Problem 41 - Advanced Engineering Mathematics

Question

The temperature at a point \((x, y)\) on a rectangular metal plate is given by \(T(x, y)=100-2 x^{2}-y^{2}\). Find the path a heatseeking particle will take, starting at \((3,4)\), as it moves in the direction in which the temperature increases most rapidly.

Step-by-Step Solution

Answer

The path is given by \( x(t) = 3 - 0.83t \) and \( y(t) = 4 - 0.55t \).

Step 1: Understanding the Problem

We have a temperature function \( T(x, y) = 100 - 2x^2 - y^2 \). We need to find the path a particle takes, starting at \((3, 4)\), that moves in the direction of the steepest ascent. This requires calculating the gradient of \( T \), which gives the direction of steepest increase.

Step 2: Calculate the Gradient

The gradient of the function \( T(x, y) \) is given by \( abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) \). Calculating each partial derivative, we have:\[ \frac{\partial T}{\partial x} = -4x, \]\[ \frac{\partial T}{\partial y} = -2y. \]Thus, \( abla T = (-4x, -2y) \).

Step 3: Evaluate the Gradient at the Starting Point

Substitute \( x = 3 \) and \( y = 4 \) into the gradient:\[ abla T (3, 4) = (-4 \times 3, -2 \times 4) = (-12, -8). \]This shows the direction of the steepest temperature increase at the given point.

Step 4: Determine the Direction of the Path

The path is taken by moving in the direction of the gradient vector \( (-12, -8) \). The direction vector can be obtained by normalizing \( (-12, -8) \). Normalization involves dividing by its magnitude, calculated as \( \sqrt{(-12)^2 + (-8)^2} = \sqrt{144 + 64} = \sqrt{208} \approx 14.42 \).

Step 5: Write the Equation of the Path

The unit direction vector is \( \left( \frac{-12}{14.42}, \frac{-8}{14.42} \right) \approx (-0.83, -0.55) \). The path is described by a parametric equation:\[ x(t) = 3 - 0.83t \]\[ y(t) = 4 - 0.55t \]

Step 6: Conclude the Path Description

The heat-seeking particle starts at \( (3, 4) \) and moves along the path defined by the equations \( x(t) = 3 - 0.83t \) and \( y(t) = 4 - 0.55t \) as \( t \) increases, moving in the direction of the steepest temperature increase in the negative x and y directions.

Key Concepts

Gradient

In calculus, the concept of a gradient is fundamental when dealing with multivariable functions.
The gradient is a vector that consists of partial derivatives of a function with respect to each independent variable. This vector points in the direction of the steepest increase of the function's value. The notation for the gradient is often represented as \( abla T \), where \( T \) is a given function of two or more variables.

• For a function \( T(x, y) \), the gradient is \( abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right) \).
• The magnitude of the gradient vector \( \|abla T\| \) gives the rate of maximum increase.

Understanding the gradient is crucial when you want to find out in which direction a particular variable affects the overall behavior of a multivariable function.

Temperature Function

A temperature function is a mathematical representation used to describe how temperature varies over a given space.In the given exercise, the temperature function is expressed as \( T(x, y) = 100 - 2x^2 - y^2 \).

• This formula indicates that temperature depends on the variables \( x \) and \( y \) and decreases as these variables increase.
• The constant \( 100 \) represents the maximum temperature at the origin \((0,0)\), while the terms \( -2x^2 \) and \( -y^2 \) indicate how the temperature decreases with distance from the center.

Temperature functions play a critical role in physical applications, enabling predictions and understanding of how heat spreads across given materials or environments.

Partial Derivatives

Partial derivatives are an essential concept in calculus, particularly when working with functions of multiple variables.A partial derivative of a function with respect to a particular variable indicates how the function changes as that variable is altered, while other variables are held constant.

• For function \( T(x, y) \), the partial derivatives are \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \).
• In the provided exercise, we found \( \frac{\partial T}{\partial x} = -4x \) and \( \frac{\partial T}{\partial y} = -2y \).

Partial derivatives step towards understanding more complex concepts like gradients and surface curvature, allowing calculus to analyze each variable's contribution individually.

Directional Derivative

Directional derivatives expand upon the concept of partial derivatives by indicating the rate of change of a function in any given direction, not just along one of the coordinate axes.To compute a directional derivative, you need a vector that defines the direction in which you are moving.

• The directional derivative in the direction of a vector \( \mathbf{v} \) for a function \( T \) is given by \( D_{\mathbf{v}}T = abla T \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|} \).
• The solution showed that the steepest ascent direction corresponds to the gradient \( abla T = (-12, -8) \) at point \((3, 4)\).
• Normalizing this gradient gives the direction of the path for a particle moving through the field.

Understanding directional derivatives is crucial in scenarios where one needs to determine how a function behaves along specific paths, such as finding a heat-seeker's route.