Problem 42 - Calculus: Early Transcendentals

Question

A sphere with radius 1 m has temperature \( 15^oC. \) It lies inside a concentric sphere with radius 2 m and temperature \( 25^oC. \) The temperature \( T(r) \) at a distance \( r \) from the common center of the spheres satisfies the differential equation \( \frac {d^2T}{dr^2} + \frac {2}{r}\frac {dT}{dr} = 0 \) If we let \( S = dT/dr, \) then \( S \) satisfies a first-order differential equation. Solve it to find an expression for the temperature \( T(r) \) between the spheres.

Step-by-Step Solution

Answer

The temperature expression is \( T(r) = \frac{10}{r} + 5 \).

Step 1: Substitute and Recast the Equation

Given that \( S = \frac{dT}{dr} \), substitute \( S \) into the differential equation: \( \frac {d^2T}{dr^2} + \frac {2}{r}\frac {dT}{dr} = 0 \). This transforms into \( \frac {dS}{dr} + \frac {2}{r}S = 0 \), a first-order differential equation.

Step 2: Solve the First-Order Differential Equation for S(r)

The equation \( \frac{dS}{dr} + \frac{2}{r}S = 0 \) can be solved using separation of variables. Rearrange to \( \frac{dS}{S} = -\frac{2}{r} dr \) and integrate both sides: \( \ln|S| = -2 \ln|r| + C \). Exponentiating gives \( S = C r^{-2} \), where \( C \) is an integration constant.

Step 3: Integrate S(r) to Find T(r)

Since \( S = \frac{dT}{dr} = C r^{-2} \), integrate to find \( T(r) \): \( T(r) = \int C r^{-2} dr = -C r^{-1} + C_1 \), where \( C_1 \) is another constant of integration.

Step 4: Apply Boundary Conditions

Use the boundary conditions: \( T(1) = 15 \) and \( T(2) = 25 \). For \( T(1) = 15 \), we get \( -C(1) + C_1 = 15 \). For \( T(2) = 25 \), \( -C(2) + C_1 = 25 \). Solving these equations, \( C = -10 \) and \( C_1 = 5 \).

Step 5: Write the Final Expression for T(r)

Using the values of \( C \) and \( C_1 \), the expression for \( T(r) \) is \( T(r) = 10 r^{-1} + 5 \).

Key Concepts

Boundary Conditions

Boundary conditions are essential when solving differential equations, as they provide the necessary information to determine the specific solution applicable to a problem. In the context of temperature distribution between two concentric spheres, the boundary conditions tell us the temperatures on the inner and outer spheres, which are critical for finding an accurate solution.

Let's consider our original problem:

• A sphere of radius 1 m has a temperature of 15°C.
• A concentric sphere of radius 2 m has a temperature of 25°C.

Knowing these temperatures at specific points helps us establish equations that include constants determined by the conditions. For instance, after deriving a general solution for temperature (\(T(r)\)), we can apply the boundary conditions to solve for any unknown constants involved.

Ultimately, these boundary conditions ensure that our solution accurately reflects the physical reality of the system, highlighting their role in shaping the specific form of the temperature function between the two spheres.

Separation of Variables

Separation of variables is a powerful mathematical technique used to simplify and solve differential equations, especially when dealing with partial differential equations. The goal is to separate the variables to allow for integration and solution finding.

In our exercise, we face a differential equation: \( \frac{dS}{dr} + \frac{2}{r}S = 0 \). By using separation of variables, we aim to manipulate the equation to isolate each variable on different sides of the equation.

• Rewriting explicitly, it becomes: \( \frac{dS}{S} = -\frac{2}{r} \, dr \).
• After this separation, we integrate both sides: \( \ln|S| = -2 \ln|r| + C \).

This technique enables the conversion of a differential equation into integrable forms, making it straightforward to solve. It is particularly useful for equations where multiplying or dividing pays off in simplifying complex mathematical expressions, aiding in the derivation of precise solutions.

Temperature Distribution

Understanding temperature distribution is crucial in thermodynamics and numerous applied sciences. It reflects how heat spreads in different materials and shapes, in this case, between two concentric spheres.

The temperature function \( T(r) \) we derived shows how the temperature changes depending on the distance from the center of the spheres:

• Initially, the equation for our temperature distribution comes from solving the differential equation \( S = \frac{dT}{dr} = C r^{-2} \).
• Integrating, we find the expression for temperature: \( T(r) = -C r^{-1} + C_1 \).

The boundary conditions help pinpoint the constants \( C \) and \( C_1 \), resulting in \( T(r) = 10 r^{-1} + 5 \). Understanding these patterns of temperature change can inform how to manage or alter the heat flow within such a spherical object.

Spherical Coordinates

Spherical coordinates are a coordinate system that is particularly useful for problems symmetrical about a point, such as the temperature distribution between spherical shells.

• In the spherical coordinate system, points are determined by a radius \(r\), and angular coordinates, which can include \(\theta\) (polar angle) and \(\phi\) (azimuthal angle).
• For problems with radial symmetry, the equations often simplify as the angular dependences can be set aside.

In our exercise, the focus is on the radial part \(r\), simplifying the spatial variables into a single-dimensional function dependent only on the distance from the center. Working in spherical coordinates streamlines the analysis and allows for the clear application of mathematical techniques like separation of variables, making it ideal for analyzing temperature distributions within these concentric spheres.