Problem 17 - Advanced Engineering Mathematics

Question

Using separation of variables to solve \(d T / d t=k\left(T-T_{m}\right)\) we get \(T(t)=T_{m}+c e^{k t} .\) Using \(T(0)=70\) we find \(c=70-T_{m},\) so \(T(t)=T_{m}+\left(70-T_{m}\right) e^{k t} .\) Using the given observations, we obtain $$\begin{array}{c} T\left(\frac{1}{2}\right)=T_{m}+\left(70-T_{m}\right) e^{k / 2}=110 \\ T(1)=T_{m}+\left(70-T_{m}\right) e^{k}=145 \end{array}$$ Then, from the first equation, \(e^{k / 2}=\left(110-T_{m}\right) /\left(70-T_{m}\right)\) and $$\begin{aligned} e^{k}=\left(e^{k / 2}\right)^{2}=\left(\frac{110-T_{m}}{70-T_{m}}\right)^{2} &=\frac{145-T_{m}}{70-T_{m}} \\ \frac{\left(110-T_{m}\right)^{2}}{70-T_{m}} &=145-T_{m} \\ 12100-220 T_{m}+T_{m}^{2} &=10150-250 T_{m}+T_{m}^{2} \\ T_{m} &=390 \end{aligned}$$ The temperature in the oven is \(390^{\circ}\).

Step-by-Step Solution

Answer

The temperature in the oven is \(390^\circ\).

Step 1: Identify the differential equation

The given differential equation is \(\frac{d T}{d t}=k\left(T-T_{m}\right)\). This is a separable differential equation, which can be solved by separating variables.

Step 2: Separate variables and integrate

Separate the variables in the equation: \(\frac{d T}{T-T_{m}}=k dt\). Integrate both sides to get \(\ln|T-T_{m}|=kt+C\), where \(C\) is the constant of integration.

Step 3: Solve for T

Exponentiate both sides to remove the logarithm: \(T-T_{m}=Ce^{kt}\). Therefore, \(T(t)=T_{m}+Ce^{kt}\).

Step 4: Determine constant c using initial condition

Given \(T(0)=70\), plug it into the equation: \(70 = T_{m} + Ce^{0k}\) leading to \(C = 70 - T_{m}\). Thus, \(T(t) = T_{m} + (70-T_{m})e^{kt}\).

Step 5: Use observations to form equations

From the two given points: \(T\left(\frac{1}{2}\right)=110\) and \(T(1)=145\), form the equations: \(T_{m} + (70-T_{m})e^{k/2} = 110\) and \(T_{m} + (70-T_{m})e^{k} = 145\).

Step 6: Solve for \(e^{k/2}\)

From the equation for \(T\left(\frac{1}{2}\right)\): \(e^{k/2} = \frac{110 - T_{m}}{70 - T_{m}}\).

Step 7: Solve for \(e^{k}\)

Squaring the result from Step 6 gives \(e^{k} = \left(\frac{110 - T_{m}}{70 - T_{m}}\right)^{2}\). This also equals the expression for \(T(1)\): \(\frac{145 - T_{m}}{70 - T_{m}}\).

Step 8: Equate and solve for \(T_{m}\)

Equating the expressions from Step 7: \(\frac{(110-T_{m})^{2}}{70-T_{m}} = 145-T_{m}\). Simplifying this gives: \(12100 - 220T_{m} + T_{m}^{2} = 10150 - 250T_{m} + T_{m}^{2}\). Solving the equation results in \(T_{m} = 390\).

Key Concepts

Differential Equations

In mathematics, differential equations are essential tools that describe how a quantity changes over time. A differential equation is an equation that involves a function and its derivatives. These equations tell us how a certain variable, often dependent on another variable, such as time, evolves.

The differential equation given in the exercise, \(\frac{dT}{dt} = k(T - T_m)\), is a separable one. This means we can solve it by separating the variables to their respective sides of the equation. In our case, we separate \(T\) and \(t\) to different sides to solve it.

Here’s a breakdown of what happens:

• We moved all terms involving \(T\) to one side and all terms involving \(t\) to the other: \(\frac{dT}{T-T_m} = k\,dt\).
• Next, we integrate both sides to find a function that connects \(T\) and \(t\): \( \ln|T-T_m| = kt + C \).
• Finally, we solve for \(T\) by exponentiating both sides and arrive at a general solution: \( T(t) = T_m + C e^{kt} \).

Differential equations like these model various phenomena, such as exponential growth and decay, oscillations, and wave propagation, in fields ranging from physics to economics.

Initial Conditions

Initial conditions are crucial for determining the specific solution of a differential equation, tailored to match real-world scenarios. They help pinpoint the unique solution from an infinite set of solutions provided by the general form of the equation.

In this exercise, the initial condition states that the temperature is 70 degrees at time zero, expressed as \(T(0) = 70\). This is used to solve for the constant \(c\) in the equation.

Here's how:

• Given that the temperature at \(t = 0\) is 70, substituting in the formula gives us: \(70 = T_m + Ce^{0}\).
• Considering \(e^0 = 1\), we find \(C = 70 - T_m\).
• With this value of \(C\), the solution becomes \(T(t) = T_m + (70 - T_m)e^{kt}\).

Using initial conditions not only helps in simplifying the equation but also in ensuring that our model precisely fits the situation we are analyzing.

Exponential Growth

Exponential growth occurs when the rate of change of a quantity is proportional to its current value. This concept is often seen in different scenarios, such as population growth, radioactive decay, and our temperature change problem.

In the equation \(T(t) = T_m + (70 - T_m)e^{kt}\), the term \(e^{kt}\) signifies exponential growth or decay, depending on the sign of \(k\). If \(k\) is positive, the function grows exponentially; if negative, it decays.

Key points about exponential growth:

• Exponential functions involve exponents and have a constant base, in this case, \(e\), which is approximately 2.718.
• They grow faster over time when the exponent \(kt\) increases, due to the multiplication effect of \(e\).
• In our oven problem, solving further for \(k\) through additional observations showed how temperature changes over time as the function increases towards the oven's constant temperature.

Understanding exponential growth helps explain not just temperature changes, but phenomena like compound interest in finance, and even the spread of diseases in epidemiology.