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Chapter 12: Problem 58

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

Short Answer

Expert verified
Furthermore, what is the directional derivative of f at point P in the direction of the vector u = <0, -1/鈭�2, -1/鈭�2>? The unit vector in the direction of maximum increase of the function f at point P is given by: $$\mathbf{u}_{max} = \left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}},0 \right\rangle$$ The rate of change of the function in this direction is: $$\text{Rate of Change}_{max} = \sqrt{10}$$ The directional derivative of f at point P in the direction of the given vector is: $$D_{\mathbf{u}}f(P) = -\frac{3}{\sqrt{2}}$$

Step by step solution

01

Compute the gradient of f

To find the gradient of f, we need to compute its partial derivatives with respect to x, y, and z. Let's first find the partial derivatives of f: \(\frac{\partial f}{\partial x} = y+z\) \(\frac{\partial f}{\partial y} = x+z\) \(\frac{\partial f}{\partial z} = x+y\) The gradient of f, denoted as \(\nabla f,\) can be written in vector notation as: $$\nabla f = \left\langle y+z, x+z, x+y \right\rangle$$
02

Evaluate the gradient at point P

To evaluate the gradient at point P, we substitute the coordinates of P into the gradient: $$P(2,-2,1) \rightarrow \nabla f_P = \left\langle-2+1, 2+1, 2-2 \right\rangle$$ Thus, the gradient of f at point P is: $$\nabla f_P=\left\langle-1,3,0\right\rangle$$
03

Find the unit vector in the direction of maximum increase at point P

The unit vector in the direction of maximum increase of f at point P is given by the normalized gradient of f at P: $$\mathbf{u}_{max} = \frac{\nabla f_P}{||\nabla f_P||}$$ First, find the magnitude of the gradient at P: $$||\nabla f_P|| = \sqrt{(-1)^2+3^2+0^2} = \sqrt{10}$$ Then, find the unit vector in the direction of maximum increase: $$\mathbf{u}_{max} = \frac{\left\langle-1,3,0\right\rangle}{\sqrt{10}} = \left\langle -\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}},0 \right\rangle$$
04

Find the rate of change in the direction of maximum increase at point P

The rate of change of the function in the direction of the maximum increase at point P is given by the magnitude of the gradient of f at P, which we already calculated: $$\text{Rate of Change}_{max} = ||\nabla f_P|| = \sqrt{10}$$
05

Find the directional derivative at point P in the direction of the given vector

To find the directional derivative at point P in the direction of the given vector \(\mathbf{u} = \left\langle 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle\), we can use the formula: $$D_{\mathbf{u}}f(P) = \nabla f_P \cdot \mathbf{u}$$ Compute the dot product of \(\nabla f_P\) and \(\mathbf{u}\): $$D_{\mathbf{u}}f(P) = \left\langle -1,3,0\right\rangle \cdot \left\langle 0, -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right\rangle = -\frac{3}{\sqrt{2}}$$ So the directional derivative of f at point P in the direction of the given vector is \(-\frac{3}{\sqrt{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Directional Derivative
The directional derivative represents how a function changes as you move in a specific direction from a given point. Understanding this concept can be quite helpful, especially when you are dealing with surfaces in three dimensions, like in our exercise with the function f(x, y, z). To find it, you start with the gradient of the function, which you can think of as a vector that points in the direction of the steepest increase. Then, you take the dot product of this gradient with the unit vector in the direction you're interested in. This is exactly what's done in step 5 of our exercise:
Unit Vector
A unit vector is indispensable in calculations because it retains the direction information but its magnitude is one, which simplifies a lot of vector arithmetic. It's like a pure 'direction' with no concern about how 'strong' or 'far' it goes. To find a unit vector, you divide a vector by its magnitude, as shown in step 3 of our exercise with the gradient vector. This provides a clean, standardized vector that can be used to direct other calculations, such as finding the directional derivative.
Rate of Change
When we speak of the 'rate of change' in mathematics, we're describing how fast a function's value is changing at a given point, and in a given direction. This sounds abstract, but think of walking up a hill; the steeper the slope, the harder you're working against gravity. In step 4 of our exercise, the rate of change in the direction of maximum increase is the magnitude of the gradient. It tells you how steep the 'hill' is at point P.
Partial Derivatives
Partial derivatives come into play when we want to assess how a function changes with respect to one variable while keeping all others constant. They are the building blocks of the gradient, and calculating them is our first step in understanding how the function behaves in multidimensional space. Just like in step 1 of the exercise, you find how f changes as only x, only y, or only z moves.
Vector Notation
Vector notation is a compact, mathematical way to describe coordinates and direction. It's impressive how much information you can fit into a simple angled bracket like <x, y, z>! This notation is used to describe the gradient in a way that further calculations, such as finding unit vectors or directional derivatives, can be comfortably performed, just as you see throughout the solution in our exercise.
Magnitude of Vector
The magnitude of a vector is a measure of its 'length', which you can think of as the amount of force it represents, or how far you would travel in that direction if the vector were a path. To find it, you take the square root of the sum of the squares of each component, a process you can see in step 3 where we calculate the magnitude ||∇ f_P|| of the gradient to then find the unit vector. It's like using the Pythagorean theorem in high-dimensional space!

