/*! 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 53 To find the extreme values of a ... [FREE SOLUTION] | 91Ó°ÊÓ

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To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=x+y\) c. \(h(x, y)=2 x^{2}+y^{2}\) Curves: i. The semicircle \(x^{2}+y^{2}=4, \quad y \geq 0\) ii. The quarter circle \(x^{2}+y^{2}=4, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=2 \cos t, y=2 \sin t\)

Short Answer

Expert verified
For \(f\): Max is \(2\sqrt{2}\), min is \(-2\) on semicircle; max is \(2\sqrt{2}\), min is \(2\) on quarter circle. For \(h\): Max is \(8\), min is \(4\) on both curves.

Step by step solution

01

Parametrize the Function

For the semicircle and quarter circle, start by parametrizing the given circle equation using the parametrization: \[ x = 2 \cos(t), \quad y = 2 \sin(t) \]where the parameter domain for the semicircle is \(t \in [0, \pi]\) and for the quarter circle is \(t \in [0, \frac{\pi}{2}]\).
02

Evaluate Function a on the Semicircle

For \(f(x, y) = x + y\), substitute the parametrizations into the function:\[ f(t) = 2 \cos(t) + 2 \sin(t) \]This simplifies to:\[ f(t) = 2(\cos(t) + \sin(t)) \]
03

Find Critical Points for Function a on the Semicircle

To find the critical points, differentiate \(f(t) = 2(\cos(t) + \sin(t))\):\[ f'(t) = 2(-\sin(t) + \cos(t)) \]Set \(f'(t) = 0\) to find critical points:\[ -\sin(t) + \cos(t) = 0 \]This simplifies to:\[ \tan(t) = 1 \]Thus, \(t = \frac{\pi}{4}\) falls within the interval \([0, \pi]\).
04

Evaluate Function a at Critical and Endpoints

Calculate \(f(t)\) at critical point \(t = \frac{\pi}{4}\) and endpoints \(t = 0, \pi\):- \( f(0) = 2 \cos(0) + 2 \sin(0) = 2 \)- \( f(\pi) = 2 \cos(\pi) + 2 \sin(\pi) = -2 \)- \( f\left(\frac{\pi}{4}\right) = 2\left(\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right)\right) = 2\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) = 2\sqrt{2} \)
05

Find Absolute Max/Min for Function a on the Semicircle

By comparing the values from Step 4, the absolute maximum is \(2\sqrt{2}\) and the absolute minimum is \(-2\).
06

Evaluate Function a on the Quarter Circle

For \(t \in [0, \frac{\pi}{2}]\), evaluate \(f(t) = 2(\cos(t) + \sin(t))\) as in Step 2. Reuse the calculation:- Check endpoints again: \( f(0) = 2 \), \( f\left(\frac{\pi}{2}\right) = 2 \). Since \(\frac{\pi}{4}\) is within this range, check it too:- Already calculated in Step 4: \( f\left(\frac{\pi}{4}\right) = 2\sqrt{2}\).
07

Find Absolute Max/Min for Function a on the Quarter Circle

For \(t \in [0, \frac{\pi}{2}]\): - Since \(2\sqrt{2} > 2\), the absolute maximum is still \(2\sqrt{2}\), while the absolute minimum is \(2\) as it does not reach \(-2\) here.
08

Evaluate Function c on the Semicircle

For \(h(x, y) = 2x^2 + y^2\), substitute the parametrizations:\[ h(t) = 2(2\cos(t))^2 + (2\sin(t))^2 \]Which simplifies to:\[ h(t) = 8\cos^2(t) + 4\sin^2(t) \]
09

Find Critical Points for Function c on the Semicircle

Differentiate \(h(t) = 8\cos^2(t) + 4\sin^2(t)\):\[ h'(t) = 16\cos(t)(-\sin(t)) + 8\sin(t)\cos(t) \]Simplify to:\[ h'(t) = -16\cos(t)\sin(t) + 8\sin(t)\cos(t) = -8\cos(t)\sin(t) \]Set \(h'(t) = 0\):\[ -8\cos(t)\sin(t) = 0 \]Thus, \(\cos(t) = 0 \) or \(\sin(t) = 0\); solutions \(t = \frac{\pi}{2}\) (since \(\cos\left(\frac{\pi}{2}\right) = 0\)) and \(t = 0, \pi\) (since \(\sin(0)=\sin(\pi)=0\)).
10

