Problem 24 - The Calculus with Analytic Geometry

Question

The temperature is \(T\) degrees at any point \((x, y, z)\) in three-dimensional space and \(T=60 /\left(x^{2}+y^{2}+z^{2}+3\right)\). Distance is measured in inches. (a) Find the rate of change of the temperature at the point \((3,-2,2)\) in the direction of the vector \(-2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k}\). (b) Find the direction and magnitude of the greatest rate of change of \(T\) at \((3,-2,2)\).

Step-by-Step Solution

Answer

a) \(\frac{240}{847}\); b) direction: \(\left(\frac{-30}{121}, \frac{20}{121}, \frac{-20}{121}\right)\), magnitude: \(\frac{10\sqrt{17}}{121}\)

Step 1 - Write down the given function

The temperature function is given by: \[ T(x, y, z) = \frac{60}{x^2 + y^2 + z^2 + 3} \]

Step 2 - Compute the gradient of T

The gradient of the function, \(abla T\), represents the rate of change of the temperature with respect to its variables. Use partial derivatives: \( \frac{\partial T}{\partial x} = \frac{-120x}{(x^2 + y^2 + z^2 + 3)^2} \), \( \frac{\partial T}{\partial y} = \frac{-120y}{(x^2 + y^2 + z^2 + 3)^2} \), \( \frac{\partial T}{\partial z} = \frac{-120z}{(x^2 + y^2 + z^2 + 3)^2} \).

Step 3 - Evaluate the gradient at the specific point

Plug the point (3, -2, 2) into the gradient components: \( abla T(3, -2, 2) = \left(\frac{-120(3)}{(3^2 + (-2)^2 + 2^2 + 3)^2}, \frac{-120(-2)}{(3^2 + (-2)^2 + 2^2 + 3)^2}, \frac{-120(2)}{(3^2 + (-2)^2 + 2^2 + 3)^2}\right) \) \( abla T(3, -2, 2) = \left(\frac{-360}{(22)^2}, \frac{240}{(22)^2}, \frac{-240}{(22)^2} \right) \) \( abla T(3, -2, 2) = \left(\frac{-360}{484}, \frac{240}{484}, \frac{-240}{484} \right) \) \( abla T(3, -2, 2) = \left(\frac{-30}{121}, \frac{20}{121}, \frac{-20}{121} \right) \)

Step 4 - Normalize the direction vector

Compute the magnitude of the direction vector \(-2 \, \mathbf{i} + 3 \, \mathbf{j} - 6 \, \mathbf{k}\): \( \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \) Normalize the direction vector: \( \mathbf{d} = \left(\frac{-2}{7}, \frac{3}{7}, \frac{-6}{7}\right) \)

Step 5 - Compute the directional derivative

The directional derivative is given by the dot product of the gradient and the normalized direction vector: \(D_\mathbf{d} T = abla T(3,-2,2) \cdot \mathbf{d}\) \( D_\mathbf{d} T = \left(\frac{-30}{121}, \frac{20}{121}, \frac{-20}{121}\right) \cdot \left(\frac{-2}{7}, \frac{3}{7}, \frac{-6}{7}\right) \) \( D_\mathbf{d} T = \frac{-30}{121} \cdot \frac{-2}{7} + \frac{20}{121} \cdot \frac{3}{7} + \frac{-20}{121} \cdot \frac{-6}{7} \) \( D_\mathbf{d} T = \frac{60}{847} + \frac{60}{847} + \frac{120}{847} \) \( D_\mathbf{d} T = \frac{240}{847} = \frac{240}{847} \)

Step 6 - Find the direction and magnitude of the greatest rate of change

The greatest rate of change of function \(T\) occurs in the direction of the gradient \(abla T\) itself. The magnitude of the gradient at the point (3, -2, 2) is: \[||abla T|| = \sqrt{\left(\frac{-30}{121}\right)^2 + \left(\frac{20}{121}\right)^2 + \left(\frac{-20}{121}\right)^2} \] \[||abla T|| = \frac{1}{121} \sqrt{900 + 400 + 400} \] \[||abla T|| = \frac{1}{121} \sqrt{1700} = \frac{\sqrt{1700}}{121} = \frac{10\sqrt{17}}{121} \]

