Problem 56 - Calculus Early Transcendentals

Question

Gradients in three dimensions Consider the following functions \(f,\) points \(P,\) and unit vectors \(\mathbf{u}\) a. Compute the gradient of \(f\) and evaluate it at \(P\). b. Find the unit vector in the direction of maximum increase of \(f\) at \(P\). c. Find the rate of change of the function in the direction of maximum increase at \(P\) d. Find the directional derivative at \(P\) in the direction of the given vector. $$f(x, y, z)=4-x^{2}+3 y^{2}+\frac{z^{2}}{2} ; P(0,2,-1) ;\left\langle 0, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right\rangle$$

Step-by-Step Solution

Answer

- The gradient at point P is \(\left\langle 0, 12, -1 \right\rangle\). 2. What is the unit vector in the direction of maximum increase of the function at point P? - The unit vector in the direction of maximum increase is \(\left\langle 0, \frac{12}{\sqrt{145}}, -\frac{1}{\sqrt{145}} \right\rangle\). 3. What is the rate of change of the function in the direction of maximum increase at point P? - The rate of change in the direction of maximum increase is \(\sqrt{145}\). 4. What is the directional derivative of the function at point P in the direction of the given vector \(\left\langle 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle\)? - The directional derivative at point P in the given direction is \(\frac{13}{\sqrt{2}}\).

Step 1: Compute the gradient of f

To find the gradient of a given function \(f(x, y, z)\), we will compute the partial derivatives of the function with respect to each variable. The gradient is given by the vector \(\nabla f = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\). So, $$\nabla f(x, y, z) = \left\langle\frac{\partial}{\partial x}\left(4 - x^2 + 3y^2 + \frac{z^2}{2}\right), \frac{\partial}{\partial y}\left(4 - x^2 + 3y^2 + \frac{z^2}{2}\right), \frac{\partial}{\partial z}\left(4 - x^2 + 3y^2 + \frac{z^2}{2}\right)\right\rangle.$$ After calculating the partial derivatives, we have: $$\nabla f(x, y, z) = \left\langle -2x, 6y, z \right\rangle.$$

Step 2: Evaluate the gradient at point P

Now, we need to find the value of the gradient at point \(P(0, 2, -1)\). Substitute \(x=0, y=2\), and \(z=-1\) into the gradient \(\nabla f(x, y, z) =\left\langle -2x, 6y, z \right\rangle\): $$\nabla f(0, 2, -1) = \left\langle -2(0), 6(2), (-1) \right\rangle = \left\langle 0, 12, -1 \right\rangle.$$

Step 3: Find the unit vector in the direction of maximum increase

The unit vector in the direction of maximum increase of a function is just the normalized gradient vector. So, we need to find the magnitude of the gradient vector at point P and then divide the gradient vector by its magnitude. The magnitude of the gradient vector \(\nabla f(0, 2, -1) =\left\langle 0, 12, -1 \right\rangle\) is given by: $$\left\lVert\left\langle 0, 12, -1 \right\rangle\right\rVert =\sqrt{(0)^2 + (12)^2 + (-1)^2} = \sqrt{145}.$$ Now, divide the gradient vector by its magnitude to get the unit vector in the direction of maximum increase of f at P: $$\text{Unit Vector} = \frac{\left\langle 0, 12, -1 \right\rangle}{\sqrt{145}} = \left\langle 0, \frac{12}{\sqrt{145}}, -\frac{1}{\sqrt{145}} \right\rangle.$$

Step 4: Find the rate of change in the direction of maximum increase

The rate of change in the direction of maximum increase is equal to the magnitude of the gradient vector at point P, which we found in step 3: $$\text{Rate of Change} = \left\lVert\left\langle 0, 12, -1 \right\rangle\right\rVert = \sqrt{145}.$$

