Problem 12 - Thomas Calculus

Question

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 \(z=2\)

Step-by-Step Solution

Answer

Rates: Volume = 2 m³/s, Surface Area = 0 m²/s, Diagonal = 0 m/s.

Step 1: Calculate the Rate of Change of Volume

The volume \( V \) of the box is given by \( V = x \, y \, z \). To find \( \frac{dV}{dt} \), apply the product rule for differentiation:\[ \frac{dV}{dt} = y \, z \, \frac{dx}{dt} + x \, z \, \frac{dy}{dt} + x \, y \, \frac{dz}{dt} \]Substitute \( x = 4, \, y = 3, \, z = 2 \), \( \frac{dx}{dt} = 1, \, \frac{dy}{dt} = -2, \, \frac{dz}{dt} = 1 \):\[ \frac{dV}{dt} = 3 \, (2) \, (1) + 4 \, (2) \, (-2) + 4 \, (3) \, (1) = 6 - 16 + 12 = 2 \]The volume is changing at a rate of \( 2 \, \text{m}^3/\text{sec} \).

Step 2: Calculate the Rate of Change of Surface Area

The surface area \( A \) of the box is given by \( A = 2(xy + yz + zx) \). Find \( \frac{dA}{dt} \) using the formula:\[ \frac{dA}{dt} = 2 \, \left( y \, \frac{dx}{dt} + x \, \frac{dy}{dt} \right) + 2 \, \left( z \, \frac{dy}{dt} + y \, \frac{dz}{dt} \right) + 2 \, \left( x \, \frac{dz}{dt} + z \, \frac{dx}{dt} \right) \]Substitute the given values:\[ \frac{dA}{dt} = 2((3)(1) + (4)(-2)) + 2((2)(-2) + (3)(1)) + 2((4)(1) + (2)(1)) \]Calculate each term:\[ = 2(3 - 8) + 2(-4 + 3) + 2(4 + 2) = 2(-5) + 2(-1) + 2(6) = -10 - 2 + 12 = 0 \]The surface area is not changing, \( \frac{dA}{dt} = 0 \, \text{m}^2/\text{sec} \).

Step 3: Calculate the Rate of Change of the Diagonal Length

The diagonal length \( s \) is given by \( s = \sqrt{x^2 + y^2 + z^2} \). To find \( \frac{ds}{dt} \), use implicit differentiation:\[ \frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \times (2x\frac{dx}{dt} + 2y\frac{dy}{dt} + 2z\frac{dz}{dt}) \]Substitute \( x = 4, \, y = 3, \, z = 2 \) and their derivatives:\[ \frac{ds}{dt} = \frac{1}{2\sqrt{4^2 + 3^2 + 2^2}} \times (2(4)(1) + 2(3)(-2) + 2(2)(1)) \]Solve for \( s \):\[ = \frac{1}{2\sqrt{29}} \times (8 - 12 + 4) = \frac{1}{2\sqrt{29}} \times 0 = 0 \]The diagonal length is not changing, \( \frac{ds}{dt} = 0 \).

Key Concepts

Derivative

Understanding the concept of a derivative is crucial when dealing with related rates problems. The derivative represents how a function changes as its input changes. In simpler words, it tells us how fast or slow something is changing at a particular point in time.
For example, if you have a function that describes how the volume of a box changes as its dimensions change, the derivative of that function gives the rate at which the volume is changing. This is what we see when calculating \( \frac{dV}{dt} \), the rate of change of volume with respect to time. In this problem, we're using derivatives to relate the changing lengths, \(x, y, z\), of the box's edges to how fast the entire volume, surface area, and diagonal length of the box are changing.
Understanding derivatives allows you to break down complex changes into simpler, understandable components. Each derivative essentially measures a different piece of the box's behavior as its shape alters over time.

Volume Change

Volume change in the context of this problem measures how quickly the entire space inside the box is increasing or decreasing. The formula for the volume of a rectangular box is \( V = x \cdot y \cdot z \).
To find the rate of volume change, or \( \frac{dV}{dt} \), the product rule is used because volume is the product of the box's three dimensions. We calculate how fast the lengths \(x\), \(y\), and \(z\) are changing to understand how these changes contribute to the overall rate of volume change.

• For \( \frac{dV}{dt} \), each term in the differentiation formula represents how one part of the box’s dimensions affects the volume.
• In this example, we plug in values for \(x\), \(y\), \(z\), and their respective rates of change to find that the volume is increasing by 2 cubic meters per second.

By calculating these rates correctly, we can predict how the box is expanding or contracting over time, which can be crucial for applications where space and storage capacity are concerned.

Surface Area Change

The rate of change of the surface area of a rectangular box is another interesting aspect to consider. The surface area \( A \) of a box is calculated as \( A = 2(xy + yz + zx) \).
To find \( \frac{dA}{dt} \), we apply the rule of differentiation for each term separately, considering the rate of change for every edge length. This results in understanding how the growth or shrinkage of each side of the box affects its overall surface area.

• In this calculation, you sum up the contributions from all three pairs of sides.
• For this example, the final calculation shows that the surface area isn't changing over time.

Despite the box's changing dimensions, the balancing act of the increasing and decreasing terms leads to a net change of 0, which means the surface area is constant at this moment. This can inform decisions where surface treatment processes, coatings, or visibility of the box are important considerations.

Diagonal Length

The diagonal length provides an interesting look into the overall size of the box. The formula to find the diagonal is \( s = \sqrt{x^2 + y^2 + z^2} \), which takes into account all the dimensions of the box.
Through implicit differentiation, the rate of change of the diagonal length \( \frac{ds}{dt} \) can be determined from the derivatives of the side lengths' squares. This involves a derivative of a square root function, which gives us a comprehensive view of how the box's changing dimensions impact its diagonal length.

• Interestingly, in our scenario, even though the box's side lengths are changing, the rate of change of the diagonal, \( \frac{ds}{dt} \), is 0.
• This means that, momentarily, the stretching and shrinking of the box balance out perfectly to keep the diagonal length constant.

Understanding diagonal length change is important in navigation, construction, and design fields, where symmetry or fixed points in space are key factors.