/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 The density is \(\rho\) slugs/ft... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 22

The density is \(\rho\) slugs/ft \(^{2}\) at any point \((x, y)\) of a rectangular plate in the \(x y\) plane and \(\rho=1 / \sqrt{x^{2}+y^{2}+3}\). (a) Find the rate of change of the density at the point \((3,2)\) in the direction of the unit vector \(\cos \frac{2}{3} \pi \mathrm{i}+\sin \frac{2}{3} \pi \mathrm{j} .\) (b) Find the direction and magnitude of the greatest rate of change of \(\rho\) at \((3,2)\).

Short Answer

Expert verified
a) \( \frac{3 - 2\sqrt{3}}{128} \) b) Direction: \( \left( -\frac{3}{64}, -\frac{2}{64} \right) \), Magnitude: \( \frac{\sqrt{13}}{64} \)

Step by step solution

01

Express the density function

Given the density function: \[ \rho = \frac{1}{\sqrt{x^2 + y^2 + 3}} \]
02

Find partial derivatives

Compute the partial derivatives of \( \rho \) with respect to \( x \) and \( y \).Let's calculate \( \frac{\partial \rho}{\partial x} \): \[ \frac{\partial \rho}{\partial x} = \frac{\partial}{\partial x} \left( \frac{1}{\sqrt{x^2 + y^2 + 3}} \right) = -\frac{x}{(x^2 + y^2 + 3)^{3/2}} \]Similarly, calculate \( \frac{\partial \rho}{\partial y} \): \[ \frac{\partial \rho}{\partial y} = \frac{\partial}{\partial y} \left( \frac{1}{\sqrt{x^2 + y^2 + 3}} \right) = -\frac{y}{(x^2 + y^2 + 3)^{3/2}} \]
03

Evaluate the partial derivatives at (3, 2)

Substitute \( x = 3 \) and \( y = 2 \) into the partial derivatives to find the gradient at the point (3, 2).\[ \left. \frac{\partial \rho}{\partial x} \right|_{(3,2)} = -\frac{3}{(3^2 + 2^2 + 3)^{3/2}} = -\frac{3}{(9 + 4 + 3)^{3/2}} = -\frac{3}{16^{3/2}} = -\frac{3}{64} \]\[ \left. \frac{\partial \rho}{\partial y} \right|_{(3,2)} = -\frac{2}{(3^2 + 2^2 + 3)^{3/2}} = -\frac{2}{(9 + 4 + 3)^{3/2}} = -\frac{2}{16^{3/2}} = -\frac{2}{64} \]
04

Form the gradient vector

The gradient vector at the point \(3,2\) is found using the partial derivatives.\[ abla \rho = \left( -\frac{3}{64}, -\frac{2}{64} \right) \]
05

Compute the directional derivative

To find the rate of change of the density at (3, 2) in the direction of the unit vector\( \cos \frac{2}{3} \pi \mathbf{i} + \sin \frac{2}{3} \pi \mathbf{j} \), calculate the directional derivative using:\[ D_{\mathbf{u}} \rho = abla \rho \cdot \mathbf{u} \]where \( \mathbf{u} \) is the unit vector.First, find \( \mathbf{u} \):\[ \cos \left( \frac{2}{3} \pi \right) = -\frac{1}{2}, \quad \sin \left( \frac{2}{3} \pi \right) = \frac{\sqrt{3}}{2} \]So,\[ \mathbf{u} = \left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right) \]Thus,\[ D_{\mathbf{u}} \rho = \left( -\frac{3}{64}, -\frac{2}{64} \right) \cdot \left( -\frac{1}{2}, \frac{\sqrt{3}}{2} \right) = \frac{3}{128} - \frac{2\sqrt{3}}{128} = \frac{3 - 2\sqrt{3}}{128} \]
06

Determine the direction and magnitude of the greatest rate of change

The direction of the greatest rate of change is in the direction of the gradient vector. The magnitude of the greatest rate of change is the magnitude of the gradient vector. Calculate the magnitude of the gradient vector:\[ ||abla \rho|| = \sqrt{\left( \frac{3}{64} \right)^2 + \left( \frac{2}{64} \right)^2} = \sqrt{\frac{9}{4096} + \frac{4}{4096}} = \sqrt{\frac{13}{4096}} = \frac{\sqrt{13}}{64} \]

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Key Concepts

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

Gradient Vector
The gradient vector, denoted as 鈭囅�, represents the direction and rate of the fastest increase of a function. For a function 蟻(x, y), the gradient vector consists of its partial derivatives with respect to x and y. So, if we have 蟻=1/鈭�(x虏 + y虏 + 3), we need to compute the partial derivatives:

鈭傁�/鈭倄 and 鈭傁�/鈭倅.

In our example, we calculate

鈭傁�/鈭倄=-x/(x虏 + y虏 + 3)^{3/2}

and

鈭傁�/鈭倅=-y/(x虏 + y虏 + 3)^{3/2}.

Plugging in the point (3, 2), we found that

鈭囅� = (-3/64, -2/64).

The gradient vector points in the direction where 蟻 increases the fastest and its magnitude reveals how steeply it increases.
Directional Derivative
The directional derivative gives the rate of change of a function in any specified direction. It's denoted by D_u蟻 and is calculated by taking the dot product of the gradient vector 鈭囅� with a unit vector u that shows the direction.

For the given problem, we were asked to find the rate of change at point (3, 2) in the direction of the unit vector given by cos(2蟺/3)i + sin(2蟺/3)j.

The unit vector u can be written as (-1/2, 鈭�3/2).

Here, we computed D_u蟻 = 鈭囅� 路 u = (-3/64, -2/64) 路 (-1/2, 鈭�3/2).

This calculation showed us the rate of change in the specific direction, which turned out to be (3 - 2鈭�3) / 128.
Rate of Change
The rate of change in the context of multivariable calculus usually measures how a function changes as we move from one point to another.

The greatest rate of change in any function happens in the direction of the gradient vector. In our problem, we were asked to find the direction and magnitude of the greatest rate of change of density 蟻 at the point (3, 2).

We determined that the direction of maximal increase is aligned with the gradient vector 鈭囅�,

(-3/64, -2/64).

The magnitude of this greatest rate of change is simply the magnitude of the gradient vector, which can be calculated using the Euclidean norm:

||鈭囅亅| = 鈭�( (3/64)虏 + (2/64)虏 ) = 鈭�(13/4096) = 鈭�13 / 64.

This value denotes how rapidly the density changes in the direction of the gradient.

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