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Chapter 13: Problem 55

Find a unit vector in the direction in which \(f\) decreases most rapidly at \(P,\) and find the rate of change of \(f\) at \(P\) in that direction. $$ f(x, y)=20-x^{2}-y^{2} ; P(-1,-3) $$

Short Answer

Expert verified
The unit vector is \( \left(\frac{-1}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right) \), and the rate of change is \(-2\sqrt{10}\).

Step by step solution

01

Compute the Gradient of the Function

The first step is to find the gradient of the function \( f(x, y) = 20 - x^2 - y^2 \). The gradient is a vector containing the partial derivatives with respect to each variable. Compute these partial derivatives:\[\frac{\partial f}{\partial x} = -2x \quad \text{and} \quad \frac{\partial f}{\partial y} = -2y\]So, the gradient is:\[ abla f(x, y) = (-2x, -2y) \]
02

Evaluate the Gradient at Point P

Next, substitute the coordinates of the point \( P(-1, -3) \) into the gradient to find its value at this point:\[abla f(-1, -3) = (-2(-1), -2(-3)) = (2, 6)\]
03

Find the Unit Vector for Most Rapid Decrease

The direction in which \( f \) decreases most rapidly is opposite to the gradient direction. Therefore, the vector in this direction is \((-2, -6)\). Convert this vector into a unit vector. First, calculate its magnitude:\[\|(-2, -6)\| = \sqrt{(-2)^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}\]Divide the vector by its magnitude to get the unit vector:\[\left(\frac{-2}{2\sqrt{10}}, \frac{-6}{2\sqrt{10}}\right) = \left(\frac{-1}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right)\]
04

Calculate the Rate of Change in the Fastest Decrease Direction

The rate of change of \( f \) in the direction of the most rapid decrease is given by the dot product of the gradient and the unit vector in that direction:\[abla f(-1, -3) \cdot \left(\frac{-1}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right) = (2, 6) \cdot \left(\frac{-1}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right)\]Evaluating the dot product:\[2 \times \frac{-1}{\sqrt{10}} + 6 \times \frac{-3}{\sqrt{10}} = \frac{-2}{\sqrt{10}} + \frac{-18}{\sqrt{10}} = \frac{-20}{\sqrt{10}} = -2\sqrt{10}\]

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Key Concepts

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

Partial Derivative
A partial derivative is a basic concept in calculus used to find the rate at which a function changes with respect to one of its variables, while keeping the other variables constant.
This is particularly useful in functions of multiple variables, such as two-variable functions like the one in our exercise: \( f(x, y) = 20 - x^2 - y^2 \).
In this scenario, when we compute the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), we're assessing how \( f \) changes when only \( x \) changes.
Similarly, \( \frac{\partial f}{\partial y} \) tells us how \( f \) changes with changes in \( y \).

These derivatives essentially provide a snapshot of the 'slope' or 'gradient' of the function at a particular point along the \( x \)-axis and \( y \)-axis.
This information is vital in understanding the behavior of multivariable functions and helps us plot the landscape of this function over a coordinate plane.
Unit Vector
A unit vector is a vector with a magnitude of 1 that points in a specific direction.
To form a unit vector, you take a vector and divide it by its magnitude or length.
In our exercise, we needed a unit vector to represent the direction in which the function \( f \) decreases most rapidly.
This direction is opposite to the direction of the gradient vector, which is \( (2, 6) \) for this exercise.
So, we took the opposite vector, \((-2, -6)\), and converted it into a unit vector.

Here鈥檚 how it鈥檚 done:
  • Calculate the magnitude of the vector \((-2, -6)\).
  • Divide each component of the vector by its magnitude.
The result is a unit vector \( \left(\frac{-1}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right) \).
This unit vector preserves the direction of \((-2, -6)\) but with a length of 1, making computations simpler without altering the direction.
Rate of Change
The rate of change is a mathematical concept that describes how one quantity changes in relation to another.
This is critical when examining how a multivariable function behaves as its inputs change.
In the context of this exercise, the rate of change of the function \( f \) at point \( P \) in the direction of the steepest decrease is akin to finding how fast \( f \) is decreasing.

