balance between the decreasing and increasing dimensions. Changes in \( y \) are perfectly counteracted by changes in \( x \) and \( z \) at that specific instance.

Diagonal Length Change

The diagonal length \( s \) of a rectangular box is given by the formula \( s = \sqrt{x^2 + y^2 + z^2} \). It represents the longest straight-line distance across the box. To determine how this diagonal changes over time, we differentiate this expression using the chain rule.

• The differentiation results in: \( \frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} (2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt}) \).


• By substituting the rate values and dimensions \( x = 4, \ y = 3, \ z = 2 \), we get a rate of \( 0 \ \text{m/sec} \).

The zero rate signifies that the length of the diagonal does not change at that precise moment due to the simultaneous actions of the dimension changes. While individual dimensions have changes, they are balanced in such a way that the overall diagonal remains constant.