Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

The temperature of a metal plate at \((x, y)\) is \(e^{-x-3 y}\) degrees. A bug is walking northeast at a rate of \(\sqrt{8}\) feet per minute (i.e., \(d x / d t=d y / d t=2\) ). From the bug's point of view, how is the temperature changing with time as it crosses the origin?

Step-by-Step Solution

Answer

The temperature is decreasing by 8 degrees per minute as the bug crosses the origin.

Step 1: Determine the Temperature Function

The temperature function given is \( T(x, y) = e^{-x - 3y} \). This function describes how temperature varies with position \( (x, y) \) on the plate.

Step 2: Find the Partial Derivatives

To find how the temperature changes with respect to time, calculate the partial derivatives of the temperature function. The partial derivative with respect to \( x \) is \( \frac{\partial T}{\partial x} = -e^{-x - 3y} \) and with respect to \( y \) is \( \frac{\partial T}{\partial y} = -3e^{-x - 3y} \).

Step 3: Apply the Chain Rule for Time Rate of Change

The rate of change of temperature with time \( \frac{dT}{dt} \) can be found using the chain rule: \( \frac{dT}{dt} = \frac{\partial T}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial T}{\partial y} \cdot \frac{dy}{dt} \).

Step 4: Substitute Known Derivatives and Values

Substitute the partial derivatives and values: \( \frac{dx}{dt} = 2 \) and \( \frac{dy}{dt} = 2 \). At the origin \( (0,0) \), the partial derivatives simplify to \( \frac{\partial T}{\partial x} = -1 \) and \( \frac{\partial T}{\partial y} = -3 \). Thus, \( \frac{dT}{dt} = (-1) \cdot 2 + (-3) \cdot 2 = -2 - 6 = -8 \).

Step 5: Conclusion About Temperature Rate

The negative sign indicates that the temperature is decreasing. Therefore, the temperature is decreasing at a rate of 8 degrees per minute as the bug moves across the origin.

Key Concepts

Temperature Function

A temperature function describes how the temperature varies at different points on an object or surface. In our exercise, this function is given by the formula \( T(x, y) = e^{-x - 3y} \). This tells us that the temperature depends on the coordinates \((x, y)\) on the metal plate. The function \( e^{-x - 3y} \) indicates an exponential decrease in temperature as either \(x\) or \(y\) increases. This behavior can be visualized as contour lines on the plate, where each line represents a constant temperature.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus. They show how a function changes with respect to one variable while keeping other variables constant. For the temperature function \( T(x, y) = e^{-x - 3y} \), we find two partial derivatives:

• \( \frac{\partial T}{\partial x} = -e^{-x - 3y} \)
• \( \frac{\partial T}{\partial y} = -3e^{-x - 3y} \)

These derivatives help us understand the rate at which temperature changes along the \(x\) and \(y\) directions. Both derivatives contain negative terms, implying that temperature decreases as \(x\) or \(y\) increases.

Chain Rule

The chain rule is a crucial concept in calculus used to find the derivative of composite functions. In the context of our exercise, it allows us to determine how the temperature varies with time as the bug moves. By applying the chain rule, we combine the partial derivatives with the rates of change of \(x\) and \(y\) to find \( \frac{dT}{dt} \), the rate of temperature change:\[\frac{dT}{dt} = \frac{\partial T}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial T}{\partial y} \cdot \frac{dy}{dt}\]This formula considers that both \(x\) and \(y\) coordinates change as the bug moves northeast, affecting the overall temperature change over time.

Rate of Change

The rate of change in this context refers to how the temperature changes over time as the bug travels across the metal plate. In our calculation, we found the rate of change to be \(-8\) degrees per minute. This negative value tells us that the temperature is decreasing.

• At the origin, where \( (x, y) = (0, 0) \), \( \frac{dx}{dt} = 2 \) and \( \frac{dy}{dt} = 2 \)
• Partial derivatives become \(-1\) for \( \frac{\partial T}{\partial x} \) and \(-3\) for \( \frac{\partial T}{\partial y} \)

By substituting these and performing the calculations, we determined that the temperature rate of change is \(-8\), indicating a decrease by 8 degrees each minute as the bug moves in its path.