Problem 64 - Calculus with Concepts in Calculus

Question

A metal plate with vertices \((1,1),(5,1),(1,3)\), and \((5,3)\) is heated by a flame at the origin, and the temperature at a point on the plate is inversely proportional to the distance from the origin. If an ant is located at the point \((3,2)\), in what direction should the ant crawl to cool the fastest?

Step-by-Step Solution

Answer

The ant should crawl in the direction \(\left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right)\) to cool fastest.

Step 1: Identify the Temperature Function

The temperature at any point \( (x, y) \) on the plate is given by: \( T(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \), where \( k \) is a constant. The temperature decreases as the distance from the origin increases.

Step 2: Determine the Gradient

The direction of steepest decrease in temperature is opposite to the gradient vector, \( abla T(x, y) \). Calculate the partial derivatives: \( \frac{\partial T}{\partial x} = \frac{-kx}{(x^2 + y^2)^{3/2}} \) and \( \frac{\partial T}{\partial y} = \frac{-ky}{(x^2 + y^2)^{3/2}} \). Then, \( abla T(x, y) = \left( \frac{-kx}{(x^2 + y^2)^{3/2}}, \frac{-ky}{(x^2 + y^2)^{3/2}} \right) \).

Step 3: Evaluate the Gradient at the Ant’s Position

Substitute the ant's position \( (3, 2) \) into the gradient: \( abla T(3, 2) = \left( \frac{-3k}{(3^2 + 2^2)^{3/2}}, \frac{-2k}{(3^2 + 2^2)^{3/2}} \right) = \left( \frac{-3k}{13^{3/2}}, \frac{-2k}{13^{3/2}} \right) \).

Step 4: Direction of Fastest Cooling

The fastest cooling occurs in the direction opposite to \( abla T(3,2) \), so the direction vector is \( \left( \frac{3}{13^{3/2}}, \frac{2}{13^{3/2}} \right) \).

Step 5: Normalize the Direction Vector

A unit vector in the direction of \( \left( 3, 2 \right) \) is obtained by dividing by its magnitude: \( \text{Magnitude} = \sqrt{3^2 + 2^2} = \sqrt{13} \). The unit vector is \( \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \). Thus, the ant should crawl in the direction \( \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \).

Key Concepts

Inverse Proportion

Inverse proportion is a relationship where one quantity increases as another decreases. In the context of the temperature function given in the problem, the temperature at a point on the plate is inversely proportional to the distance from the origin. This means that as you move away from the origin, the temperature decreases. An inverse proportional relationship is typically represented by a formula like \( y = \frac{k}{x} \), where \( y \) decreases as \( x \) increases and \( k \) is a constant.
This concept is crucial to understanding the temperature function \( T(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \), where the temperature \( T \) decreases as \( x^2 + y^2 \), a measure of distance from the origin, increases. Here, \( k \) is a constant specific to the setup of the problem.

Partial Derivatives

Partial derivatives are used to find the rate of change of a function with respect to one variable while holding other variables constant. This is particularly useful for functions of several variables, like our temperature function \( T(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \).

• In the original exercise, we compute the partial derivative with respect to \( x \), \( \frac{\partial T}{\partial x} = \frac{-kx}{(x^2 + y^2)^{3/2}} \). This tells us how the temperature changes as we move along the \( x \)-axis, holding \( y \) constant.
• Similarly, \( \frac{\partial T}{\partial y} = \frac{-ky}{(x^2 + y^2)^{3/2}} \) gives the temperature change along the \( y \)-axis.

The gradient \( abla T(x, y) = \left( \frac{-kx}{(x^2 + y^2)^{3/2}}, \frac{-ky}{(x^2 + y^2)^{3/2}} \right) \) combines these partial derivatives, indicating the direction of the steepest increase in temperature.

Unit Vector

A unit vector is a vector with a magnitude of 1. It is used to indicate direction without affecting magnitude. When we want to represent a direction using a vector, we often convert it to a unit vector.
After calculating the gradient vector in this problem, \( \left( \frac{-3k}{13^{3/2}}, \frac{-2k}{13^{3/2}} \right) \), we need to convert it to a unit vector to determine the direction the ant should move.

• The magnitude of the vector \((3, 2)\) is \( \sqrt{3^2 + 2^2} = \sqrt{13} \).
• The unit vector becomes \( \left( \frac{3}{\sqrt{13}}, \frac{2}{\sqrt{13}} \right) \), indicating the ant's movement direction.

This unit vector ensures that the ant moves in the accurate direction for the fastest decrease in temperature, regardless of how fast it moves.

Temperature Function

The temperature function in the problem describes how temperature varies at different points on the heated metal plate. It is given by \( T(x, y) = \frac{k}{\sqrt{x^2 + y^2}} \). Here, \( k \) is a constant that relates to the intensity of the flame at the origin.
The temperature function falls under the inverse proportion category because as the distance from the origin \( \sqrt{x^2 + y^2} \) grows, the temperature decreases. This function is a specific instance of how temperature might change in relation to a heat source as you move further away.

• We use this function to determine the gradient, which helps us know how the temperature changes at each point.
• The gradient then leads us to finding the direction of steepest temperature decrease, indicating where the ant should crawl to cool down quickly.

This function is at the core of understanding how the flame at the origin influences the plate's temperature distribution.