Problem 15 - Advanced Engineering Mathematics

Question

Assume that \(d T / d t=k(T-100)\) so that \(T=100+c e^{k t} .\) If \(T(0)=20^{\circ}\) and \(T(1)=22^{\circ},\) then \(c=-80\) and \(k=\ln (39 / 40)\) so that \(T(t)=90^{\circ},\) which implies \(t=82.1\) seconds. If \(T(t)=98^{\circ}\) then \(t=145.7\) seconds.

Step-by-Step Solution

Answer

The times for reaching 90°C and 98°C are 82.1 seconds and 145.7 seconds, respectively.

Step 1: Understand the Given Information

The problem provides the differential equation \( \frac{dT}{dt} = k(T-100) \) and its solution as \( T = 100 + ce^{kt} \). We know from this that \( T(0) = 20 \) and \( T(1) = 22 \). It also states that \( c = -80 \) and \( k = \ln\left(\frac{39}{40}\right) \).

Step 2: Verify Given Conditions

At \( T(0) = 20 \), we substitute into the general solution: \( 20 = 100 + c \times e^{k \times 0} = 100 + c \). Solving this gives \( c=-80 \), which matches the condition given. Additionally, for \( T(1) = 22 \), substitute in: \( 22 = 100 - 80 \times e^{k} \).

Step 3: Solve for k Constant

From \( 22 = 100 - 80 \times e^{k} \), rearrange to find \( e^{k} = \frac{78}{80} = \frac{39}{40} \). Thus \( k = \ln\left(\frac{39}{40}\right) \), verifying the given value of \( k \).

Step 4: Find Time for Specific Temperature 90°C

Set \( T(t) = 90 \) in the equation \( T = 100 - 80e^{kt} \). Thus, \( 90 = 100 - 80e^{kt} \) implies \(80e^{kt} = 10 \) which simplifies to \( e^{kt} = \frac{1}{8} \). Solve \( kt = \ln\left(\frac{1}{8}\right) \).

Step 5: Calculate Time for T(t)=90°C

With \( k = \ln\left(\frac{39}{40}\right) \), substitute to find \( t \): \( t = \frac{\ln\left(\frac{1}{8}\right)}{\ln\left(\frac{39}{40}\right)} = 82.1 \) seconds.

Step 6: Find Time for Specific Temperature 98°C

Set \( T(t) = 98 \) in the equation \( T = 100 - 80e^{kt} \). Thus \( 98 = 100 - 80e^{kt} \) implies \(80e^{kt} = 2 \), so \( e^{kt} = \frac{1}{40} \). Solve for \( t \): \( kt = \ln\left(\frac{1}{40}\right) \).

Step 7: Calculate Time for T(t)=98°C

With \( k = \ln\left(\frac{39}{40}\right) \), substitute to find \( t \): \( t = \frac{\ln\left(\frac{1}{40}\right)}{\ln\left(\frac{39}{40}\right)} = 145.7 \) seconds.

Key Concepts

Exponential Growth and Decay

Exponential growth and decay describe processes where a quantity increases or decreases at a rate proportional to its current value. This concept is widely used in physics, biology, finance, and many other fields. In the context of the given exercise, the temperature change model is a classic example of exponential decay. This is because the rate of change of temperature is proportional to the temperature difference from a fixed value (100°C in this case). In the equation \(\frac{dT}{dt} = k(T - 100)\), the solution is given as \(T(t) = 100 + ce^{kt}\). Here:

• \(k\) is the decay constant and determines how fast the temperature changes.
• \(c\) represents the initial deviation from the equilibrium temperature (100°C).

This equation models how the temperature approaches 100°C over time, either cooling down or heating up depending on the initial conditions and signs of \(k\) and \(c\). In our exercise, \(c\) is negative, indicating an initial temperature drop from the equilibrium.

Initial Value Problems

Initial value problems (IVPs) are a category of differential equations accompanied by specified values at the beginning of the interval of interest. These problems are crucial because they let us find specific solutions that reflect the real-world situations described. In our textbook problem, this involves the differential equation and initial conditions \(T(0) = 20\) and \(T(1) = 22\).To solve an IVP, follow these steps:

• Start with the differential equation and integrate or apply given formulas to find the general solution.
• Use the initial conditions to determine any constants in your solution.

In our task, once the general solution \(T(t) = 100 + ce^{kt}\) was known, we could substitute initial conditions to solve for the constants \(c\) and \(k\). These values help us build a particular solution that accurately predicts temperature over time.

Natural Logarithms

Natural logarithms, expressed as \( \ln(x) \), play a key role in solving exponential growth and decay problems. Natural logarithms are used to undo exponential functions, making it easier to solve for unknowns, especially when they appear as exponents.In our exercise, we encounter \(k = \ln\left(\frac{39}{40}\right)\), a step necessary for understanding how fast the temperature changes. The differential equation contains terms with \(e^{kt}\), an exponential function. By using natural logarithms, we can isolate \(t\) when solving for specific times where the temperature meets certain criteria, such as \(90°C\) or \(98°C\).Understanding \(\ln(x)\):

• It is the power to which \(e\) (Euler's number) must be raised to yield \(x\).
• It simplifies calculations related to exponential functions, especially when derivating or integrating.

Thus, natural logarithms are not just a mathematical tool but a necessity for solving real-world problems involving processes that change exponentially, as with our temperature example.