/*! 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 35 The temperature at any point \((... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 35

The temperature at any point \((x, y)\) of a flat plate is \(T\) degrees and \(T=54-\frac{2}{3} x^{2}-4 y^{2} .\) If distance is measured in feet, find the rate of change of the temperature with respect to the distance moved along the plate in the directions of the positive \(x\) and \(y\) axes, respectively, at the point \((3,1)\).

Short Answer

Expert verified
The rate of change of temperature at \( (3,1) \) is -4 in the \(x\) direction and -8 in the \(y\) direction.

Step by step solution

01

Understand the Given Formula

The temperature at any point \(x, y\) on the plate is expressed as \(T = 54 - \frac{2}{3} x^2 - 4 y^2\). Our goal is to find the rate of change of temperature with respect to distance moved in the directions of the positive \(x\) and \(y\) axes at the point \( (3,1) \).
02

Find the Partial Derivative with Respect to x

To find the rate of change of temperature in the positive \(x\) direction, calculate the partial derivative of \(T\) with respect to \(x\).\[ \frac{\text{鈭倉T}{\text{鈭倉x} = \frac{\text{鈭倉}{\text{鈭倉x} \bigg(54 - \frac{2}{3}x^2 - 4y^2 \bigg) = -\frac{4}{3} x \]
03

Evaluate the Partial Derivative with Respect to x at (3,1)

Substitute \(x = 3\) into the partial derivative we found.\[ \frac{\text{鈭倉T}{\text{鈭倉x}\bigg|_{(3,1)} = -\frac{4}{3} (3) = -4 \]
04

Find the Partial Derivative with Respect to y

To find the rate of change of temperature in the positive \(y\) direction, calculate the partial derivative of \(T\) with respect to \(y\).\[ \frac{\text{鈭倉T}{\text{鈭倉y} = \frac{\text{鈭倉}{\text{鈭倉y} \bigg(54 - \frac{2}{3}x^2 - 4y^2 \bigg) = -8y \]
05

Evaluate the Partial Derivative with Respect to y at (3,1)

Substitute \(y = 1\) into the partial derivative we found.\[ \frac{\text{鈭倉T}{\text{鈭倉y}\bigg|_{(3,1)} = -8 (1) = -8 \]

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Key Concepts

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

rate of change
In mathematics, the rate of change measures how a quantity varies with respect to another quantity. When dealing with temperature on a flat plate, we might want to know how quickly the temperature changes as we move in certain directions. For example, moving along the positive x or y direction. By computing the partial derivatives of the temperature function, we capture these rates of change.
temperature distribution
In this exercise, the temperature distribution on a flat plate is given by a quadratic equation: \[ T = 54 - \frac{2}{3} x^{2} - 4 y^{2} \]. This equation tells us the temperature at any point \(x, y\) on the plate.

The constants and terms of the equation represent how the temperature varies across the plate. For instance, the term \( - \frac{2}{3} x^2 \) decreases temperature as \( x \) increases, while the term \( - 4 y^2 \) decreases temperature as \( y \) increases.
analytic geometry
Analytic geometry helps us understand geometric shapes using algebra. In this context, it helps us explore the temperature distribution across the plate. By plotting points and curves described by the temperature function, we can see how temperature changes over the plate鈥檚 surface.

For example, evaluating the function at multiple points and graphing the results helps us visualize temperature gradients, which are regions where temperature changes rapidly.
partial derivative evaluation
Partial derivatives are essential for finding how a function changes with one variable while keeping others constant. For the given temperature function, we compute partial derivatives with respect to \( x \) and \( y \).

Here鈥檚 how:
  • Find the partial derivative of \( T \) with respect to \( x \):
    \[ \frac{\partial T}{\partial x} = - \frac{4}{3} x \]

  • Evaluate this derivative at the point \( (3,1) \):
    \[ \left. \frac{\partial T}{\partial x} \right|_{(3,1)} = -4 \]

  • Find the partial derivative of \( T \) with respect to \( y \):
    \[ \frac{\partial T}{\partial y} = -8 y \]

  • Evaluate this derivative at the point \( (3,1) \):
    \[ \left. \frac{\partial T}{\partial y} \right|_{(3,1)} = -8 \]
These steps reveal the temperature changes at \( (3,1) \):
- Changing \( x \) reduces the temperature by 4 degrees per foot.
- Changing \( y \) reduces the temperature by 8 degrees per foot.

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Most popular questions from this chapter

In Exercises 8 through 17, determine the region of continuity of \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the region of continuity of \(f\). $$ f(x, y)= \begin{cases}\frac{\sin (x+y)}{x+y} & \text { if } x+y \neq 0 \\ 1 & \text { if } x+y=0\end{cases} $$

In Exercises 15 through 18, show that \(u(x, y)\) satisfies the equation $$ \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0 $$ which is known as Laplace's equation in \(R^{2}\). $$ u(x, y)=\ln \left(x^{2}+y^{2}\right) $$

In Exercises 1 through 4 , find the indicated partial derivative by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) before differentiating. $$ u=x^{2}-y^{2} ; x=3 r-s ; y=r+2 s ; \frac{\partial u}{\partial r} ; \frac{\partial u}{\partial s} $$

A company has contracted to manufacture 10,000 closed wooden crates having dimensions \(3 \mathrm{ft}, 4 \mathrm{ft}\), and \(5 \mathrm{ft}\). The cost of the wood to be used is \(5 \notin\) per square foot. If the machines that are used to cut the pieces of wood have a possible error of \(0.05 \mathrm{ft}\) in each dimension, find approximately, by using the total differential, the greatest possible error in the estimate of the cost of the wood.

In Exercises 30 through 33 , we show that a function may be differentiable at a point even though it is not continuously differentiable there. Hence, the conditions of Theorem \(19.5 .5\) are sufficient but not necessary for differentiability. The function \(f\) in these exercises is defined by In Exercises 30 through 33 , we show that a function may be differentiable at a point even though it is not continuously differentiable there. Hence, the conditions of Theorem \(19.5 .5\) are sufficient but not necessary for differentiability. The function \(f\) in these exercises is defined by $$ f(x, y)= \begin{cases}\left(x^{2}+y^{2}\right) \sin \frac{1}{\sqrt{x^{2}+y^{2}}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases} $$ $$ \text { Find } \Delta f(0,0) \text {. } $$
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