Problem 31 - Calculus Early Transcendentals

Question

The temperature \(T\) in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point \((1,2,2)\) is \(120^{\circ}\) . (a) Find the rate of change of \(T\) at \((1,2,2)\) in the direction toward the point \((2,1,3) .\) (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.

Step-by-Step Solution

Answer

(a) Rate of change of temperature is \(-\frac{40\sqrt{3}}{3}\). (b) Direction of greatest increase is toward the origin.

Step 1: Understanding Inverse Proportionality

The problem states that the temperature \(T\) is inversely proportional to the distance \(r\) from the origin. This can be mathematically expressed as \( T = \frac{k}{r} \), where \( k \) is a constant to be determined.

Step 2: Calculate the Distance

The point \((1,2,2)\) is located at a distance \( r \) from the origin. Calculate this distance using the formula for the Euclidean distance: \( r = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \).

Step 3: Determine the Constant k

We know that \( T = 120 \) at the point \((1,2,2)\). Using the formula \( T = \frac{k}{r} \) and substituting \( T = 120 \) and \( r = 3 \), we solve for \( k \) as follows: \( 120 = \frac{k}{3} \Rightarrow k = 360 \).

Step 4: Write the Temperature Function

We now have the expression for the temperature: \( T = \frac{360}{r} \). Recall that \( r = \sqrt{x^2 + y^2 + z^2} \), so \( T = \frac{360}{\sqrt{x^2 + y^2 + z^2}} \).

Step 5: Calculate the Gradient of T

The gradient \( abla T \) represents the direction of greatest change of temperature. Calculate \( abla T \) using partial derivatives: \(abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) \).

Step 6: Find Partial Derivatives

Calculate the partial derivatives of \( T \):- \( \frac{\partial T}{\partial x} = -\frac{360x}{(x^2 + y^2 + z^2)^{3/2}} \)- \( \frac{\partial T}{\partial y} = -\frac{360y}{(x^2 + y^2 + z^2)^{3/2}} \)- \( \frac{\partial T}{\partial z} = -\frac{360z}{(x^2 + y^2 + z^2)^{3/2}} \)

Step 7: Evaluate the Gradient at (1,2,2)

Substitute \((1,2,2)\) into the gradient expression:\[ abla T = \left( -\frac{360 \times 1}{27}, -\frac{360 \times 2}{27}, -\frac{360 \times 2}{27} \right) = (-\frac{40}{3}, -\frac{80}{3}, -\frac{80}{3}) \].

Step 8: Determine the Direction Vector

The direction vector toward the point \((2,1,3)\) from \((1,2,2)\) is given by \( \mathbf{v} = (2-1, 1-2, 3-2) = (1, -1, 1) \).

Step 9: Find Rate of Change Using Directional Derivative

The rate of change of \( T \) in the direction of \( \mathbf{v} \) is given by the dot product of \( abla T \) and the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \). First, find \( \mathbf{u} = \frac{1}{\sqrt{3}}(1, -1, 1) \). Then, compute the directional derivative: \( abla T \cdot \mathbf{u} = (-\frac{40}{3}, -\frac{80}{3}, -\frac{80}{3}) \cdot \frac{1}{\sqrt{3}}(1, -1, 1) = -\frac{40\sqrt{3}}{3} \).

Step 10: Show Direction of Greatest Increase

Observe that the temperature function \( T\) has its gradient \(abla T\) in the form of \(-k \cdot \vec{r} \), where \(\vec{r} = (x, y, z)\). This indicates that the direction of greatest increase in temperature is toward the origin since the gradient declines as you move away from the origin.

Key Concepts

Directional Derivative

The directional derivative represents how a function changes as you move in a specific direction. In this context, the problem involves finding how temperature changes in the direction of a given vector. Here, the temperature at point \((1,2,2)\) on the metal ball is 120°C, and we want to know how it changes as we move towards point \((2,1,3)\).

To calculate this rate of change, we find the dot product of the gradient of the temperature function and a unit vector in the chosen direction.

• Start by calculating the gradient \(abla T\), which gives the direction of greatest change in temperature.
• Obtain the direction vector \(\mathbf{v}\) by subtracting the coordinates of the initial point from the target point, resulting in \((1, -1, 1)\).
• Form the unit vector \(\mathbf{u}\) by normalizing \(\mathbf{v}\).
• Finally, compute the directional derivative using \(abla T \cdot \mathbf{u}\).

This method allows us to precisely quantify the temperature change rate as we move towards the specified direction.

Inverse Proportionality

Inverse proportionality indicates a relationship where one value decreases as another value increases. In this exercise, the temperature \(T\) of the metal ball is inversely proportional to the distance \(r\) from the center of the ball. This means that as you move further from the center, the temperature reduces, following the equation \(T = \frac{k}{r}\), where \(k\) is a constant.

In steps to determine \(k\):

• Find the Euclidean distance \(r\) from the origin, where the known temperature \(T\) is given.
• Substitute \(T\) and \(r\) values to solve for the constant \(k\).

This concept helps us understand how temperature behaves as one moves away from or towards the ball's center.

Euclidean Distance

Euclidean distance is a measure of the 'straight-line' distance between two points in Euclidean space. It's crucial for calculating gradients and determining directions. In our problem, it helps find the exact distance of a point on the surface of the ball from its center.

The formula to calculate the Euclidean distance \(r\) from the origin to any point \((x,y,z)\) is\[ r = \sqrt{x^2 + y^2 + z^2} \].

Steps involved:

• Square the differences between the corresponding coordinates.
• Sum these squared values.
• Take the square root of the result to find \(r\).

Using this, the distance was calculated as 3 for the point \((1,2,2)\), vital for determining both the temperature function and the gradient.

Partial Derivatives

Partial derivatives are used to understand how a multivariable function changes as one of its variables changes, while the others are held constant. They are an integral part of finding the gradient of a scalar field, like temperature in this exercise.

For our temperature function \(T = \frac{360}{\sqrt{x^2 + y^2 + z^2}}\), the partial derivatives help in constructing the gradient \(abla T\).

• To calculate, differentiate \(T\) with respect to \(x\), \(y\), and \(z\) individually.
• Use the chain rule to handle the square root expression effectively.
• Combine these derivatives into the gradient vector \((\frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z})\).

These derivatives show how the temperature varies in different directions, which is central to understanding how the temperature changes at any point on the ball's surface.