When calculating the rate of change for the steepest direction of decrease, we use the formula for the dot product between the gradient vector and the unit direction vector:
  • The dot product gives a scalar that represents the magnitude of the rate of change in that direction.
  • For this exercise, it gives us \( -2\sqrt{10} \), indicating not only the direction but also how rapidly the function \( f \) descends.
This value is particularly useful in optimization problems, where understanding how fast things are changing can help steer towards optimal solutions.

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Most popular questions from this chapter

Given \(f(x, y)=x^{3} y^{5}-2 x^{2} y+x,\) find $$\begin{array}{lll}{\text { (a) } f_{x x y}} & {\text { (b) } f_{y x y}} & {\text { (c) } f_{y y y}}\end{array}$$

A point moves along the intersection of the plane \(y=3\) and the surface \(z=\sqrt{29-x^{2}-y^{2}}\) At what rate is \(z\) changing with respect to \(x\) when the point is at \((4,3,2) ?\)

Solve using Lagrange multipliers. Find the points on the surface \(x y-z^{2}=1\) that are closest to the origin.

Show that the function satisfies the heat equation $$\frac{\partial z}{\partial t}=c^{2} \frac{\partial^{2} z}{\partial x^{2}} \quad(c>0, \text { constant })$$ $$\text { (a) } z=e^{-t} \sin (x / c) \quad \text { (b) } z=e^{-t} \cos (x / c)$$

A common problem in experimental work is to obtain a mathematical relationship \(y=f(x)\) between two variables \(x\) and \(y\) by "fitting" a curve to points in the plane that correspond to experimentally determined values of \(x\) and \(y,\) say $$ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) $$ The curve \(y=f(x)\) is called a mathematical model of the data. The general form of the function \(f\) is commonly determined by some underlying physical principle, but sometimes it is just determined by the pattern of the data. We are concerned with fitting a straight line \(y=m x+b\) to data. Usually, the data will not lie on a line (possibly due to experimental error or variations in experimental conditions), so the problem is to find a line that fits the data "best" according to some criterion. One criterion for selecting the line of best fit is to choose \(m\) and \(b\) to minimize the function $$ g(m, b)=\sum_{i=1}^{n}\left(m x_{i}+b-y_{i}\right)^{2} $$ This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometrically, \(\left|m x_{i}+b-y_{i}\right|\) is the vertical distance between the data point \(\left(x_{i}, y_{i}\right)\) and the line \(y=m x+b\) These vertical distances are called the residuals of the data points, so the effect of minimizing \(g(m, b)\) is to minimize the sum of the squares of the residuals. In these exercises, we will derive a formula for the regression line. The purpose of this exercise is to find the values of \(m\) and \(b\) that produce the regression line. (a) To minimize \(g(m, b),\) we start by finding values of \(m\) and \(b\) such that \(\partial g / \partial m=0\) and \(\partial g / \partial b=0 .\) Show that these equations are satisfied if \(m\) and \(b\) satisfy the conditions $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right) m+\left(\sum_{i=1}^{n} x_{i}\right) b=\sum_{i=1}^{n} x_{i} y_{i} $$ \(\left(\sum_{i=1}^{n} x_{i}\right) m+n b=\sum_{i=1}^{n} y_{i}\) (b) Let \(\bar{x}=\left(x_{1}+x_{2}+\cdots+x_{n}\right) / n\) denote the arithmetic average of \(x_{1}, x_{2}, \ldots, x_{n} .\) Use the fact that $$ \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq 0 $$ to show that $$ n\left(\sum_{i=1}^{n} x_{i}^{2}\right)-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geq 0 $$ with equality if and only if all the \(x_{i}\) 's are the same. (c) Assuming that not all the \(x_{i}\) 's are the same, prove that the equations in part (a) have the unique solution $$ \begin{aligned} m=& \frac{n \sum_{i=1}^{n} x_{i} y_{i}-\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}}{n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2}} \\ b=& \frac{1}{n}\left(\sum_{i=1}^{n} y_{i}-m \sum_{i=1}^{n} x_{i}\right) \end{aligned} $$ [Note: We have shown that \(g\) has a critical point at these values of \(m\) and \(b\). In the next exercise we will show that \(g\) has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line \(y=m x+b\) is the regression line for these values of \(m \text { and } b .]\)
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