Problem 21 - Technical Mathematics with Calculus

Question

The temperature above its surroundings of an iron casting initially at \(2005^{\circ} \mathrm{F}\) will be $$T=2005 e^{-0.0620 t}$$ after \(t\) seconds. Find the time for the casting to be at a temperature of \(500^{\circ} \mathrm{F}\) above its surroundings.

Step-by-Step Solution

Answer

Calculate the time \(t\) to be approximately \(t = \frac{\ln \left(\frac{500}{2005}\right)}{-0.0620}\) seconds.

Step 1 - Establish the Temperature Function

We are given the temperature function in the form of an exponential decay model: \(T = 2005e^{-0.0620t}\), where \(T\) is the temperature above the surroundings in degrees Fahrenheit, and \(t\) is the time in seconds.

Step 2 - Set the Temperature Function to 500 Degrees

To find the time when the temperature is 500 degrees above its surroundings, set \(T = 500\): \[500 = 2005e^{-0.0620t}\].

Step 3 - Isolate the Exponential Term

Divide both sides of the equation by 2005 to isolate the exponential term: \[\frac{500}{2005} = e^{-0.0620t}\].

Step 4 - Take the Natural Logarithm of Both Sides

Taking the natural logarithm of both sides will allow us to solve for \(t\): \[\ln \left(\frac{500}{2005}\right) = \ln(e^{-0.0620t})\].

Step 5 - Use Properties of Logarithms

Use the property of logarithms that \(\ln(e^x) = x\) to simplify the right side of the equation: \[\ln \left(\frac{500}{2005}\right) = -0.0620t\].

Step 6 - Solve for t

Divide both sides of the equation by -0.0620 to solve for \(t\): \[t = \frac{\ln \left(\frac{500}{2005}\right)}{-0.0620}\].

Step 7 - Calculate the Time

Calculate the value of \(t\) using a calculator: \[t \approx \frac{\ln \left(\frac{500}{2005}\right)}{-0.0620}\].

Key Concepts

Natural Logarithm

Understanding the natural logarithm is essential for solving equations involving exponential decay. Quite simply, the natural logarithm, denoted as \(\ln\), is the inverse operation of taking an exponential function with base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. When we take the natural logarithm of an exponential function, we effectively 'undo' the exponentiation. For instance, \(\ln(e^x) = x\). This property becomes particularly useful when we need to isolate a variable in the exponent of an exponential expression, as was needed in the temperature decay problem.

In practical terms, if you see an equation like \(e^x = b\), you would take the natural logarithm of both sides to find that \(x = \ln(b)\), thereby isolating \(x\) and making \(b\) much simpler to work with. This process is a key step in solving many problems in physics, finance, and other fields where growth or decay is described exponentially.

Temperature Decay

Temperature decay can be expressed mathematically as an exponential decay process. This concept describes how the temperature of an object changes over time as it cools down or heats up towards the temperature of its surroundings. Exponential decay models like \(T = T_0e^{-kt}\), where \(T_0\) is the initial temperature difference from the surroundings, \(k\) is the decay constant, and \(t\) is time, are used to represent this phenomenon.

Interpreting the variables in our exercise, \(2005^\circ F\) represents the initial temperature above the surroundings and \(e^{-0.0620t}\) shows how this temperature difference decreases exponentially over time. The decay constant, 0.0620 in this case, affects how rapidly the temperature approaches the surrounding temperature. The bigger the decay constant, the faster the decay of temperature. This mathematical model helps us predict when the iron casting will reach a certain temperature, which is crucial in many industrial processes.

Isolating the Exponential Term

Isolating the exponential term in an equation is a critical step towards finding a solution. When dealing with equations involving exponential decay, such as \(T = T_0e^{-kt}\), isolating the exponential term \(e^{-kt}\) enables us to apply logarithmic functions to determine the value of \(t\).

In our exercise, we begin by equating the temperature function to 500, representing the desired temperature above the surroundings. The equation \(500 = 2005e^{-0.0620t}\) quickly becomes \(\frac{500}{2005} = e^{-0.0620t}\) after dividing both sides by 2005. This step isolates the exponential term. By taking the natural logarithm of both sides, we can then harness the property \(\ln(e^x) = x\) and proceed to solve for \(t\), as the variable is no longer in the exponent. This process exemplifies a common strategy to tackle a wide range of problems in algebra and calculus that involve exponential functions.