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Most popular questions from this chapter

Find the maximum value of \(x_{1}+x_{2}+x_{3}+x_{4}\) subject to the condition that \(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}=16\).

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) In two dimensions, the motion of an ideal fluid (an incompressible and irrotational fluid) is governed by a velocity potential \(\varphi .\) The velocity components of the fluid, \(u\) in the \(x\) -direction and \(v\) in the \(y\) -direction, are given by \(\langle u, v\rangle=\nabla \varphi .\) Find the velocity components associated with the velocity potential \(\varphi(x, y)=\sin \pi x \sin 2 \pi y\).

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). \(u(x, t)=A e^{-\alpha^{2} t} \cos a x,\) for any real numbers \(a\) and \(A\)

An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for \(z=f(x, y),\) the Laplacian is \(z_{x x}+z_{y y} .\) Determine the Laplacian in polar coordinates using the following steps. a. Begin with \(z=g(r, \theta)\) and write \(z_{x}\) and \(z_{y}\) in terms of polar coordinates (see Exercise 64). b. Use the Chain Rule to find \(z_{x x}=\frac{\partial}{\partial x}\left(z_{x}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. c. Use the Chain Rule to find \(z_{y y}=\frac{d}{\partial y}\left(z_{y}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. d. Combine parts (b) and (c) to show that $$z_{x x}+z_{y y}=z_{r r}+\frac{1}{r} z_{r}+\frac{1}{r^{2}} z_{\theta \theta}$$

The output \(Q\) of an economic system subject to two inputs, such as labor \(L\) and capital \(K,\) is often modeled by the Cobb-Douglas production function \(Q(L, K)=c L^{a} K^{b} .\) Suppose \(a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=1\). a. Evaluate the partial derivatives \(Q_{L}\) and \(Q_{K}\). b. Suppose \(L=10\) is fixed and \(K\) increases from \(K=20\) to \(K=20.5 .\) Use linear approximation to estimate the change in \(Q\). c. Suppose \(K=20\) is fixed and \(L\) decreases from \(L=10\) to \(L=9.5 .\) Use linear approximation to estimate the change in \(\bar{Q}\). d. Graph the level curves of the production function in the first quadrant of the \(L K\) -plane for \(Q=1,2,\) and 3. e. Use the graph of part (d). If you move along the vertical line \(L=2\) in the positive \(K\) -direction, how does \(Q\) change? Is this consistent with \(Q_{K}\) computed in part (a)? f. Use the graph of part (d). If you move along the horizontal line \(K=2\) in the positive \(L\) -direction, how does \(Q\) change? Is this consistent with \(Q_{L}\) computed in part (a)?
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