Problem 39 - Calculus Early Transcendentals

Question

The length \(\ell,\) width \(w,\) and height \(h\) of a box change with time. At a certain instant the dimensions are \(\ell=1 \mathrm{m}\) and \(w=h=2 \mathrm{m},\) and \(\ell\) and \(w\) are increasing at a rate of 2 \(\mathrm{m} / \mathrm{s}\) while \(h\) is decreasing at a rate of 3 \(\mathrm{m} / \mathrm{s}\) . At that instant find the rates at which the following quantities are changing. (a) The volume (b) The surface area (c) The length of a diagonal

Step-by-Step Solution

Answer

(a) 6 m³/s; (b) 10 m²/s; (c) 0 m/s.

Step 1: Define the Variables

Identify the given variables and their rates of change. We have the lengths \( \ell = 1 \, \mathrm{m}, w = 2 \, \mathrm{m}, h = 2 \, \mathrm{m} \). The rates of change are \( \frac{d\ell}{dt} = 2 \, \mathrm{m/s}, \frac{dw}{dt} = 2 \, \mathrm{m/s}, \frac{dh}{dt} = -3 \, \mathrm{m/s} \).

Step 2: Rate of Change of Volume

The volume \( V \) of the box is given by \( V = \ell \cdot w \cdot h \). To find the rate of change of the volume, differentiate with respect to time \( t \): \[ \frac{dV}{dt} = \ell \cdot w \cdot \frac{dh}{dt} + \ell \cdot h \cdot \frac{dw}{dt} + w \cdot h \cdot \frac{d\ell}{dt} \]. Fill in the values: \( \frac{dV}{dt} = 1 \cdot 2 \cdot (-3) + 1 \cdot 2 \cdot 2 + 2 \cdot 2 \cdot 2 = -6 + 4 + 8 = 6 \, \mathrm{m^3/s} \).

Step 3: Rate of Change of Surface Area

The surface area \( S \) is given by \( S = 2(\ell w + \ell h + wh) \). Differentiate with respect to \( t \): \( \frac{dS}{dt} = 2( w \cdot \frac{d\ell}{dt} + h \cdot \frac{d\ell}{dt} + \ell \cdot \frac{dw}{dt} + w \cdot \frac{dh}{dt} + \ell \cdot \frac{dh}{dt} + h \cdot \frac{dw}{dt} ) \). Substituting the values gives: \( \frac{dS}{dt} = 2( 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 2 + 2 \cdot (-3) + 1 \cdot (-3) + 2 \cdot 2 ) = 2(4 + 4 + 2 - 6 - 3 + 4) = 2 \cdot 5 = 10 \, \mathrm{m^2/s} \).

Step 4: Rate of Change of Length of Diagonal

The length of a diagonal \( D \) is \( D = \sqrt{\ell^2 + w^2 + h^2} \). To find \( \frac{dD}{dt} \), use the chain rule: \( \frac{dD}{dt} = \frac{1}{2\sqrt{\ell^2 + w^2 + h^2}} (2\ell \frac{d\ell}{dt} + 2w \frac{dw}{dt} + 2h \frac{dh}{dt}) \). Substituting the given values: \( D = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \) and \( \frac{dD}{dt} = \frac{1}{6}(2 \cdot 1 \cdot 2 + 2 \cdot 2 \cdot 2 + 2 \cdot 2 \cdot (-3)) = \frac{1}{6}(4 + 8 - 12) = 0 \, \mathrm{m/s} \).

Key Concepts

Volume Rate of Change

In this exercise, we are asked to find the rate of change of the volume of a box with respect to time. The volume of a box, or a rectangular prism, is calculated by multiplying its length, width, and height, meaning the equation is given by \( V = \ell \cdot w \cdot h \). Because the box’s dimensions are changing over time, we need to use calculus to determine how fast the volume changes.
To do this, we take the time derivative of the volume equation:

• \( \frac{dV}{dt} = \ell \cdot w \cdot \frac{dh}{dt} + \ell \cdot h \cdot \frac{dw}{dt} + w \cdot h \cdot \frac{d\ell}{dt} \)

This formula represents how changes in length, width, and height each contribute to the rate at which the volume changes. Let's break it down:

• \( \ell \cdot w \cdot \frac{dh}{dt} \) accounts for the change in volume due to changes in height.
• \( \ell \cdot h \cdot \frac{dw}{dt} \) accounts for changes in width.
• \( w \cdot h \cdot \frac{d\ell}{dt} \) accounts for changes in length.

Plugging in the given values, where \( \ell = 1 \ \mathrm{m}, w = 2 \ \mathrm{m}, h = 2 \ \mathrm{m} \) with their respective rates, results in a volume change of \( 6 \, \mathrm{m}^3/\mathrm{s} \). This indicates an increase in the box's volume over time.

Surface Area Rate of Change

The surface area of a box is associated with the sum of the areas of all its sides. For a box with dimensions \( \ell \), \( w \), and \( h \), the surface area \( S \) is expressed as:

• \( S = 2(\ell w + \ell h + wh) \)

As the dimensions change, each surface area component contributes to the overall rate of change of the surface area. To find this rate, the derivative with respect to time is taken:

• \( \frac{dS}{dt} = 2( w \cdot \frac{d\ell}{dt} + h \cdot \frac{d\ell}{dt} + \ell \cdot \frac{dw}{dt} + w \cdot \frac{dh}{dt} + \ell \cdot \frac{dh}{dt} + h \cdot \frac{dw}{dt} ) \)

This derivative consolidates the individual changes from each dimension:

• The terms involving \( \ell, w, h \) represent the contribution from each face of the box.
• Each rate of change term (e.g., \( \frac{d\ell}{dt} \)) is multiplied by the dimensions of the face it affects, doubled due to symmetry.

After substituting the known values for \( \ell, w, \) and \( h \) and their rates, we find that the surface area is changing at \( 10 \, \mathrm{m}^2/\mathrm{s} \). This calculation reflects how the shrinking and expanding sides interact over time to alter total area.

Length of Diagonal Rate of Change

The diagonal of a box is the cavernous link connecting opposite corners, visually representing the longest length within the box. Mathematically, if the dimensions of the box are \( \ell, w, h \), its diagonal \( D \) is calculated as:

• \( D = \sqrt{\ell^2 + w^2 + h^2} \)

As the box's dimensions shift, how quickly the diagonal length changes is explored by differentiating \( D \) with respect to time \( t \). The derivative uses the chain rule:

• \( \frac{dD}{dt} = \frac{1}{2\sqrt{\ell^2 + w^2 + h^2}} (2\ell \frac{d\ell}{dt} + 2w \frac{dw}{dt} + 2h \frac{dh}{dt}) \)

This equation factors in how each dimension influences diagonal length by utilizing each dimension's square and its respective rate:

• The prefactor adjusts the rate by the initial diagonal length \( D \).
• The contributing factor from each dimension, scaled by its rate of change, highlights individual impacts on the diagonal.

When applying our given dimensions \( \ell, w, h \) and their changing rates, the derivative simplifies to zero, meaning that—even as the dimensions change—the diagonal's length remains constant at this instant.