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The temperature at a point \((x, y)\) on a metal plate is \(T(x, y)=4 x^{2}-4 x y+y^{2} .\) An ant on the plate walks around the circle of radius 5 centered at the origin. What are the highest and lowest temperatures encountered by the ant?

Short Answer

Expert verified
The highest temperature is 100 and the lowest is 0.

Step by step solution

01

Identification of Function and Constraints

Identify that the temperature function is \(T(x, y) = 4x^2 - 4xy + y^2\) and that the ant is constrained to move along a circle described by \(x^2 + y^2 = 25\).
02

Setup Lagrange Multiplier Method

We will use the method of Lagrange multipliers to find the extreme temperatures. Define the Lagrangian as \(\mathcal{L}(x, y, \lambda) = 4x^2 - 4xy + y^2 + \lambda(x^2 + y^2 - 25)\).
03

Compute Partial Derivatives

Take the partial derivatives: \(\frac{\partial \mathcal{L}}{\partial x} = 8x - 4y + 2\lambda x\), \(\frac{\partial \mathcal{L}}{\partial y} = -4x + 2y + 2\lambda y\), and \(\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 25\). Set these partial derivatives to zero for the system of equations.
04

Solve System of Equations

From \(\frac{\partial \mathcal{L}}{\partial x} = 0\) : \(8x - 4y + 2\lambda x = 0\), simplify to \(4x(2+\lambda) = 4y\). From \(\frac{\partial \mathcal{L}}{\partial y} = 0\) : \(-4x + 2y + 2\lambda y = 0\), simplify to \(-4x = y(2+\lambda)\). Solve these equations along with \(x^2 + y^2 = 25\) to find the critical points.
05

Calculate Potential Extreme Points

Using the simplified equations \(y = x(2+\lambda)\), test critical points found by substitution into the circle equation \(x^2 + y^2 = 25\). Solve the resulting quadratic equations.
06

Evaluate Temperature at Critical Points

Plug each potential \((x, y)\) point into the temperature function \(T(x, y) = 4x^2 - 4xy + y^2\) to find corresponding temperatures. Compare these values to determine highest and lowest temperatures.

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Key Concepts

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

Temperature Distribution
In the analysis of the temperature distribution across a metal plate, we have a specified temperature formula, namely, \(T(x, y) = 4x^2 - 4xy + y^2\). This function shows how temperature varies depending on the point \(x, y\) on the plate. As the formula suggests, each point on the plate has a specific temperature determined by its \(x\ ext{ and }y\) coordinates. Understanding this temperature distribution is crucial for many applications, such as material science and thermodynamics.
In practice, by analyzing the function, we notice it is a quadratic form, indicating that the temperature changes consistently across the plane, forming parabolic contours. This can signal hot and cold regions on the plate. Also, transformations involving coordinates could help visualize how temperature patterns shift, which is essential in practical situations where temperature control is necessary, such as cooking or metallurgy.
In this context, an ant walking on a circular path—defined by the equation \(x^2 + y^2 = 25\)—can be used to explore the range of temperatures experienced across the circle. This means that we are interested in discovering how the changes in x and y, constrained by the circle, influence the observed temperatures.
Extreme Value
The concept of extreme values revolves around finding the highest and lowest points within a specific range. In our exercise, using the given temperature function, we want to determine what temperatures an ant might encounter at various points around a predefined circle. This is known as finding the local (and potentially global) extrema of the function subject to a constraint.
To solve this problem, we employ the method of Lagrange multipliers. It's a powerful technique used to find the local extrema of functions subject to constraints. Here, our constraint is the circle \(x^2 + y^2 = 25\). By combining both the temperature function and the constraint, we establish a Lagrangian function, where we introduce an additional variable, \(\lambda\).
  • The goal is to differentiate the Lagrangian with respect to each variable, which provides a system of equations.
  • Solving these equations gives us potential critical points, the candidates where these extreme temperatures can occur.
Further analysis of these points informs us about the highest and lowest temperatures experienced by the ant, revealing the range of conditions across the metal plate.
Constraint Optimization
Constraint optimization involves adjusting one or more variables to achieve the best outcome, considering certain limitations. In the textbook exercise, we are optimizing the temperature function \(T(x, y)\) with respect to the constraint that defines a circle. The Lagrange multiplier method serves precisely this purpose and simplifies finding maxima and minima under specific conditions.
The essence of constraint optimization here is to balance the original function with the constraint effectively. Setting up the Lagrangian:\[\mathcal{L}(x, y, \lambda) = 4x^2 - 4xy + y^2 + \lambda(x^2 + y^2 - 25)\]
we incorporate the circle constraint into the temperature function through the factor \(\lambda\).
  • By differentiating \(\mathcal{L}\) with respect to \(x\), \(y\), and \(\lambda\), we derive conditions that must be satisfied for optimal solutions.
  • This process transforms the problem of finding boundless extrema into a solvable set of equations that form boundaries along the circle.
Through this optimization, we discover optimal x and y values where extreme temperatures occur, illustrating how mathematical tools can be used to solve real-world engineering and scientific challenges.