Key Concepts

Gradient Vector

The gradient vector is an essential concept in multivariable calculus. For any function in three-dimensional space, such as our temperature function, the gradient vector, denoted as \(abla T\), indicates the direction and rate of the steepest ascent. It consists of the partial derivatives of the function with respect to each variable:

• \(\frac{\partial T}{\partial x}\)
• \(\frac{\partial T}{\partial y}\)
• \(\frac{\partial T}{\partial z}\)

For the given temperature function \( T(x, y, z) = \frac{60}{x^2 + y^2 + z^2 + 3} \), the gradient at any point is calculated as:

• \( \frac{\partial T}{\partial x} = \frac{-120x}{(x^2 + y^2 + z^2 + 3)^2} \)
• \( \frac{\partial T}{\partial y} = \frac{-120y}{(x^2 + y^2 + z^2 + 3)^2} \)
• \( \frac{\partial T}{\partial z} = \frac{-120z}{(x^2 + y^2 + z^2 + 3)^2} \)

By evaluating these partial derivatives at the point \((3, -2, 2)\), we find the gradient vector at that point.

Directional Derivative

To find the rate of change of a function in a specific direction, we use the directional derivative. It combines the gradient vector and a given direction vector. The directional derivative represents how fast the temperature changes in that specified direction.
Let's say we have a direction vector \(\mathbf{d} = -2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}\). First, we normalize this vector to obtain its unit vector. The magnitude of \(\mathbf{d}\) is \( \sqrt{(-2)^2 + 3^2 + (-6)^2} = 7 \), giving the normalized direction vector \(\mathbf{d} = \left( \frac{-2}{7}, \frac{3}{7}, \frac{-6}{7} \right) \).
The directional derivative \( D_\mathbf{d} T \) is then the dot product of the gradient at \((3, -2, 2)\) and this normalized direction vector. Calculating this dot product will provide the rate of temperature change at the point in the given direction.

Partial Derivatives

Partial derivatives measure the rate of change of a function concerning one variable, keeping others constant. They are fundamental when working with functions of multiple variables.
For the temperature function \( T(x, y, z) = \frac{60}{x^2 + y^2 + z^2 + 3} \), the partial derivatives are:

• \( \frac{\partial T}{\partial x} = \frac{-120x}{(x^2 + y^2 + z^2 + 3)^2} \)
• \( \frac{\partial T}{\partial y} = \frac{-120y}{(x^2 + y^2 + z^2 + 3)^2} \)
• \( \frac{\partial T}{\partial z} = \frac{-120z}{(x^2 + y^2 + z^2 + 3)^2} \)

They help in constructing the gradient vector and are crucial for finding the rate of change in specific directions and analyzing the function's behavior at different points.

Three-dimensional Space

Calculations involving functions like the temperature function \(T(x, y, z)\) take place in three-dimensional space. Concepts like the gradient vector and directional derivative are extended into this 3D realm.
Each point \((x, y, z)\) represents a position in 3D space, and the function \(T\) provides the temperature at these points. When calculating derivatives or evaluating functions in this space, understanding spatial coordinates and vector operations is essential.
Functions and calculations in 3D space are often visualized using geometric interpretations, which can aid in comprehending how changes occur in different directions.

Temperature Function Analysis

Analyzing the temperature function involves multiple steps: computing partial derivatives, finding the gradient, and evaluating changes in temperature at specific points.
The temperature function here is \( T(x, y, z) = \frac{60}{x^2 + y^2 + z^2 + 3} \). This type of function, with variables in the denominator, typically has a temperature decreasing with increasing distance from the origin.
Once we know the gradient, we can determine the greatest rate of change's direction and magnitude. Here, finding it at \((3, -2, 2)\) and calculating the directional derivative in a given direction involves:

• Generating the gradient vector at the point
• Normalizing the direction vector
• Calculating the dot product for the directional derivative
• Finding the magnitude of the gradient for the steepest temperature change.