Evaluate Function c at Critical and Endpoints

Calculate \(h(t)\) at critical points \(t=0, \frac{\pi}{2}, \pi\):- \( h(0) = 8(\cos^2(0)) + 4(\sin^2(0)) = 8 \)- \( h\left(\frac{\pi}{2}\right) = 8(\cos^2(\frac{\pi}{2})) + 4(\sin^2(\frac{\pi}{2})) = 4 \)- \( h(\pi) = 8(\cos^2(\pi)) + 4(\sin^2(\pi)) = 8 \)
11

Find Absolute Max/Min for Function c on the Semicircle

By comparing these values, the absolute maximum is \(8\) and the absolute minimum is \(4\).
12

Evaluate Function c on the Quarter Circle

Check the same derivations for the range \(t \in [0, \frac{\pi}{2}]\):- Values need only be checked at this interval's endpoints and the critical point \(t = \frac{\pi}{2}\), which is: - \( h(0) = 8 \) and \( h\left(\frac{\pi}{2}\right) = 4 \).
13

Find Absolute Max/Min for Function c on the Quarter Circle

Thus for the quarter-circle, absolute maximum remains \(8\) and absolute minimum is \(4\).

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Key Concepts

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

Parametric Equations
In mathematics, parametric equations are a way to define the coordinates of points on a curve using one or more parameters. Let's consider the semicircle and quarter circle with the equation \(x^2 + y^2 = 4\). By using the parametric equations \(x = 2 \cos(t)\) and \(y = 2 \sin(t)\), each point on these curves can be represented as the parameter \(t\) changes.
- For the semicircle, the parameter \(t\) ranges from \([0, \pi]\), allowing us to trace points only where \(y \ge 0\).
- For the quarter circle, \(t\) takes values from \([0, \pi/2]\), ensuring points are where both \(x \ge 0\) and \(y \ge 0\).This parametrization is particularly useful as it simplifies complex curves into a single variable function, making computations like finding derivatives and critical points more manageable.
Chain Rule
The Chain Rule is a fundamental calculus concept, especially useful when dealing with functions of several variables through a parameter. If a function \(f(x, y)\) is defined along a curve parameterized by \(t\), then \(f\) becomes a function \(f(t)\). The derivative of \(f\) with respect to \(t\) can be expressed using the Chain Rule:\[\frac{d f}{d t} = \frac{\partial f}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial f}{\partial y} \cdot \frac{d y}{d t}\]This enables us to find points where the derivative is zero, indicating potential maxima or minima.
- For example, the function \(f(x, y) = x + y\) when parametrized becomes \(f(t) = 2(\cos(t) + \sin(t))\), allowing us to take \(\frac{d f}{d t}\) and analyze where it equals zero, thus identifying critical points on the curve.
Using the Chain Rule simplifies tracing how derivations change along parameterized paths.
Absolute Maximum and Minimum
Finding the absolute maximum and minimum values of a function on a given curve involves looking at critical points, where the derivative \(df/dt = 0\), and endpoints of the domain. In the context of our semicircle and quarter circle curves, these are important to determine specific highest or lowest values that a function reaches.
- For function \(f(x, y) = x + y\), the critical point \(t = \pi/4\) gives a maximum, whereas endpoints like \(t = 0\) or \(t = \pi\) provide bounded values of the function.
- For function \(h(x, y) = 2x^2 + y^2\), analyzing endpoints and critical points within the domain finds that the function's maximum and minimum values occur similarly.
Identifying these values is essential in understanding the behavior of a function constrained within a particular domain, especially when it comes to applications in engineering and physics.
Semicircle and Quarter Circle
Semicircles and quarter circles are specific formations of the circle equation \(x^2 + y^2 = 4\). These segments are frequently analyzed because they represent practical geometric shapes found in many real-world contexts.- The **semicircle** is defined within \(y \ge 0\) by the parameter range \(t \in [0, \pi]\). This section includes points from the top half of a circle, forming a clear boundary from one side to another.- The **quarter circle** confines \(x \, \text{and} \, y\) to both non-negative quadrants using \(t \in [0, \pi/2]\), resulting in one-fourth of the circle's full area, situated in the first quadrant.These constraints are paramount in solving optimization problems where only a portion of a complete circle is considered. Analyzing functions over these segmented geometries involves understanding that their symmetries and boundaries limit function behavior specifically therein.

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