Problem 2 - Applied Calculus

Question

Find the area under \(P=100(0.6)^{t}\) between \(t=0\) and \(t=8.\)

Step-by-Step Solution

Answer

The area under the curve is approximately -42.735.

Step 1: Understand the Problem

The problem asks us to find the area under the curve given by the function \(P = 100(0.6)^t\) from \(t = 0\) to \(t = 8\). This involves calculating the definite integral of the function over the specified interval.

Step 2: Set Up the Integral

To find the area under the curve, we need to compute the definite integral of \(P(t)\) from \(t = 0\) to \(t = 8\). This can be set up as: \[ \int_{0}^{8} 100(0.6)^t \,dt \]

Step 3: Recognize the Integral of an Exponential Function

The function \(P(t) = 100(0.6)^t\) can be expressed in terms of the base \(e\) by noting that \(0.6 = e^{\ln(0.6)}\). Thus, rewrite the function as: \[ 100 e^{t \cdot \ln(0.6)} \] This transforms the expression allowing the use of the formula for the integral of an exponential function.

Step 4: Integrate the Function

Using the formula \(\int e^{ax} \,dx = \frac{1}{a} e^{ax} + C\) for integration, where \(a = \ln(0.6)\), the integral becomes:\[ \int 100 e^{t \ln(0.6)} \, dt = \frac{100}{\ln(0.6)} e^{t \ln(0.6)} \bigg|_{0}^{8} \] Simplifying this expression, we note that \(e^{t \ln(0.6)} = (0.6)^t\).

Step 5: Evaluate the Definite Integral

Substitute the limits into the evaluated integral:\[ \frac{100}{\ln(0.6)} \left[ (0.6)^8 - (0.6)^0 \right] = \frac{100}{\ln(0.6)} \left[ (0.6)^8 - 1 \right] \] Calculate the values of \((0.6)^8 = 0.016777216\) and \((0.6)^0 = 1\).

Step 6: Compute the Final Result

Subtract these values: \[ \frac{100}{\ln(0.6)} \left[ 0.016777216 - 1 \right] = \frac{100}{\ln(0.6)} \times (-0.983222784)\] Simplifying further, we find the numerical value for \(\frac{100}{\ln(0.6)}\) and multiply it by \(-0.983222784\) to get the area. Using a calculator, compute this value to get the approximate area under the curve.

Key Concepts

Exponential Growth

Exponential growth is a fundamental concept in mathematics and many applied fields such as biology, economics, and physics. It describes a process of increase that happens at a rate proportional to the value at any given time.
For example, when you have a function like \(P(t) = 100(0.6)^t\), it represents exponential growth with a base less than 1, which is actually a decay.
This particular form \(0.6^t\) shows a rapid decrease over time, which is known as exponential decay.

• The function starts from a value and decreases exponentially.
• It is expressed as \(P(t) = P_0 a^t\) where \(P_0 \) is the initial value and \(a\) is the base, less than 1 for decay.
• The decay factor, \(0.6\), indicates the speed of decay.

This type of growth or decay is opposite to linear growth, where the change is constant over time. Understanding this concept is key to interpreting the changes described by these functions.

Area Under the Curve

The area under the curve represents the integral of a function over a specified interval. This is an important concept in calculus, as it allows us to quantify the accumulated value of a function like \(P(t) = 100(0.6)^t\) from a specific start point to an endpoint.
For the problem at hand, we look at the area under \(P(t)\) between \(t = 0\) and \(t = 8\). Calculating this area gives insight into the total value accumulated by the function over that time.

• The area under the curve is found using definite integrals.
• It is a measure of the total accumulation or decay of the function over the interval.
• This process is useful, for example, in calculating distances, quantities, or probabilities.

The concept of area under a curve is widely used in various real-world applications, like determining the total profit or loss over time when dealing with growth or decay models.

Integration Techniques

Integration is a critical technique in calculus used to calculate the area under the curve of a function. In this context, we need to integrate the function \(P(t) = 100(0.6)^t\).One helpful technique is recognizing the form of the function. Notice that the expression \( (0.6)^t\) can be rewritten as \(e^{t \, \ln(0.6)}\) using logarithmic properties.
This transformation allows us to apply the known formula for integrating exponential functions:

• Applying the formula \(\int e^{ax} \,dx = \frac{1}{a} e^{ax} + C\).
• Identify \(a\) as \(-\ln(0.6)\) to transform the function for easy integration.
• Evaluate the integral over the specified limits to find the definite area under the curve.

Through these steps, the computation of definite integrals becomes manageable, allowing us to solve complex problems involving exponential functions with greater ease and accuracy.