Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

The temperature in degrees Celsius on a metal plate in the \(x y\)-plane is given by \(T(x, y)=4+2 x^{2}+y^{3}\). What is the rate of change of temperature with respect to distance (measured in feet) if we start moving from \((3,2)\) in the direction of the positive \(y\)-axis?

Step-by-Step Solution

Answer

The rate of change of temperature is 12 degrees Celsius per foot.

Step 1: Identify the Gradient

The gradient of the temperature function \( T(x, y) = 4 + 2x^2 + y^3 \) is given by \( abla T(x, y) = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) \). Calculate the partial derivatives to find the gradient.

Step 2: Calculate Partial Derivatives

The partial derivative with respect to \(x\) is \( \frac{\partial T}{\partial x} = 4x \), and the partial derivative with respect to \(y\) is \( \frac{\partial T}{\partial y} = 3y^2 \). Hence, \( abla T(x, y) = (4x, 3y^2) \).

Step 3: Evaluate Gradient at Point

Substitute \((3, 2)\) into the gradient: \( abla T(3, 2) = (4 \times 3, 3 \times 2^2) = (12, 12) \).

Step 4: Determine Direction Vector

Since the movement is in the positive \(y\)-axis direction, the direction vector \( \mathbf{u} \) is \( \langle 0, 1 \rangle \).

Step 5: Calculate Directional Derivative

The directional derivative of \(T\) in the direction of \(\mathbf{u}\) is given by \( abla T \cdot \mathbf{u} = (12, 12) \cdot (0, 1) = 12 \). This is the rate of change of temperature.

Key Concepts

Partial Derivatives

Partial derivatives are a key tool in calculus used to find the rate of change of a multivariable function with respect to one variable while keeping others constant.
When dealing with a function like the temperature function given by \( T(x, y) = 4 + 2x^2 + y^3 \), we can understand how changing \(x\) affects temperature by computing the partial derivative with respect to \(x\), denoted as \( \frac{\partial T}{\partial x} \). Similarly, for the \(y\)-axis, we find \( \frac{\partial T}{\partial y} \).
This essentially breaks down a multivariable function into simpler, single-variable components:

• For \( \frac{\partial T}{\partial x} \), we differentiate \( T \) with respect to \( x \), treating \( y \) as a constant, resulting in \( 4x \).
• For \( \frac{\partial T}{\partial y} \), we differentiate \( T \) with respect to \( y \), treating \( x \) as a constant, resulting in \( 3y^2 \).

Gradient Vector

The gradient vector is like a compass that points in the direction of the steepest rate of increase of a function. For a function \( T(x, y) \), the gradient \( abla T(x, y) \) consists of the partial derivatives and is expressed as \( \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) \).
This tells us how fast the function changes in different directions from a given point.

• In our problem, the gradient vector of \( T(x, y) = 4 + 2x^2 + y^3 \) is \( (4x, 3y^2) \).
• Evaluating this at the point \( (3, 2) \), we substitute \( x = 3 \) and \( y = 2 \) into the gradient expression, giving us \( (12, 12) \).

These values indicate how temperature changes with slight shifts in \(x\) and \(y\).

Rate of Change

The rate of change in a specific direction is called the directional derivative. It measures how fast the function value changes as you move in that direction, hinting at the velocity of change.
For the temperature function, this involves using both the gradient vector and the direction vector. The dot product of these two vectors gives the directional derivative:

• Given the gradient vector \( abla T(3, 2) = (12, 12) \).
• With the direction vector being \( \langle 0, 1 \rangle \), indicating movement along the positive \( y \)-axis.
• The rate of change is then computed as \( (12, 12) \cdot (0, 1) = 12 \).

This 12 indicates that the temperature increases by 12 degrees Celsius per unit of distance moved along the positive \( y \)-axis.

Temperature Function

A temperature function describes how temperature varies across different points in space. In this exercise, \( T(x, y) = 4 + 2x^2 + y^3 \) represents a temperature distribution over a metal plate in the \(xy\)-plane.
Analyzing such a function helps us understand how temperature changes spatially:

• The constant \(4\) suggests a base temperature level across the plate.
• The term \(2x^2\) indicates that temperature increases as \(x\) increases, contributing more to the temperature the further along the \(x\)-axis you go.
• Likewise, \(y^3\) indicates changes along the \(y\)-axis, with steeper increases as \(y\) grows.

Studying these variations gives insights into where the temperature will be higher or lower based on the plate's position. This helps in practical applications, such as ensuring materials don't overheat.