Step 5: Find the directional derivative in the direction of the given vector

The directional derivative of a function at a point in a given direction is the dot product of the gradient vector and the unit vector of the given direction. So, the directional derivative at point \(P(0, 2, -1)\) in the direction of the given vector \(\left\langle 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle\) is: $$\left\langle 0, 12, -1 \right\rangle \cdot \left\langle 0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle = (0)(0) + (12)(\frac{1}{\sqrt{2}}) + (-1)(-\frac{1}{\sqrt{2}}) = \frac{13}{\sqrt{2}}.$$

Key Concepts

Understanding Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when working with functions of multiple variables. Essentially, a partial derivative represents the rate of change of a function with respect to one of its variables, while holding the other variables constant. For a function \(f(x, y, z)\), the partial derivative with respect to \(x\) is denoted \( \frac{\partial f}{\partial x} \). This tells us how the function \(f\) changes as \(x\) changes, independently of changes in \(y\) or \(z\).

To compute partial derivatives, follow these steps:

• Differentiation with respect to a chosen variable while treating other variables as constants.
• Apply standard differentiation rules: power rule, product rule, etc., only focusing on terms involving the variable for which you differentiate.

In the given problem, the gradient \( abla f = \langle -2x, 6y, z \rangle \) is composed of the partial derivatives of \(f(x, y, z)\) with respect to \(x\), \(y\), and \(z\). Partial derivatives are the building blocks of the gradient, each capturing how the function changes along one individual dimension.

Directional Derivative Explained

A directional derivative provides information about how a function changes as you move in a specific direction from a given point. It's essentially a generalization of a partial derivative, which only considers the axes-aligned directions. The directional derivative of a function \(f\) at a point \(P\) in the direction of a vector \(\mathbf{v}\) is computed using:

\[ D_\mathbf{v}f = abla f \cdot \mathbf{u} \]
where \(abla f\) is the gradient of the function, and \(\mathbf{u}\) is the unit vector in the direction of \(\mathbf{v}\).

The dot product \(abla f \cdot \mathbf{u}\) effectively projects the gradient onto the direction of \(\mathbf{u}\), representing the rate of change of \(f\) in that particular direction.

• The computation involves finding the gradient and the unit vector form of the direction vector.
• It reveals how steeply, and in what direction, the function \(f\) increases or decreases as we move from point \(P\) towards the direction of \(\mathbf{v}\).

In our problem, the given directional derivative was \(\frac{13}{\sqrt{2}}\), indicating a specific rate of change of the function in the direction of the provided vector.

What is a Unit Vector?

Unit vectors play a critical role in the calculation of directional derivatives and in understanding vector directions in general. A unit vector is simply a vector with a magnitude (or length) of exactly one. The unit vector maintains the direction of the original vector, but scales its length to one.

To find a unit vector in the direction of a given vector \(\mathbf{v} = \langle a, b, c \rangle \), you would:

• Calculate the magnitude of the vector: \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \).
• Divide each component of the vector by its magnitude: \( \mathbf{u} = \frac{1}{\| \mathbf{v} \|} \langle a, b, c \rangle \).

In the exercise, the unit vector \(\langle 0, \frac{12}{\sqrt{145}}, -\frac{1}{\sqrt{145}} \rangle\) serves as the direction of maximum increase of the function \(f\) at point \(P\). By converting the gradient at \(P\) into a unit vector, we aligned it along its direction without regard to its original magnitude, enabling clear computation of the directional derivative.

Understanding Rate of Change

The rate of change is a broad concept that quantifies how a certain variable or quantity changes with respect to another. In the context of calculus and three-dimensional functions, we often refer to both partial derivatives and directional derivatives as measures of rate of change.

For this specific exercise, we focused on:

• Partial derivatives, which show the rate of change along individual coordinate axes.
• The directional derivative, indicating how rapidly the function value changes as one moves in a specific direction.

The exercise symbolizes how the function \(f\) changes at point \(P = (0,2,-1)\), particularly exploring its gradient's directions of maximum rate of change.

The rate of change in the direction of maximum increase at point \(P\) was \(\sqrt{145}\), simply the magnitude of the gradient at that point. This indicates the steepest path on the function's surface from the point \(P\), ensuring clarity on where the function increases most significantly.