Problem 58 - 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. $$f(x, y, z)=x y+y z+x z+4 ; P(2,-2,1) ;\left\langle 0,-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right\rangle$$

Step-by-Step Solution

Answer

Based on the provided solution, the gradient of f at point P is $\nabla f(2,-2,1) = \langle -1, 3, 0\rangle$. The unit vector in the direction of maximum increase of f at P is $\left\langle -\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}, 0 \right\rangle$. The rate of change of f in the direction of maximum increase is $\sqrt{10}$. Finally, the directional derivative at P in the direction of the given vector is $-\frac{3}{\sqrt{2}}$.

Step 1: Find the gradient of f

To find the gradient of the function \(f(x,y,z) = xy+yz+xz+4\), we need to compute the partial derivatives of f with respect to x, y, and z. $$ \frac{\partial f}{\partial x} = y + z; \quad \frac{\partial f}{\partial y} = x + z; \quad \frac{\partial f}{\partial z} = x + y $$

Step 2: Evaluate the gradient at point P

Now that we have the partial derivatives, let's evaluate them at the given point \(P(2,-2,1)\): $$ \left(\frac{\partial f}{\partial x}(2,-2,1), \frac{\partial f}{\partial y}(2,-2,1), \frac{\partial f}{\partial z}(2,-2,1)\right) = (-2+1,2+1,-2+2) = (-1,3,0) $$ So, the gradient of f at point P is \(\nabla f(2,-2,1) = \langle -1, 3, 0\rangle\).

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

The unit vector in the direction of maximum increase of f at point P is the normalized gradient vector. We can normalize the gradient by dividing it by its magnitude: $$ \frac{\nabla f}{\lVert\nabla f\rVert} = \frac{\left\langle -1, 3, 0 \right\rangle}{\sqrt{(-1)^2 + (3)^2 + (0)^2}} = \frac{\left\langle -1, 3, 0 \right\rangle}{\sqrt{10}} = \left\langle -\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}, 0 \right\rangle $$

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

The rate of change of f in the direction of maximum increase is the magnitude of the gradient vector: $$ \lVert\nabla f\rVert = \sqrt{(-1)^2 + (3)^2 + (0)^2} = \sqrt{10} $$

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

The directional derivative of f at P in the direction of the given vector \(\mathbf{u} = \left\langle 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right\rangle\) is the dot product of the gradient vector and the given vector: $$ D_{\mathbf{u}} f(P) = \nabla f(P) \cdot \mathbf{u} = \left\langle -1, 3, 0 \right\rangle \cdot \left\langle 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right\rangle = -1\cdot 0 + 3\cdot \left(-\frac{1}{\sqrt{2}}\right) + 0\cdot \left(-\frac{1}{\sqrt{2}}\right) = - \frac{3}{\sqrt{2}} $$ So the directional derivative at P in the direction of the given vector is \(-\frac{3}{\sqrt{2}}\).

Key Concepts

Directional Derivative

The directional derivative tells us how a function changes as we move in a specific direction. Imagine placing yourself at a point on a surface, and then walking in a particular direction. The directional derivative helps you understand the rate at which the height of the surface would change as you walk.

To calculate it, we first need the gradient vector of the function at the given point. The gradient points in the direction of steepest increase. However, we're often interested in how the function changes in other directions.

To find the directional derivative, we use a dot product between the gradient and the direction vector. If the direction vector isn't already a unit vector, it must be normalized first. This ensures that the rate of change is correctly measured per unit distance.

In summary, the directional derivative indicates the rate of change of a function in any specified direction, and is fundamental in approximating how functions behave across different directions.

Rate of Change

The rate of change in the context of calculus and gradients shows how a particular function evolves as its variables change. The concept of rate of change is key in understanding and predicting behaviors in various fields like physics, economics, and engineering.

When talking about functions depending on several variables, identifying the rate of change can become complex. However, the gradient provides a powerful tool to simplify this process. It captures all the rates of change of a multivariable function and points towards the steepest ascent.

For a function of three variables, the rate of change in the direction of the greatest increase (or steepest complete climb) at a particular point is the magnitude of the gradient vector. This is intuitive: imagine standing on a three-dimensional hill, the steepest path upwards gives you the highest rate of height increase.

Another key aspect is how this rate can be calculated in directions other than the steepest. Here, knowing how to compute the directional derivative is crucial. It provides the rate of change in any desired direction.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus to understand how a multivariable function changes as each variable is independently tweaked. If you have a function of several variables, like our function of three variables: \(f(x, y, z)\), partial derivatives focus on how the function changes when just one variable at a time is modified, while keeping others constant.

To find a partial derivative with respect to a variable, you consider the function as if all other variables are constants. This action "freezes" all variables except the one you are differentiating.

For instance, in calculating the partial derivative of \(f(x, y, z) = xy + yz + xz + 4\) with respect to \(x\), we consider \(y\) and \(z\) as constants. Hence, the partial derivatives are:

• \(\frac{\partial f}{\partial x} = y + z\)
• \(\frac{\partial f}{\partial y} = x + z\)
• \(\frac{\partial f}{\partial z} = x + y\)

This allows us to construct the gradient vector, which ultimately guides computations like the directional derivatives and helps determine the rate of change.