Problem 62 - Calculus Early Transcendentals

Question

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?

Step-by-Step Solution

Answer

The rate of change of the density as a function of the parameter \(t\) is given by: $$ \frac{d\rho}{dt} = -4\sin^2 t + 4\cos^2 t $$ b. What are the points on the edge of the plate where the density is maximum? There are two points on the edge of the plate where the density is maximum: $$ P_{1}(x_{1}, y_{1}) = (\sqrt{2}, \sqrt{2}) $$ $$ P_{2}(x_{2}, y_{2}) = (-\sqrt{2}, -\sqrt{2}) $$

Step 1: Find the partial derivatives of the density function with respect to x and y

To find the rate of change of density with respect to \(t\), we need to evaluate the partial derivatives with respect to \(x\) and \(y\). $$ \frac{\partial\rho}{\partial x} = \frac{\partial(4+xy)}{\partial x} = y $$ $$ \frac{\partial\rho}{\partial y} = \frac{\partial(4+xy)}{\partial y} = x $$

Step 2: Compute the rate of change of x and y with respect to t

Since the edge of the plate is described by the parametric equations \(x=2 \cos t\) and \(y=2 \sin t\), we need to find the rate of change of \(x\) and \(y\) with respect to \(t\). $$ \frac{dx}{dt} = \frac{d(2\cos t)}{dt} = -2\sin t $$ $$ \frac{dy}{dt} = \frac{d(2\sin t)}{dt} = 2\cos t $$

Step 3: Compute the total rate of change of density with respect to t

Next, we can find the total rate of change of density with respect to \(t\) using the chain rule. $$ \frac{d\rho}{dt} = \frac{\partial\rho}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial\rho}{\partial y} \cdot \frac{dy}{dt} = y(-2\sin t) + x(2\cos t) $$ Substitute the parametric equations of the edge \(x=2\cos t, y=2\sin t\) back in: $$ \frac{d\rho}{dt} = (2\sin t)(-2\sin t) + (2\cos t)(2\cos t) = -4\sin^2 t + 4\cos^2 t $$

Step 4: Determine the point(s) where the density is maximum

To find the maximum density, we need to find the points where the rate of change of density is equal to 0. So, we will set \(d\rho/dt = 0\) and get: $$ 0 = -4\sin^2 t + 4\cos^2 t $$ Divide both sides by 4: $$ 0 = \cos^2 t - \sin^2 t $$ Notice that \(\cos^2 t - \sin^2 t = \cos(2t)\), so: $$ 0 = \cos(2t) $$ This equation is satisfied when \(2t = (2n+1)\pi/2\), where \(n \in \mathbb{Z}\). Considering \(0 \leq t \leq 2\pi\), we obtain two values of \(t\): \(t_1 = \pi/4\) and \(t_2 = 5\pi/4\). Now, we calculate the coordinates of these points using the parametric equations for the edge of the plate: $$ P_{1}(x_{1}, y_{1}) = (2\cos\frac{\pi}{4}, 2\sin\frac{\pi}{4}) = (\sqrt{2}, \sqrt{2}) $$ $$ P_{2}(x_{2}, y_{2}) = (2\cos\frac{5\pi}{4}, 2\sin\frac{5\pi}{4}) = (-\sqrt{2}, -\sqrt{2}) $$ These are the two points where the density is highest on the edge of the plate.

Key Concepts

Density Function

In calculus, a density function refers to a mathematical expression that describes how density is distributed across a surface or within a volume. Let's explore the concept with an example.
In our exercise, the density function is given by \( \rho(x, y) = 4 + xy \). This specifies how the density varies at any point \((x, y)\) on the plate.
Density functions are crucial in many real-world applications, such as physics for calculating material distributions and probabilities in statistics. A good understanding of density functions involves evaluating how they interact with other variables and change across different areas or volumes.

Parametric Equations

Parametric equations are equations that express coordinates \((x, y)\) as functions of a parameter, often denoted by \( t \). This representation is beneficial for describing curves or boundaries that are not easily expressed in standard \( y = f(x) \) form.
In the context of our example, the circular edge of the plate is parameterized by:

• \( x = 2 \cos t \)
• \( y = 2 \sin t \)

For \( 0 \leq t \leq 2 \pi \), these equations trace out the edge of the plate. The benefit of using parametric equations is the ease with which they can model complex curves in a straightforward manner by simply varying \( t \). In calculus problems, particularly those involving rates of change, parametric equations are commonly used for their clarity and precision.

Partial Derivatives

Partial derivatives are an essential concept in calculus when dealing with functions of multiple variables. They represent the rate of change of a function with respect to one variable while keeping the others constant.
In this exercise, the density function \( \rho(x, y) = 4 + xy \) is a two-variable function. The partial derivatives are:

• \( \frac{\partial \rho}{\partial x} = y \)
• \( \frac{\partial \rho}{\partial y} = x \)

These derivatives help in calculating how the density changes with slight changes in \( x \) and \( y \) while treating each separately. Understanding and computing partial derivatives allow us to explore how small variations in inputs affect the overall behavior of the function.

Chain Rule

The chain rule is a fundamental tool in calculus, used for finding the derivative of a composite function. It allows us to differentiate a function concerning one variable by considering its dependency on intermediate variables.
In this example, the chain rule helps us find the rate of change of density \( \rho \) with respect to the parameter \( t \). We have:
\[\frac{d\rho}{dt} = \frac{\partial\rho}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial\rho}{\partial y} \cdot \frac{dy}{dt}\]This formula combines the partial derivatives and the derivatives of the parametric equations with respect to \( t \).

• \( \frac{dx}{dt} = -2 \sin t \)
• \( \frac{dy}{dt} = 2 \cos t \)

Substitute these into the formula, and we can evaluate the total rate of change of the density along the edge of the plate. The chain rule is used extensively in calculus and provides a powerful way to deal with functions dependent on multiple interrelated variables.