Problem 13 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14, a function \(f\), a point \(P\), and a unit vector \(\mathbf{U}\) are given. Find (a) the gradient of \(f\) at \(P\), and (b) the rate of change of the function value in the direction of \(\mathrm{U}\) at \(P\). \(f(x, y, z)=y^{2}+z^{2}-4 x z ; P=(-2,1,3) ; \mathbf{U}=\frac{2}{7} \mathbf{i}-\frac{6}{\mathbf{j}} \mathbf{j}+\frac{3}{7} \mathbf{k}\)

Step-by-Step Solution

Answer

The gradient of \( f \) at \( P \) is \( (-12, 2, 14) \). The rate of change in the direction of \( \mathbf{U} \) at \( P \) is \( \frac{6}{7} \).

Step 1 - Calculate the partial derivatives

First, find the partial derivatives of the function \(f(x, y, z)\) with respect to \(x\), \(y\), and \(z\). These represent the components of the gradient vector \( abla f \).\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (y^{2}+z^{2}-4xz) = -4z \]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (y^{2}+z^{2}-4xz) = 2y \]\[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z} (y^{2}+z^{2}-4xz) = 2z - 4x \]

Step 2 - Evaluate the gradient at point \(P(-2, 1, 3)\)

Next, substitute the coordinates of point \(P\) into the partial derivatives to find the gradient at \(P\).\[ \frac{\partial f}{\partial x} (-2, 1, 3) = -4(3) = -12 \]\[ \frac{\partial f}{\partial y} (-2, 1, 3) = 2(1) = 2 \]\[ \frac{\partial f}{\partial z} (-2, 1, 3) = 2(3) - 4(-2) = 6 + 8 = 14 \]So the gradient \( abla f \) at \( P \) is:\[ abla f (-2, 1, 3) = (-12, 2, 14) \]

Step 3 - Confirm unit vector \( \mathbf{U} \)

The given unit vector is \( \mathbf{U} = \frac{2}{7} \mathbf{i} - \frac{6}{7} \mathbf{j} + \frac{3}{7} \mathbf{k} \). We check the magnitude to ensure it is a unit vector.\[ \left|\mathbf{U}\right| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(-\frac{6}{7}\right)^2 + \left(\frac{3}{7}\right)^2} \]\[ \left|\mathbf{U}\right| = \sqrt{\frac{4}{49} + \frac{36}{49} + \frac{9}{49}} = \sqrt{\frac{49}{49}} = 1 \]Since the magnitude is 1, \( \mathbf{U} \) is indeed a unit vector.

Step 4 - Calculate the directional derivative

The rate of change of the function \(f\) in the direction of vector \( \mathbf{U} \) at point \( P \) is given by the dot product of the gradient \( abla f \) at \( P \) and \( \mathbf{U} \).\[ D_{\mathbf{U}}f = abla f \cdot \mathbf{U} \]\[ D_{\mathbf{U}}f = (-12, 2, 14) \cdot \left(\frac{2}{7}, -\frac{6}{7}, \frac{3}{7}\right) \]\[ D_{\mathbf{U}}f = -12 \left(\frac{2}{7}\right) + 2 \left(-\frac{6}{7}\right) + 14 \left(\frac{3}{7}\right) \]\[ D_{\mathbf{U}}f = \frac{-24}{7} + \frac{-12}{7} + \frac{42}{7} \]\[ D_{\mathbf{U}}f = \frac{-24 - 12 + 42}{7} = \frac{6}{7} \]

Key Concepts

Partial Derivatives

Partial derivatives allow us to understand how a multivariable function changes as one of the variables changes, while keeping the others constant. These derivatives are fundamental in calculus and are widely used in different fields like physics and engineering.
To find partial derivatives, we differentiate the function concerning each variable separately. For a function like \(f(x, y, z)\), we get three partial derivatives:
\[ \frac{\partial f}{\partial x} \text{,} \frac{\partial f}{\partial y} \text{,} \frac{\partial f}{\partial z} \]
These derivatives form the basis of the gradient vector.

In the provided exercise, we have the function \(f(x, y, z) = y^2 + z^2 - 4xz\). We find the partial derivatives:
\[ \frac{\partial f}{\partial x} = -4z \]
\[ \frac{\partial f}{\partial y} = 2y \]
\[ \frac{\partial f}{\partial z} = 2z - 4x \]
This reveals how the function \(f\) changes as \(x, y \), or \(z\) individually varies.

Gradient Vector

The gradient vector is a crucial concept in multivariable calculus. It combines all partial derivatives into a single vector, representing the direction and rate of steepest ascent for the function.

The gradient vector, \(abla f\), for a function \(f(x, y, z)\), is written as:
\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]

In the exercise, we evaluated the gradient at point \( P(-2, 1, 3) \):
\[abla f (-2,1,3) = (-12, 2, 14)\]
This means that at point \(P\), the direction of the steepest ascent (most rapid increase) of the function is along the vector \((-12, 2, 14)\). Additionally, the gradient vector's magnitude at this point tells us the rate of change in that direction.
This becomes particularly useful when calculating directional derivatives.

Directional Derivative

The directional derivative extends the concept of the gradient. It measures the rate of change of a function in any specified direction, not just along the coordinate axes.
To compute the directional derivative, we need:
\(1.\) The gradient vector(
\(2.\) A unit vector \(\mathbf{U}\) giving the direction
The formula for the directional derivative is:
\[ D_{\mathbf{U}}f = abla f \cdot \mathbf{U} \]

In this exercise, we have the gradient \(abla f (-2, 1, 3) = (-12, 2, 14)\), and a unit vector:
\( \mathbf{U} = \frac{2}{7} \mathbf{i} - \frac{6}{7} \mathbf{j} + \frac{3}{7} \mathbf{k} \)
Combining these using the dot product:
\[ D_{\mathbf{U}}f = (-12, 2, 14) \cdot \left( \frac{2}{7}, -\frac{6}{7}, \frac{3}{7} \right) = \frac{6}{7} \]
Hence, the rate of change of the function \(f\) in the direction of \(\mathbf{U}\) at point \(P\) is \( \frac{6}{7} \). This gives us a detailed and precise understanding of how \(f\) behaves not just at point \(P\), but specifically in the direction of vector \(\mathbf{U}\).
This concept is widely used in fields like optimization, physics, and computer graphics where understanding the effect of direction on a function's change is crucial.