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Chapter 12: Problem 62

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

Short Answer

Expert verified
Question: Compute the directional derivative of the function \(f(x, y, z) = \frac{x - z}{y - z}\) at the point P(3, 2, -1) in the direction of the vector \(\frac{1}{3} \langle 1, 2, -1\rangle\). Answer: The directional derivative of the function at point P(3, 2, -1) in the direction of the vector \(\frac{1}{3} \langle 1, 2, -1\rangle\) is \(-\frac{1}{3}\).

Step by step solution

01

Compute the Gradient of the Function

To compute the gradient of the function, we will find the partial derivatives with respect to each variable (x, y, and z) in the function. The gradient is represented by \(\nabla f(x, y, z)\). $$ \nabla f(x, y, z) = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle $$ Now, let's compute the partial derivatives: $$ \frac{\partial f}{\partial x} = \frac{1}{y-z}\ ,\ \frac{\partial f}{\partial y} = -\frac{x-z}{(y-z)^2}\ , \ \frac{\partial f}{\partial z} = -\frac{1}{y-z} +\frac{x-z}{(y-z)^2} $$ So, the gradient of f is: $$ \nabla f(x, y, z) = \left\langle \frac{1}{y-z}, -\frac{x-z}{(y-z)^2}, -\frac{1}{y-z} + \frac{x-z}{(y-z)^2} \right\rangle $$
02

Evaluate the gradient at P

Now, let's evaluate the gradient at point P(3,2,-1): $$ \nabla f(3, 2, -1) = \left\langle \frac{1}{2-(-1)}, -\frac{3-(-1)}{(2-(-1))^2}, -\frac{1}{2-(-1)} + \frac{3-(-1)}{(2-(-1))^2} \right\rangle $$ After computing the values, $$ \nabla f(3, 2, -1) = \left\langle 1, -1, 0 \right\rangle $$
03

The unit vector in the direction of maximum increase of \(f\)

The unit vector in the direction of maximum increase of \(f\) is the unit vector of the gradient. So, let's find the magnitude of the gradient first. $$ ||\nabla f(3, 2, -1)|| = \sqrt{(1)^2 + (-1)^2 + (0)^2} = \sqrt{2} $$ Now, divide the gradient by its magnitude to obtain the unit vector. $$ \frac{\nabla f(3, 2, -1)}{||\nabla f(3, 2, -1)||} = \frac{\left\langle 1, -1, 0 \right\rangle}{\sqrt{2}} = \left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle $$
04

Rate of change of the function in the direction of maximum increase at P

The rate of change of the function in the direction of the maximum increase at point P is equal to the magnitude of the gradient evaluated at the point. $$ ||\nabla f(3, 2, -1)|| = \sqrt{2} $$
05

Directional derivative at P in the direction of the given vector

The directional derivative at P in the direction of the given vector u is given by \(D_{\mathbf{u}}f = \nabla f(3,2,-1) \cdot \mathbf{u}\). So let's evaluate it. $$ D_{\mathbf{u}}f = \left\langle 1, -1, 0 \right\rangle \cdot \left\langle \frac{1}{3}, \frac{2}{3},-\frac{1}{3} \right\rangle = \frac{1}{3} - \frac{2}{3} + 0 $$ After the computation, $$ D_{\mathbf{u}}f = -\frac{1}{3} $$

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Key Concepts

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

Partial Derivatives
When dealing with multivariable functions, we often fix one variable and look at how the function changes with respect to another. This is where the concept of partial derivatives comes in. Partial derivatives measure how a function changes as one of the variables changes while the others are held constant. For a function like \( f(x, y, z) \), you would find the partial derivative with respect to \( x \) by considering \( y \) and \( z \) as constants.
For example:\[ \frac{\partial f}{\partial x} = \frac{1}{y-z} \]focuses purely on how \( f \) changes with \( x \), assuming \( y \) and \( z \) remain fixed.
  • \( \frac{\partial f}{\partial y} \) involves holding \( x \) and \( z \) constant and differentiating with respect to \( y \).
  • \( \frac{\partial f}{\partial z} \) similarly fixes \( x \) and \( y \), considering only shifts in \( z \).
The partial derivatives together form the gradient, a vector highlighting how a function changes in each direction from a given point.
Directional Derivative
Imagine you are hiking up a mountain and want to calculate how steep the path is in a particular direction. The directional derivative serves this purpose by measuring the rate of change in any specified direction.
Instead of just focusing on one variable, this derivative applies the gradient of a function to any chosen direction.
The directional derivative is computed using the dot product of the gradient and a unit direction vector:\[ D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \]This formula tells us how quickly the function changes in the direction of vector \( \mathbf{u} \).
To find this, make sure \( \mathbf{u} \) is a unit vector. Normalize it if necessary by dividing the vector by its own magnitude.
  • It allows you to find the steepest path uphill - parallel to the gradient.
  • On the contrary, it can pinpoint the direction of maximum decrease as well by pointing opposite to the gradient.
This mathematical tool is essential for understanding the behavior of functions in multiple dimensions, guiding how they change as you move in various directions.
Rate of Change
In mathematics, the rate of change describes how quickly one quantity varies in relation to another.
For functions of multiple variables, such as \( f(x, y, z) \), we use the gradient to determine rates of change.
The gradient, denoted as \( abla f \), provides a vector that points in the direction of the steepest increase of the function value. Its magnitude gives the maximum rate of change at a point.\[||abla f||\]This tells you how fast the function is increasing in the steepest direction.
  • It is useful in forecasting how a function behavior evolves nearby a particular point.
  • The magnitude of the gradient evaluates how intense the increase or decrease is at that point.
Practical applications include understanding topographic maps or visualizing the rate of temperature change across different regions. Understanding the rate of change properly provides insights into how dynamic systems behave and transform over space.

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Most popular questions from this chapter

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)?

Consider the following functions \(f\). a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\). d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at (0,0). e. Explain why Theorems 12.5 and 12.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$

Imagine a string that is fixed at both ends (for example, a guitar string). When plucked, the string forms a standing wave. The displacement \(u\) of the string varies with position \(x\) and with time \(t .\) Suppose it is given by \(u=f(x, t)=2 \sin (\pi x) \sin (\pi t / 2),\) for \(0 \leq x \leq 1\) and \(t \geq 0\) (see figure). At a fixed point in time, the string forms a wave on [0, 1]. Alternatively, if you focus on a point on the string (fix a value of \(x\) ), that point oscillates up and down in time. a. What is the period of the motion in time? b. Find the rate of change of the displacement with respect to time at a constant position (which is the vertical velocity of a point on the string). c. At a fixed time, what point on the string is moving fastest? d. At a fixed position on the string, when is the string moving fastest? e. Find the rate of change of the displacement with respect to position at a constant time (which is the slope of the string). f. At a fixed time, where is the slope of the string greatest?

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 .) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

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