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 decreases at 8 degrees per minute as the bug crosses the origin.

Step 1: Define the Problem

We are given a temperature function of the form \(T(x, y) = e^{-x - 3y}\) and the rate of movement of a bug in the northeast direction. Our task is to find how the temperature changes over time as the bug moves, specifically as it crosses the origin \((0, 0)\).

Step 2: Identify Rates of Change

The bug is moving northeast, meaning both its \(x\) and \(y\) coordinates increase at the same rate. We are given \(\frac{dx}{dt} = \frac{dy}{dt} = 2\).

Step 3: Partial Derivatives

Calculate the partial derivatives \(\frac{\partial T}{\partial x}\) and \(\frac{\partial T}{\partial y}\) at the origin \((0, 0)\). The partial derivative with respect to \(x\) is \(-e^{-x - 3y}\), and with respect to \(y\) is \(-3e^{-x - 3y}\).

Step 4: Evaluate Partial Derivatives at Origin

Substitute \((x, y) = (0, 0)\) into the partial derivatives: \(\frac{\partial T}{\partial x}(0, 0) = -e^{0} = -1\), and \(\frac{\partial T}{\partial y}(0, 0) = -3e^{0} = -3\).

Step 5: Use the Chain Rule for Total Derivative

We use the chain rule to find \(\frac{dT}{dt}\). The chain rule states that \(\frac{dT}{dt} = \frac{\partial T}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial T}{\partial y} \cdot \frac{dy}{dt}\).

Step 6: Substitute and Calculate

Substitute the values into the chain rule expression: \(\frac{dT}{dt} = (-1) \cdot 2 + (-3) \cdot 2 = -2 - 6 = -8\).

Step 7: Interpret the Result

The temperature is decreasing at a rate of \(8\) degrees per minute as the bug moves northeast across the origin.

Key Concepts

Partial Derivatives

In the realm of calculus differentiation, partial derivatives are crucial when dealing with functions of multiple variables. A partial derivative helps us understand how a function changes as one of the variables is altered, keeping others constant.

• For a given function, say a temperature function defined as \( T(x, y) = e^{-x - 3y} \), the partial derivative with respect to \( x \) \( \left( \frac{\partial T}{\partial x} \right) \) gives insight into how the temperature changes as only \( x \) changes.
• Similarly, the partial derivative with respect to \( y \) \( \left( \frac{\partial T}{\partial y} \right) \), provides the rate of change of temperature as only \( y \) varies.

When evaluating at a point, such as the origin \((0,0)\), these derivatives give specific rates of change relevant to that location. For our scenario, \( \frac{\partial T}{\partial x}(0, 0) = -1 \) and \( \frac{\partial T}{\partial y}(0, 0) = -3 \). This tells us that, independently, temperature changes at these rates for small movements in x or y.

Chain Rule

The chain rule in calculus differentiation is a powerful tool for finding the rate of change of a function that depends on multiple interrelated variables. In the context of this exercise, where we have a moving bug affecting two dimensions simultaneously, the chain rule becomes indispensable.

• It allows us to compute the total derivative \( \frac{dT}{dt} \) of temperature with respect to time, even when both \( x \) and \( y \) are changing.
• By using the formula \( \frac{dT}{dt} = \frac{\partial T}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial T}{\partial y} \cdot \frac{dy}{dt} \), we can incorporate known rates of change \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) into the calculation.

For our problem, substituting known values yields \( \frac{dT}{dt} = (-1) \cdot 2 + (-3) \cdot 2 = -8 \). Thus, the temperature changes at an overall rate of \(-8\) degrees per minute as the bug travels northeast.

Rates of Change

Understanding rates of change, especially in how one variable impacts another over time, is foundational to calculus differentiation. In this exercise, we look at how the coordinates \( x \) and \( y \) shift as time progresses.

• The bug's movement northeast implies that both \( x \) and \( y \) coordinates increase at 2 units per minute, i.e., \( \frac{dx}{dt} = \frac{dy}{dt} = 2 \).
• This movement directly affects the temperature function \( T(x, y) \), requiring us to calculate how these positional changes translate to a temperature change over time.

This concept helps us transition from abstract change in position to quantitative change in temperature, as demonstrated by the computed \( \frac{dT}{dt} = -8 \) degrees per minute, indicating a decreasing temperature.

Temperature Gradient

The temperature gradient describes how temperature varies with position and is a vector composed of its partial derivatives. It points in the direction of the greatest rate of increase of the temperature.

• For the function \( T(x, y) = e^{-x - 3y} \), the gradient \( abla T \) at the origin is given by the vector \((-1, -3) \).
• The magnitude and direction of this gradient tell us how rapidly and in what direction temperature increases as one moves across the plate.

Despite the bug moving northeast, it encounters a temperature gradient that points in the direction of maximum decrease. Thus, as the bug moves, it follows a path with decreasing temperature, confirming the negative rate of temperature change computed earlier.