Problem 33 - Calculus Early Transcendentals

Question

Suppose that over a certain region of space the electrical potential \(V\) is given by \(V(x, y, z)=5 x^{2}-3 x y+x y z\) . (a) Find the rate of change of the potential at \(P(3,4,5)\) in the direction of the vector \(\mathbf{v}=\mathbf{i}+\mathbf{j}-\mathbf{k} .\) (b) In which direction does \(V\) change most rapidly at \(P ?\) (c) What is the maximum rate of change at \(P ?\)

Step-by-Step Solution

Answer

(a) \( \frac{32}{\sqrt{3}} \); (b) \( \langle 38, 6, 12 \rangle \); (c) \( 2 \sqrt{406} \)."

Step 1: Delineate the Problem

We're tasked with finding how the electrical potential function \( V(x, y, z) = 5x^2 - 3xy + xyz \) changes at a specific point and in certain directions. We need to find (a) the rate of change in a given direction, (b) the direction of steepest ascent, and (c) the maximum rate of change at the point \( P(3, 4, 5) \).

Step 2: Compute the Gradient

The gradient \( abla V \) of the function \( V \) is the vector of partial derivatives. Find the partial derivatives:\( \frac{\partial V}{\partial x} = 10x - 3y + yz \)\( \frac{\partial V}{\partial y} = -3x + xz \)\( \frac{\partial V}{\partial z} = xy \)Hence, the gradient is:\[ abla V = \langle 10x - 3y + yz, -3x + xz, xy \rangle \]

Step 3: Evaluate the Gradient at P

Substitute \( P(3, 4, 5) \) into \( abla V \):\[ \left. abla V \right|_{(3, 4, 5)} = \langle 10(3) - 3(4) + (4)(5), -3(3) + (3)(5), (3)(4) \rangle \]This simplifies to:\[ abla V = \langle 30 - 12 + 20, -9 + 15, 12 \rangle \]Resulting in:\[ abla V = \langle 38, 6, 12 \rangle \]

Step 4: Find Rate of Change in Direction of v

The rate of change of \( V \) in the direction of vector \( \mathbf{v} = \langle 1, 1, -1 \rangle \) is given by the dot product of the gradient and the unit vector in the direction of \( \mathbf{v} \):First, find the unit vector \( \mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} \):\[ \| \mathbf{v} \| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \]\[ \mathbf{u} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right\rangle \]Then calculate the dot product:\[ abla V \cdot \mathbf{u} = \langle 38, 6, 12 \rangle \cdot \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right\rangle \]\[ = \frac{38}{\sqrt{3}} + \frac{6}{\sqrt{3}} - \frac{12}{\sqrt{3}} \]\[ = \frac{38 + 6 - 12}{\sqrt{3}} = \frac{32}{\sqrt{3}} \]

Step 5: Determine Direction of Maximum Increase

The direction in which \( V \) changes most rapidly is in the direction of its gradient.At \( P \), this direction is given by the gradient vector:\[ abla V = \langle 38, 6, 12 \rangle \]

Step 6: Calculate Maximum Rate of Change

The maximum rate at which \( V \) changes at \( P \) is the magnitude of the gradient.Calculate this magnitude:\[ \| abla V \| = \sqrt{38^2 + 6^2 + 12^2} \]\[ = \sqrt{1444 + 36 + 144} = \sqrt{1624} \]\[ = 2\sqrt{406} \]

Step 7: Conclusion

The solution to each part of the exercise is as follows:(a) The rate of change of the potential at \( P \) in the direction of \( \mathbf{v} \) is \( \frac{32}{\sqrt{3}} \).(b) The direction in which \( V \) changes most rapidly at \( P \) is \( \langle 38, 6, 12 \rangle \).(c) The maximum rate of change at \( P \) is \( 2 \sqrt{406} \).

Key Concepts

Gradient Vector

In mathematics, the Gradient Vector is a crucial concept when dealing with multivariable functions. Simply put, the gradient of a function represents the direction and rate of the steepest ascent. It is a vector composed of partial derivatives, providing insights into how a function changes with respect to each variable.

For a function like the electrical potential function, \( V(x, y, z)=5x^2 - 3xy + xyz \), the gradient \( abla V \) is calculated as \( \langle \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}, \frac{\partial V}{\partial z} \rangle \). Each component of the gradient originates from taking the partial derivative of the function with respect to one variable at a time, while keeping the other variables constant.

• \( \frac{\partial V}{\partial x} = 10x - 3y + yz \)
• \( \frac{\partial V}{\partial y} = -3x + xz \)
• \( \frac{\partial V}{\partial z} = xy \)

These calculations show how the function changes with small adjustments in \( x \), \( y \), and \( z \). Evaluating the gradient at a specific point gives the direction of the fastest increase in the function's value at that point.

Maximum Rate of Change

The Maximum Rate of Change of a function at a point is the magnitude of the gradient vector evaluated at that point. It signifies the largest possible increase of the function with an infinitesimal movement from the point. Essentially, it tells us how fast the function is climbing, helping to understand the growth tendency.

For our function \( V(x, y, z) \), to find this maximum rate at point \( P(3,4,5) \), we calculate the magnitude of the gradient vector \( abla V \). The magnitude is determined as follows:

\[\| abla V \| = \sqrt{38^2 + 6^2 + 12^2} = \sqrt{1624} = 2\sqrt{406}\]

The magnitude, \( 2\sqrt{406} \), represents how fast the potential changes when moving in the precise direction of the gradient. This gives us the maximum rate of change, which is always positive and highlights the most rapid change from the specific point.

Rate of Change

The Rate of Change in a specific direction for multivariable functions adds another dimension to understanding function behavior. It measures how fast a function changes as one moves along a particular vector direction. In our exercise, we calculated the rate at which the potential \( V \) changes in the direction of a vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} - \mathbf{k} \).

To find this, we used the dot product of the gradient vector at point \( P(3,4,5) \) and the unit vector in the direction of \( \mathbf{v} \). Transform \( \mathbf{v} \) into a unit vector \( \mathbf{u} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right\rangle \), and perform the dot product:

\[abla V \cdot \mathbf{u} = \frac{32}{\sqrt{3}}\]
This result tells us how rapidly the potential changes specifically in the vector's direction \( \mathbf{v} \). Calculating and understanding this directional rate of change is essential especially in fields like physics and engineering.

Partial Derivatives

Partial Derivatives are a key building block when exploring functions of more than one variable. They provide a way to understand how the function's output changes as we vary only one input at a time.

In this context, the function \( V(x, y, z) \) has three variables \( x, y, \) and \( z \). The partial derivative with respect to \( x \), \( \frac{\partial V}{\partial x} \), describes the function's sensitivity to small changes in \( x \) while keeping \( y \) and \( z \) constant. Similarly, \( \frac{\partial V}{\partial y} \) and \( \frac{\partial V}{\partial z} \) examined the impacts of alterations in \( y \) and \( z \), respectively.

• \( \frac{\partial V}{\partial x} = 10x - 3y + yz \)
• \( \frac{\partial V}{\partial y} = -3x + xz \)
• \( \frac{\partial V}{\partial z} = xy \)

These derivatives help form the gradient vector, an essential tool in analyzing the function's behavior. Understanding partial derivatives is crucial for tasks like optimization and modeling in several scientific and engineering disciplines. They lay the groundwork for deeper insights into how systems change in multidimensional spaces.