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Chapter 10: Problem 13

Rod When a red-hot steel rod is plunged in a bath of water that is kept at a constant temperature \(10^{\circ} \mathrm{C},\) the temperature of the rod at time \(t, f(t),\) satisfies the differential equation $$y^{\prime}=k[10-y]$$ where \(k>0\) is a constant of proportionality. Determine \(f(t)\) if the initial temperature of the rod is \(f(0)=350^{\circ} \mathrm{C}\) and \(k=.1.\)

Short Answer

Expert verified
The temperature function of the rod is \( f(t) = 10 - 340e^{-0.1t} \).

Step by step solution

01

- Write the Differential Equation

The given differential equation is \[ y' = k(10 - y) \]
02

- Substitute Constants

Substitute the given value of \(k\) into the differential equation: \[ y' = 0.1(10 - y) \]
03

- Separate Variables

Separate variables by dividing both sides by \(10 - y\) and multiplying both sides by \(dt\): \[ \frac{dy}{10 - y} = 0.1 dt \]
04

- Integrate Both Sides

Integrate both sides to find \(y\). For the left side, use the integral of \(\frac{1}{10 - y}\); for the right side, integrate \(0.1\) to get: \[ \int \frac{1}{10 - y} dy = \int 0.1 dt \] This results in: \[ -\ln|10 - y| = 0.1t + C \]
05

- Solve for Constant of Integration

Use the initial condition \( f(0) = 350 \) to find \(C\). At \( t = 0 \), \[ -\ln|10 - 350| = C \] This simplifies to: \[ C = -\ln| -340 | = ln(340).\]
06

- Rewrite Expression with Initial Condition

Combine the constant \( C \) with the logarithmic expression to write: \[ -\ln|10 - y| = 0.1t + \ln(340) \]
07

- Solve Exponential Equation

Use exponentiation to clear the logarithm: \[ |10 - y| = e^{-0.1t} \cdot 340 \]
08

- Solve for y

Solve for \( y \) to find the expression: \[ 10 - y = 340e^{-0.1t} \] Thus, \[ y = 10 - 340e^{-0.1t} \]

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Key Concepts

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

Initial Value Problem
An initial value problem (IVP) involves finding a function that satisfies a given differential equation along with a specified value at a particular time. In our exercise, the rod's temperature starts at 350掳C at time zero. This initial condition, denoted as \( f(0) = 350 \), helps to determine the unique solution of the differential equation. By solving the IVP, we ensure that the function correctly models the system right from the starting point.
Variable Separation
In many differential equations, we can separate variables to simplify the equation. This means expressing the equation in a form where functions of one variable are on one side and functions of the other variable are on the other side. For the given differential equation \( y' = 0.1(10 - y) \), we rearrange terms to get:

\[ \frac{dy}{10 - y} = 0.1 \, dt \]
By separating the variables, we can then individually integrate both sides and solve the equation more straightforwardly. Variable separation is an essential method in solving many first-order differential equations.
Exponential Decay
Exponential decay occurs when a quantity decreases at a rate proportional to its current value. In this problem, the temperature of the rod cools down exponentially over time. After separating variables and integrating, we reach:

\[ -\begin{pmatrix}\text{ln}|10 - y|\begin{pmatrix} = 0.1t + \text{constant} \]
Solving for \( y \), we get an equation in the form of \( e^{-\text{constant}}\), which describes exponential decay. Here, the temperature decreases from the initial condition, approaching the bath's temperature asymptotically.
Integration
Integration helps in finding the function from its derivative. In our differential equation:

\[ y' = 0.1(10 - y) \]
we separate variables and integrate both sides:

\[ \int{\frac{1}{10 - y} \, dy} = \int{0.1 \, dt} \]
These integrals yield \( -\text{ln}|10 - y| \) on the left and \( 0.1t \) on the right.
After integrating, we solve for the constant using the initial condition to get the final temperature function.

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