Problem 61 - 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. $$f(x, y, z)=\ln \left(1+x^{2}+y^{2}+z^{2}\right) ; P(1,1,-1) ;\left\langle\frac{2}{3}, \frac{2}{3},-\frac{1}{3}\right\rangle$$

Step-by-Step Solution

Answer

#Problem# A function is given by: $$f(x, y, z)=\ln \left(1+x^{2}+y^{2}+z^{2}\right)$$ Consider the point \(P(1, 1, -1)\). 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 \(\mathbf{u} = \langle \frac{2}{3}, \frac{2}{3}, -\frac{1}{3}\rangle\). #Solution# a) The gradient of \(f\) at point \(P\) is given by: \(\nabla{f}(1,1,-1) = \langle \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \rangle\). b) The unit vector in the direction of maximum increase is given by: \(\mathbf{u}_{max} = \langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}} \rangle\). c) The rate of change of the function in the direction of maximum increase at \(P\) is \(\frac{\sqrt{3}}{2}\). d) The directional derivative of \(f\) at \(P\) in the direction of the given vector is \(D_{\mathbf{u}}f(1,1,-1) = 0\).

Step 1: Compute the partial derivatives

To find the gradient of \(f\), we must find the partial derivatives with respect to \(x\), \(y\), and \(z\). Let us compute the partial derivative with respect to x: $$\frac{\partial f}{\partial x} = \frac{2x}{1+x^2+y^2+z^2}$$ Similarly, we find the partial derivative with respect to y and z: $$\frac{\partial f}{\partial y} = \frac{2y}{1+x^2+y^2+z^2}$$ $$\frac{\partial f}{\partial z} = \frac{2z}{1+x^2+y^2+z^2}$$

Step 2: Evaluate the gradient of \(f\) at point P

Next, we must evaluate the gradient at the given point \(P(1,1,-1)\). We substitute these values into the partial derivatives calculated in step 1: $$\left(\frac{\partial f}{\partial x}\right)_{P} = \frac{2(1)}{1+(1)^2+(1)^2+(-1)^2} = \frac{2}{4} = \frac{1}{2}$$ $$\left(\frac{\partial f}{\partial y}\right)_{P} = \frac{2(1)}{1+(1)^2+(1)^2+(-1)^2} = \frac{2}{4} = \frac{1}{2}$$ $$\left(\frac{\partial f}{\partial z}\right)_{P} = \frac{2(-1)}{1+(1)^2+(1)^2+(-1)^2} = \frac{-2}{4} = -\frac{1}{2}$$ Therefore, the gradient of \(f\) at point \(P\) is: $$\nabla{f}(1,1,-1) = \langle \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \rangle$$

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

The unit vector in the direction of maximum increase is the gradientof the function divided by its magnitude. So, first find the magnitude of the gradient: $$|\nabla{f}(1,1,-1)| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}+\frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$ Now divide the gradient by the magnitude to find the unit vector in the direction of maximum increase: $$\mathbf{u}_{max} = \frac{\nabla{f}(1,1,-1)}{|\nabla{f}(1,1,-1)|} = \frac{\langle \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \rangle}{\frac{\sqrt{3}}{2}} = \langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}} \rangle$$

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

The rate of change of the function in the direction of maximum increase at \(P\) is the dot product of the gradient and the maximum increase unit vector: $$\nabla{f}(1,1,-1) \cdot \mathbf{u}_{max}= \langle \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \rangle \cdot \langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ,-\frac{1}{\sqrt{3}} \rangle = \frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}} = \boxed{\frac{\sqrt{3}}{2}}$$

Step 5: Find the directional derivative at \(P\) in the given direction

The given direction is \(\mathbf{u} = \langle\frac{2}{3}, \frac{2}{3},-\frac{1}{3}\rangle\). To find the directional derivative of \(f\) at \(P\) in this direction, we calculate the dot product of the gradient and the given vector: $$\nabla{f}(1,1,-1) \cdot \mathbf{u}= \langle \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \rangle \cdot \langle\frac{2}{3}, \frac{2}{3}, -\frac{1}{3}\rangle = \frac{1}{3}-\frac{1}{3} half$$ Therefore, the directional derivative of \(f\) at \(P\) in the direction of the given vector is \(D_{\mathbf{u}}f(1,1,-1) = \boxed{0}\).

Key Concepts

Understanding Partial Derivatives

Imagine exploring a mountainous terrain, where the scenery changes not only as you move forward and backward but also as you move sideways or vertically. In calculus, partial derivatives help us understand how a function changes in multiple directions. Specifically, for functions of several variables, partial derivatives measure how the function changes in response to changes in one specific variable, while keeping the others constant.

For example:

• The partial derivative with respect to \(x\) is denoted by \(\frac{\partial f}{\partial x}\) and tells us how \(f\) changes when you slightly change \(x\), keeping the other variables \(y\) and \(z\) constant.
• Similarly, \(\frac{\partial f}{\partial y}\) and \(\frac{\partial f}{\partial z}\) help us understand the change in \(f\) when varying \(y\) or \(z\), respectively.

These derivatives are critical in finding the gradient of a function, which is a vector that points in the direction of the greatest rate of increase of the function. Calculating partial derivatives is the first step in understanding how a function behaves in a complicated, multi-dimensional space.

Unpacking the Directional Derivative

In daily life, we often want to know not just how something changes globally but how it changes in a specific direction. The directional derivative is a mathematical concept that extends this intuitive idea to functions of several variables.

The directional derivative represents the rate at which a function changes at a point \(P\) in the direction of a given vector \(\mathbf{u}\). The process involves taking the dot product of the gradient of the function at that point and the vector \(\mathbf{u}\). This can be visualized as projecting the vector along the direction of interest.

• Given a vector \(\mathbf{u}\), which is a direction in the space, the directional derivative tells us how steep or flat the slope is in that direction.
• A directional derivative of zero means the function does not change in the given direction, which was the result in our exercise with the given vector.

Understanding this concept is crucial, especially in fields like physics and machine learning, where knowing the direction of change helps in predicting system behaviors or optimizing processes.

Interpreting Rate of Change

The rate of change is a fundamental concept akin to velocity in physics, indicating how quickly something is changing over time or space. In the context of multivariable calculus, it tells us how a function \(f\) is changing around a specific point \(P\) as you move in a particular direction.

The maximum rate of change occurs in the direction of the gradient vector. This is because the gradient points in the direction of steepest ascent, meaning it tells us how to adjust \(x\), \(y\), and \(z\) to increase \(f\) most rapidly.

• The magnitude of the gradient itself gives the maximum rate of change.
• In simpler terms, if you were hiking up a hill, the gradient would tell you the steepest path to reach the top fastest.

The exercise also demonstrates how to determine this rate using a straightforward calculation: the magnitude of the gradient vector. This is instrumental in various fields like economics for optimizing profits or engineering for stress analysis in materials.