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The density of a thin circular plate of radius 2 is given by \(\rho(x, y)=4+x y .\) The edge of the plate is described by the parametric equations \(x=2 \cos t, y=2 \sin t\) for \(0 \leq t \leq 2 \pi\) a. Find the rate of change of the density with respect to \(t\) on the edge of the plate. b. At what point(s) on the edge of the plate is the density a maximum?

Short Answer

Expert verified
a. The rate of change of the density with respect to t on the edge of the plate is given by the derivative of the density function with respect to t, which is: $$\frac{d\rho}{dt} = 4\left(\cos^2{t} - \sin^2{t}\right).$$ b. The points on the edge of the plate where the density is maximum are: \((\sqrt{2},\sqrt{2}),\ (-\sqrt{2},\sqrt{2}),\ (-\sqrt{2},-\sqrt{2}),\ \text{and}\ (\sqrt{2},-\sqrt{2})\).

Step by step solution

01

Write the Density Function on the Edge of the Circle in Terms of t

To find the density on the edge of the plate, we need to plug the parametric equations \(x = 2\cos{t}\) and \(y = 2\sin{t}\) into the given density function \(\rho(x, y) = 4 + xy\). Let \(\rho(t) = 4 + (2\cos{t})(2\sin{t})\). Simplifying this expression, we get: $$\rho(t) = 4 + 4\cos{t}\sin{t}.$$
02

Find the Derivative of the Density Function with respect to t

Now, we will differentiate the function \(\rho(t)\) with respect to \(t\) to find the rate of change of density on the edge of the plate. $$\frac{d\rho}{dt} = \frac{d}{dt}(4 + 4\cos{t}\sin{t}).$$ Using the product and chain rule, we have: $$\frac{d\rho}{dt} = 4\left(-\sin{t}\sin{t} + \cos{t}\cos{t}\right) = 4\left(\cos^2{t} - \sin^2{t}\right).$$
03

Find the Critical Points of the Density Function

To find the points where the density is maximum on the edge of the plate, we need to find the critical points of the derivative of the density function with respect to \(t\). To do this, we will set \(\frac{d\rho}{dt} = 0\) and solve for \(t\). $$4\left(\cos^2{t} - \sin^2{t}\right) = 0$$ Then, $$\cos^2{t} = \sin^2{t}.$$ We know that \(\cos^2{t} + \sin^2{t} = 1\). So, $$2\sin^2{t} = 1$$ Solving for \(\sin{t}\), we get: $$\sin{t} = \pm\frac{1}{\sqrt{2}}.$$ The corresponding values of \(t\) are \(t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \text{and}\ \frac{7\pi}{4}\).
04

Find the Maximum Density Points on the Edge of the Plate

Finally, we will find the maximum density points on the edge of the plate by plugging the critical points found in Step 3 back into the parametric equations. For \(t = \frac{\pi}{4}\), $$x = 2\cos{\frac{\pi}{4}} = \sqrt{2}, \quad y = 2\sin{\frac{\pi}{4}} = \sqrt{2}.$$ For \(t = \frac{3\pi}{4}\), $$x = 2\cos{\frac{3\pi}{4}} = -\sqrt{2}, \quad y = 2\sin{\frac{3\pi}{4}} = \sqrt{2}.$$ For \(t = \frac{5\pi}{4}\), $$x = 2\cos{\frac{5\pi}{4}} = -\sqrt{2}, \quad y = 2\sin{\frac{5\pi}{4}} = -\sqrt{2}.$$ For \(t = \frac{7\pi}{4}\), $$x = 2\cos{\frac{7\pi}{4}} = \sqrt{2}, \quad y = 2\sin{\frac{7\pi}{4}} = -\sqrt{2}.$$ So the points on the edge of the plate where the density is maximum are \((\sqrt{2},\sqrt{2}),\ (-\sqrt{2},\sqrt{2}),\ (-\sqrt{2},-\sqrt{2}),\ \text{and}\ (\sqrt{2},-\sqrt{2})\).

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Key Concepts

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

Parametric Equations
Parametric equations are a powerful tool in calculus, used to describe the coordinates of points along a curve using one or more parameters. Unlike traditional functions which provide y as a function of x, parametric equations define both x and y in terms of another variable, often denoted as t. This is particularly useful when dealing with complex curves, such as circles and ellipses, which are not functions in the conventional sense because they fail the vertical line test.

For example, in our exercise, the edge of the circular plate is characterized by the parametric equations \(x=2 \cos t, y=2 \sin t\) for \(0 \leq t \leq 2 \pi\). The parameter t here represents the angle in radians, and as it varies, the equations provide the x and y coordinates of a point on the circle's edge. This representation is especially helpful when we want to calculate properties, such as density, along the curve.
Rate of Change
The rate of change is a central concept in calculus, referring to how a quantity changes with respect to changes in another quantity. It is the mathematical expression of a concept that can be colloquially understood as 'speed' or 'velocity', but it applies to a wide variety of situations beyond physical movement. In calculus, we use derivatives to quantify the rate of change.

In the context of our exercise, the rate of change represents how the density of the plate changes as we move along the edge of the circle. After expressing the density function \(\rho(t)\) in terms of the parameter t, the derivative \(\frac{d\rho}{dt}\) provides the rate of change of density with respect to t. By finding the derivative, we gain insight into how the density varies as we traverse the edge, which is essential for determining where the density reaches a maximum.
Maximum Density Points
Identifying the maximum density points on a curve involves locating the points at which the density has its highest value. These are crucial for understanding the distribution of mass or other quantities across a given object, like our circular plate. To find these points, one typically calculates the derivative of the density function and then identifies the values where this derivative is zero, since maxima (and minima) occur at critical points where the rate of change is zero.

In our case, we differentiate the density function with respect to the parameter t to determine where the rate of change equals zero. The critical points identified provide the angles at which the density is at its maximum on the edge of the plate. By substituting these back into the parametric equations, we pinpoint the exact coordinates on the plate where the maximum density can be found.
Critical Points
Critical points are locations on a graph where the function's derivative is zero or undefined; these points are potential candidates for local maxima, minima, or points of inflection. In other words, they are essential in determining where a function changes its behavior or trend. To find critical points, one typically sets the derivative equal to zero and solves for the variable in question.

In our exercise, we looked for the critical points of the density function's derivative, \(\frac{d\rho}{dt}\), to discover where the function's rate of change is zero. This process revealed specific values of t at which the critical points occur. Understanding critical points is pivotal for optimization problems in calculus, such as maximizing the density of a material or minimizing the cost of production in real-world scenarios.

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Most popular questions from this chapter

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