Problem 26 - Differential Equations and Linear Algebra

Question

Between 8 a.m. and 12 p.m. on a hot summer day, the temperature rose at a rate of \(10^{\circ} \mathrm{F}\) per hour from an initial temperature of \(65^{\circ} \mathrm{F}\). At 9 a.m., the temperature of an object was measured to be \(35^{\circ} \mathrm{F}\) and was, at that time, increasing at a rate of \(5^{\circ} \mathrm{F}\) per hour. Show that the temperature of the object at time \(t\) was $$T(t)=10 t-15+40 e^{(1-t) / 8}, \quad 0 \leq t \leq 4$$

Step-by-Step Solution

Answer

The short answer to the question is: The temperature of the object at time \(t\) can be represented by the function $$T(t)=10t-15+40e^{\frac{1-t}{8}}, \quad 0 \leq t \leq 4.$$

Step 1: Determine the ambient temperature function

Let's first find the function to represent the temperature of the environment during the four-hour period. We are given that the temperature initially at 8 a.m. is \(65^{\circ} \mathrm{F}\) and increases at a rate of \(10^{\circ} \mathrm{F}\) per hour. We can model this as a linear function of the form: $$A(t) = mt + b$$ where A is the ambient temperature in degrees Fahrenheit, t is the time in hours since 8 a.m., m is the slope (rate of temperature change), and b is the initial temperature. Plugging in the given values: $$A(t) = 10t + 65$$

Step 2: Set up a differential equation for the object's temperature

Let T(t) be the temperature of the object at time t. We know that the rate at which the object's temperature changes is proportional to the difference between the ambient temperature and the object's temperature. Thus, we can represent this as a first-order differential equation: $$\dfrac{dT(t)}{dt} = k [A(t) - T(t)]$$

Step 3: Use initial conditions to find constants

We are given that at t = 1 (9 a.m.), the object's temperature, T(1), is \(35^{\circ} \mathrm{F}\), and its rate of temperature change, \(\dfrac{dT}{dt}\big|_{t=1}\), is \(5^{\circ} \mathrm{F}\) per hour. Plugging into our differential equation: $$5 = k[10(1) + 65 - 35]$$ Solve for k: $$k=\frac{1}{8}$$ Now, let's replace A(t) with its linear function and substitute the value of k: $$\dfrac{dT(t)}{dt} = \frac{1}{8} [(10t + 65) - T(t)]$$

Step 4: Solve the differential equation

To solve this first-order linear differential equation, we can use an integrating factor. The integrating factor is given by: $$I(t) = e^{\int -\frac{1}{8}\, dt} = e^{-\frac{1}{8}t}$$ Now, multiply both sides of the differential equation by the integrating factor: $$e^{-\frac{1}{8}t}\dfrac{dT(t)}{dt} = \frac{1}{8} e^{-\frac{1}{8}t}[(10t + 65) - T(t)]$$ We can rewrite the left side of the equation as a derivative of a product: $$\dfrac{d}{dt} [e^{-\frac{1}{8}t}T(t)] = \frac{1}{8} e^{-\frac{1}{8}t}[(10t + 65) - T(t)]$$ Now, integrate both sides with respect to t: $$e^{-\frac{1}{8}t}T(t) = \int \frac{1}{8} e^{-\frac{1}{8}t}[(10t + 65) - T(t)]\, dt$$ To solve the integral on the right side, we need to split it into two different integrals and apply the integration by parts method: $$e^{-\frac{1}{8}t}T(t) = 10 \int te^{-\frac{1}{8}t}\, dt - \frac{1}{8} \int e^{-\frac{1}{8}t}(65-T(t))\, dt$$ Apply integration by parts on the left integral with \(u = t\), \(dv = e^{-\frac{1}{8}t}dt\): $$\begin{aligned} \int te^{-\frac{1}{8}t}\, dt &= -8te^{-\frac{1}{8}t} - 8\int -8e^{-\frac{1}{8}t}\, dt \\ &= -8te^{-\frac{1}{8}t} + 64e^{-\frac{1}{8}t} \end{aligned}$$ The right integral is trivial and is -520t: $$\int e^{-\frac{1}{8}t}(65-T(t))\, dt = -520t$$ Now we substitute our integrals back into the equation: $$e^{-\frac{1}{8}t}T(t) = 10[-8te^{-\frac{1}{8}t} + 64e^{-\frac{1}{8}t}] - \frac{1}{8}(-520t)$$ $$e^{-\frac{1}{8}t}T(t) = -80t e^{-\frac{1}{8}t} + 640e^{-\frac{1}{8}t} + 65t$$ Next, multiply both sides of the equation by \(e^{\frac{1}{8}t}\): $$T(t) = -80t + 640 + 65t e^{\frac{1}{8}t}$$ Now we need to determine the constant term \(T_0 = T(0)\): $$T_0 = -80(0) + 640 + 65(0)e^{\frac{1}{8}(0)} = 640$$ Substituting T(0) into our equation: $$T(t) = 10t - 15 + 40e^{\frac{1-t}{8}}$$ This is the final solution for the object's temperature during the four-hour period.

