Problem 9 - Calculus Early Transcendentals

Question

\(7-10\) (a) Find the gradient of \(f\) (b) Evaluate the gradient at the point \(P .\) (c) Find the rate of change of \(f\) at \(P\) in the direction of the vector \mathbf{u} . $$f(x, y, z)=x e^{2 y z}, \quad P(3,0,2), \quad \mathbf{u}=\left\langle\frac{2}{3},-\frac{2}{3}, \frac{1}{3}\right\rangle$$

Step-by-Step Solution

Answer

Gradient at P: \(\langle 1, 12, 0 \rangle\), Rate of change: \(-\frac{22}{3}\).

Step 1: Find the Partial Derivatives

To find the gradient of the function \( f(x, y, z) = x e^{2yz} \), we need its partial derivatives with respect to \( x, y, \) and \( z \).\The partial derivative of \( f \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^{2yz} \).\The partial derivative of \( f \) with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2xz e^{2yz} \).\The partial derivative of \( f \) with respect to \( z \) is \( \frac{\partial f}{\partial z} = 2xy e^{2yz} \).

Step 2: Form the Gradient Vector

The gradient of the function \( f(x, y, z) = x e^{2yz} \) is given by the vector of its partial derivatives:\\[ abla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \left\langle e^{2yz}, 2xz e^{2yz}, 2xy e^{2yz} \right\rangle \].

Step 3: Evaluate the Gradient at Point P

Now, we evaluate the gradient at the point \( P(3, 0, 2) \).\\[ abla f(3, 0, 2) = \left\langle e^{(2)(0)(2)}, 2(3)(2)e^{(2)(0)(2)}, 2(3)(0)e^{(2)(0)(2)} \right\rangle = \langle 1, 12, 0 \rangle \].

Step 4: Find the Unit Vector in Given Direction

The given vector \( \mathbf{u} \) is \( \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle \). It is already a unit vector since its magnitude is 1. Confirm by checking: \( \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2} = 1 \).

Step 5: Calculate the Directional Derivative

The rate of change of \( f \) at \( P \) in the direction of \( \mathbf{u} \) is given by the dot product of the gradient and the unit vector.\\[ D_{\mathbf{u}} f = abla f(3, 0, 2) \cdot \mathbf{u} = \langle 1, 12, 0 \rangle \cdot \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle \].\The result is \( \left(1 \cdot \frac{2}{3}\right) + \left(12 \cdot -\frac{2}{3}\right) + \left(0 \cdot \frac{1}{3}\right) = \frac{2}{3} - 8 = -\frac{22}{3} \).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus when working with functions of multiple variables. They represent how a function changes as each variable changes, while keeping the other variables constant.
When you take the partial derivative of a function, you differentiate it with respect to one of its variables. This process helps us understand how changes in that particular variable affect the overall function.
For example, consider the function \( f(x, y, z) = x e^{2yz} \). We can find its partial derivatives as follows:

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^{2yz} \).


• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2xz e^{2yz} \).


• The partial derivative with respect to \( z \) is \( \frac{\partial f}{\partial z} = 2xy e^{2yz} \).

These derivatives provide insight into the behavior of the function along each axis, aiding in forming the gradient vector, which is the next step.

Directional Derivative

The directional derivative provides a way to find the rate of change of a function in any given direction, not just along the coordinate axes. It is particularly useful in fields like physics and engineering, where changes in the direction of a function are essential.
To calculate the directional derivative, you first need the gradient vector of the function, which is a vector composed of all its partial derivatives. This gradient vector provides the direction of the greatest rate of increase of the function.
In our exercise, the gradient at point \( P(3,0,2) \) is \( \langle 1, 12, 0 \rangle \). To find the directional derivative in the direction of a unit vector \( \mathbf{u} = \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle \), we compute the dot product of the gradient and \( \mathbf{u} \):

• The calculation: \( D_{\mathbf{u}} f = \langle 1, 12, 0 \rangle \cdot \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle \).


• This results in \( \frac{2}{3} - 8 = -\frac{22}{3} \).

Hence, the directional derivative gives how quickly the function increases or decreases in that specified direction.

Rate of Change

In calculus, the rate of change helps us understand how a quantity evolves as another variable changes. This is crucial in scenarios where such relationships dynamically alter over time or space.
The concept of the rate of change applies to single-variable functions as the derivative. For multivariable functions, like in our exercise, the directional derivative is used to determine the rate at which the function changes in a particular direction.

Practical Importance

Finding the rate of change is practical in many real-life situations:

• In physics, to assess how speed or velocity changes with time.


• In economics, to analyze how demand shifts with pricing strategies.


• In meteorology, to predict how weather patterns evolve over terrain.

In our example, at point \( P(3,0,2) \), the rate of change in the direction of the vector \( \mathbf{u} \) is \( -\frac{22}{3} \), indicating how the function value decreases as we move along \( \mathbf{u} \). This negative value shows a decrease in the function relative to that direction.