Problem 21 - Applied Calculus

Question

A yam is put in a \(200^{\circ} \mathrm{C}\) oven and heats up according to the differential equation \(\frac{d H}{d t}=-k(H-200), \quad\) for \(k\) a positive constant. (a) If the yam is at \(20^{\circ} \mathrm{C}\) when it is put in the oven, solve the differential equation. (b) Find \(k\) using the fact that after 30 minutes the temperature of the yam is \(120^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The yam's temperature over time is given by \( H(t) = -180 e^{-\frac{\ln(3)}{30} t} + 200 \). The constant \( k \) is \( \frac{\ln(3)}{30} \).

Step 1: Understand the Differential Equation

We are given a first-order linear differential equation of the form \( \frac{dH}{dt} = -k(H-200) \). This equation describes how the temperature \( H \) of the yam changes over time \( t \) due to the surrounding oven temperature of \( 200^{\circ} \mathrm{C} \).

Step 2: Solve the Homogeneous Equation

The homogeneous part of the differential equation is \( \frac{dH}{dt} = -kH \). We solve this by separating variables: \( \frac{1}{H} dH = -k dt \) then integrating both sides, we get \( \ln |H| = -kt + C_1 \) leading to the solution \( H(t) = C e^{-kt} \).

Step 3: Solve the Particular Solution

We assume a particular solution of the form \( H_p(t) = 200 \). Since the derivative of a constant is zero, substituting \( H_p(t) = 200 \) in the differential equation yields \( 0 = -k(200-200) \), which satisfies the equation.

Step 4: Combine Solutions

The general solution is the sum of the homogeneous and particular solutions: \( H(t) = C e^{-kt} + 200 \).

Step 5: Apply Initial Conditions

We use the initial condition \( H(0) = 20 \) to solve for \( C \): \( 20 = C + 200 \) which implies \( C = -180 \). Thus, the specific solution is \( H(t) = -180 e^{-kt} + 200 \).

Step 6: Determine the Constant k

Use the condition \( H(30) = 120 \): Substituting \( t = 30 \) and \( H(30) = 120 \) into the solution gives \( 120 = -180 e^{-30k} + 200 \). Solving for \( k \), we get \( 60 = 180 e^{-30k} \), leading to \( e^{-30k} = \frac{1}{3} \). Thus, \( -30k = \ln(\frac{1}{3}) \), and \( k = -\frac{\ln(\frac{1}{3})}{30} \).

Step 7: Simplify k

Simplifying the expression for \( k \), we have \( k = \frac{\ln(3)}{30} \).

Key Concepts

First-Order Differential Equations

Differential equations describe how a quantity changes with respect to another, often time. A first-order differential equation involves the first derivative of a function. In our example, the equation \(\frac{dH}{dt} = -k(H-200)\) defines how the temperature \(H\) of the yam changes over time. It is termed 'first-order' because the highest derivative is the first derivative. In first-order differential equations, the purpose is to determine the original function based on its rate of change.

• First-order equations can describe physical phenomena such as heat transfer, velocity, or population dynamics.
• They are often defined with respect to time, making them essential for temporal analysis of various systems.
• Since they involve the first derivative only, they are generally easier to solve than higher-order equations.

Separation of Variables

Separation of Variables is a technique to solve differential equations. It's particularly effective for equations that can be rewritten so that each variable and its differential are on opposite sides of the equation. In the context of our yam problem, the differential equation \(\frac{dH}{dt} = -k(H-200)\) can be separated by dividing both sides by \((H-200)\) and multiplying both sides by \(dt\), yielding \(\frac{1}{H-200} dH = -k dt\). This equation can now be integrated separately on both sides:

• This method simplifies the equation, breaking it down into integrable parts.
• The goal is to transfer all terms involving one variable to one side, and all terms with the other variable to the opposite side.
• After integration, determining constants is crucial, which is done via initial conditions.

Using Separation of Variables allows us to find a general solution by solving these separate parts.

Initial Value Problems

Initial value problems (IVP) involve finding a function that satisfies a differential equation and meets specified initial conditions. For our problem, the initial condition is that the yam's temperature is \(20^{\circ} \mathrm{C}\) at \(t=0\). This provides specific information about the solution at a particular point, making it possible to determine the unknown constants in the general solution.

• IVPs are essential in predicting future behavior based on current knowledge.
• They are widely used in physics, engineering, biology, and economics, where initial states are well defined.
• The main challenge is accurately capturing this initial state to ensure reliable predictions.

By applying the initial condition \(H(0) = 20\), we can solve for the constant \(C\) in our solution, ensuring that the specific behavior of the yam’s heating process is captured accurately.

Exponential Functions

Exponential functions are vital in solving differential equations, particularly those that model growth and decay such as our yam heating problem. The form \( H(t)= C e^{-kt} + 200 \) illustrates exponential decay, characterized by a constant base raised to a variable exponent. In this equation:

• The term \(C e^{-kt}\) represents the variable part of the function that changes with time.
• The parameter \(k\) influences the rate of decay — larger \(k\) values mean faster changes.
• The constant \(200\) is the stable, equilibrium temperature of the oven, shaping the function's behavior as it approaches a steady state.

Understanding exponential functions helps in grasping how solutions evolve over time, crucial for applications in natural and engineered systems.