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 solution is \( H(t) = 200 - 180e^{-kt} \); \( k = -\frac{1}{30} \ln\left(\frac{4}{9}\right) \).

Step 1: Identify the Differential Equation

The given differential equation is \( \frac{dH}{dt} = -k(H-200) \), where \( H(t) \) is the temperature of the yam at time \( t \), and \( k \) is a constant.

Step 2: Separate Variables

Rearrange the equation to separate the variables: \( \frac{dH}{H-200} = -k \, dt \). This sets up the equation for integration.

Step 3: Integrate Both Sides

Integrate both sides of the equation:* Left side: \( \int \frac{dH}{H-200} = \ln|H-200| + C_1 \)* Right side: \( \int (-k) \, dt = -kt + C_2 \). Combine the constants \( C_1 \) and \( C_2 \) into a single constant \( C = C_2 - C_1 \).

Step 4: Solve for H(t)

Exponential both sides: \( e^{\ln|H-200|} = e^{-kt + C} \)This simplifies to:\( |H-200| = e^{C}e^{-kt} \). Let \( A = e^{C} \), a new constant, so \( H - 200 = \pm A e^{-kt} \). Since temperature won't exceed 200\( ^{\circ} \)C initially, assume \( H - 200 = A e^{-kt} \).

Step 5: Apply Initial Condition

When \( t = 0 \), \( H = 20 \). Substitute these values into the equation:\( 20 - 200 = A e^{0} \)\( -180 = A \), so \( A = -180 \). Therefore, \( H(t) = 200 - 180e^{-kt} \).

Step 6: Solve for k Using Additional Condition

Given that \( H = 120 \) at \( t = 30 \), substitute into the equation:\( 120 = 200 - 180e^{-30k} \).Rearrange: \( 180e^{-30k} = 80 \).Therefore, \( e^{-30k} = \frac{80}{180} = \frac{4}{9} \).Take natural logarithm: \( -30k = \ln\left(\frac{4}{9}\right) \).Solve for \( k \):\( k = -\frac{1}{30} \ln\left(\frac{4}{9}\right) \).

Key Concepts

Integration

Integration is a fundamental concept in calculus associated with accumulation, such as finding areas under curves or determining the anti-derivative of a function. In differential equations, integration helps move from the rate of change back to original quantities.

When solving the differential equation in our problem, \[ \frac{dH}{dt} = -k(H-200), \] we integrated both sides to find a relationship involving the function \( H(t) \).

• The left side \( \int \frac{dH}{H-200} \) was integrated, giving us \( \ln|H-200| + C_1 \) as the result.
• The right side of the equation, \( \int -k \, dt \), integrates to \(-kt + C_2 \) because when integrating a constant like \(-k\), you simply multiply it by the variable of integration, \( t \).

This step is crucial as it translates the rate of change into an equation that describes the behavior over time, allowing us to solve for the unknown function \( H(t) \).

Separation of Variables

Separation of Variables is a technique used to solve certain differential equations, useful when we can rearrange the equation into the form where one side contains only one variable and the other side, only the other variable. For our differential equation \( \frac{dH}{dt} = -k(H-200) \), we can separate variables as:

• Put everything involving \( H \) on one side: \( \frac{dH}{H-200} \).
• Move everything with \( t \) to the other side: \( -k \, dt \).

This step transforms the differential equation into a more solveable form. By integrating both sides after separation, each side becomes dependent on only its own variable, allowing us to find a solution that satisfies the original equation. This method is efficient for simple separable differential equations like the one we have here.

Initial Value Problem

An initial value problem (IVP) is a differential equation accompanied by conditions specified at the outset, which provide specific information for solving the equation. The specified initial condition helps determine the constant of integration, ensuring a unique solution.

For this scenario, the initial temperature of the yam is \( 20^{\circ} \mathrm{C} \) when \( t = 0 \). From our general solution to the equation, we derived:\[ H(t) = 200 - 180e^{-kt}. \]Substituting the initial condition, \( H(0) = 20 \), we get:\[ 20 = 200 - 180e^{0}. \]This lets us solve for \( A = -180 \), allowing us to tailor the solution to our specific situation. Initial conditions transform a family of potential solutions into a specific, applicable answer.

Exponential Growth and Decay

Exponential growth and decay are models that describe how quantities increase or decrease relative to their current total. They are characterized by equations where the rate of change is proportional to the current state.

The differential equation \( \frac{dH}{dt} = -k(H-200) \) represents exponential decay since we're subtracting from a constant backbone of \( 200 \). In this setup, the temperature \( H(t) \) approaches \( 200^{\circ} \mathrm{C} \) as \( t \) increases, modeling how the yam's temperature stabilizes to the oven temperature over time.

• We visually understand this as the yam heating up rapidly at first, then slowing down its temperature rise as it nears the oven's \( 200^{\circ} \mathrm{C} \).
• Our specific solution, \( H(t) = 200 - 180e^{-kt} \), illustrates that the rate of heating decreases exponentially with time until the yam reaches equilibrium.

This concept is widely applicable beyond cooking; for example, it also describes processes such as radioactive decay and cooling of hot objects in environments. Understanding exponential models is essential for comprehending how these phenomena behave over time.