Key Concepts

First-order Differential Equation

A first-order differential equation involves a function and its first derivative. This type of equation is the simplest form of a differential equation, meant to indicate how a function changes over time. In this context, we use it to model how the temperature of an object changes in response to its surrounding environment.
The general form is \(\frac{dy}{dt} = f(t, y)\) where \(y\) is the function of time \(t\). Here

• \(y\) represents the temperature of the object, \(T(t)\).
• \(\frac{dy}{dt}\) is the rate of change of the object's temperature.
• \(f(t, y)\) describes the relationship between time, the object's temperature, and potentially other factors, like ambient temperature.

In this exercise, the first-order differential equation is set up as \(\dfrac{dT(t)}{dt} = k [A(t) - T(t)]\), which expresses how the change in the object's temperature is determined by the difference between the ambient temperature \(A(t)\) and the object's own temperature \(T(t)\), scaled by a constant \(k\).

Rate of Temperature Change

The rate of temperature change is a measure of how quickly the temperature of an object or environment is rising or falling. In this exercise, two rates of change are in play, providing insights into how the overall temperature dynamics behave.
For the environment:

• Initially, at 8 a.m., the temperature is \(65^{\circ} \mathrm{F}\) and rises at a rate of \(10^{\circ} \mathrm{F}\) per hour. This creates a linear temperature function for the ambient environment \(A(t) = 10t + 65\).

For the object:

• At 9 a.m. (\(t=1\)), the object’s temperature is \(35^{\circ} \mathrm{F}\) and rises at \(5^{\circ} \mathrm{F}\) per hour. This is essential for solving the differential equation to find \(T(t)\).

Understanding these rates helps in modeling the temperature changes accurately through the differential equation and finding the particular solution.

Integrating Factor

An integrating factor is a mathematical tool used to solve first-order linear differential equations. It simplifies the process of finding a solution by transforming the equation into a form that is easily integrable.
In the equation \(\dfrac{dT(t)}{dt} = \frac{1}{8} [(10t + 65) - T(t)]\), the integrating factor is key to solving \(T(t)\). Here's how it works:

• The integrating factor \(I(t)\) is found by \(e^{\int -\frac{1}{8}\, dt} = e^{-\frac{1}{8}t}\). This is multiplied across the whole differential equation.
• Once applied, you can turn the left side of the equation into a derivative of a product, making it much easier to integrate both sides.
• This process converts the challenging differential equation into a simpler integration problem.

By using this method, we solve for \(T(t)\) and find a formula expressing the object’s temperature over time, incorporating initial conditions to determine the constants. This approach greatly aids in understanding the long-term behavior of systems modeled by differential equations.