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Most popular questions from this chapter

Find the maximum value that \(f(x, y, z)=x^{2}+2 y-z^{2}\) can have on the line of intersection of the planes \(2 x-y=0\) and \(y+z=0.\)

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right)\) then$$\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L$$ Use this result to support your answers to the questions. Does knowing that \(|\cos (1 / y)| \leq 1\) tell you anything about $$\lim _{(x, y) \rightarrow(0,0)} x \cos \frac{1}{y} ?$$ Give reasons for your answer.

Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I\), you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$\begin{array}{l} x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2 \\ 0 \leq v \leq 2 \pi \end{array}$$

Can you conclude anything about \(f(a, b)\) if \(f\) and its first and second partial derivatives are continuous throughout a disk centered at the critical point \((a, b)\) and \(f_{x x}(a, b)\) and \(f_{y y}(a, b)\) differ in sign? Give reasons for your answer.

If you cannot make any headway with \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) in rectangular coordinates, try changing to polar coordinates. Substitute \(x=r \cos \theta, y=r \sin \theta,\) and investigate the limit of the resulting expression as \(r \rightarrow 0 .\) In other words, try to decide whether there exists a number \(L\) satisfying the following criterion: $$|r|<\delta \Rightarrow|f(r, \theta)-L|<\epsilon$$ If such an \(L\) exists, then $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=\lim _{r \rightarrow 0} f(r \cos \theta, r \sin \theta)=L$$ For instance, $$\lim _{(x, y) \rightarrow(0,0)} \frac{x^{3}}{x^{2}+y^{2}}=\lim _{r \rightarrow 0} \frac{r^{3} \cos ^{3} \theta}{r^{2}}=\lim _{r \rightarrow 0} r \cos ^{3} \theta=0$$ To verify the last of these equalities, we need to show that Equation (1) is satisfied with \(f(r, \theta)=r \cos ^{3} \theta\) and \(L=0 .\) That is, we need to show that given any \(\epsilon>0,\) there exists a \(\delta>0\) such that for all \(r\) and \(\theta\) $$|r|<\delta \Rightarrow\left|r \cos ^{3} \theta-0\right|<\epsilon$$ since $$\left|r \cos ^{3} \theta\right|=|r|\left|\cos ^{3} \theta\right| \leq|r| \cdot 1=|r|$$ the implication holds for all \(r\) and \(\theta\) if we take \(\delta=\epsilon\) In contrast, $$\frac{x^{2}}{x^{2}+y^{2}}=\frac{r^{2} \cos ^{2} \theta}{r^{2}}=\cos ^{2} \theta$$takes on all values from 0 to 1 regardless of how small \(|r|\) is, so that \(\lim _{(x, y) \rightarrow(0,0)} x^{2} /\left(x^{2}+y^{2}\right)\) does not exist. In each of these instances, the existence or nonexistence of the limit as \(r \rightarrow 0\) is fairly clear. Shifting to polar coordinates does not always help, however, and may even tempt us to false conclusions. For example, the limit may exist along every straight line (or ray) \(\theta=\) constant and yet fail to exist in the broader sense. Example 5 illustrates this point. In polar coordinates, \(f(x, y)=\left(2 x^{2} y\right) /\left(x^{4}+y^{2}\right)\) becomes $$f(r \cos \theta, r \sin \theta)=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\sin ^{2} \theta}$$for \(r \neq 0 .\) If we hold \(\theta\) constant and let \(r \rightarrow 0,\) the limit is \(0 .\) On the path \(y=x^{2},\) however, we have \(r \sin \theta=r^{2} \cos ^{2} \theta\) and $$\begin{aligned} f(r \cos \theta, r \sin \theta) &=\frac{r \cos \theta \sin 2 \theta}{r^{2} \cos ^{4} \theta+\left(r \cos ^{2} \theta\right)^{2}} \\ &=\frac{2 r \cos ^{2} \theta \sin \theta}{2 r^{2} \cos ^{4} \theta}=\frac{r \sin \theta}{r^{2} \cos ^{2} \theta}=1 \end{aligned}$$ Find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$f(x, y)=\tan ^{-1}\left(\frac{|x|+|y|}{x^{2}+y^{2}}\right)$$

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