Problem 18 - Applied Calculus

Question

Show that, for any constant \(P_{0},\) the function \(P=P_{0} e^{t}\) satisfies the equation $$\frac{d P}{d t}=P$$

Step-by-Step Solution

Answer

The function satisfies the equation because \( \frac{dP}{dt} = P_{0} e^{t} = P \).

Step 1: Differentiate the Function

We have the function \( P = P_{0} e^{t} \). To show that it satisfies the equation \( \frac{dP}{dt} = P \), we first need to find the derivative of \( P \) with respect to \( t \). The derivative of \( e^{t} \) is itself; hence, the derivative of \( P \) is: \[ \frac{dP}{dt} = \frac{d}{dt}(P_{0} e^{t}) = P_{0} \cdot \frac{d}{dt}(e^{t}) = P_{0} e^{t} \]

Step 2: Compare with the Original Function

Now that we have the derivative \( \frac{dP}{dt} = P_{0} e^{t} \), let's compare this with the original function \( P = P_{0} e^{t} \). We can observe that:- The derivative \( \frac{dP}{dt} \) is exactly equal to \( P \).- Thus, the equation \( \frac{dP}{dt} = P \) holds true for this function.

Key Concepts

Exponential Growth

Exponential growth refers to a process in which a quantity increases rapidly over time at a rate proportional to its current value. This concept is crucial in many fields, including biology, finance, and physics. When a population of organisms grows, or when money in an investment compounds, we often see exponential growth.

This type of growth is described mathematically by functions that include the constant base of the natural logarithm, often denoted as \( e \). The function \( P = P_{0} e^{t} \) perfectly captures exponential growth. Here, \( P_{0} \) is the initial amount or the population at time \( t = 0 \), and \( t \) represents time.

Exponential growth means:

• The rate of growth is proportional to the current quantity.
• Growth accelerates over time, leading to very large numbers in surprisingly short periods.

Understanding exponential growth helps students grasp how quickly changes can occur, whether in natural settings, financial investments, or technological advancements.

Derivative

In mathematics, a derivative represents the rate at which a function changes as its input changes. It's a core concept in calculus, expressed as \( \frac{dP}{dt} \) in our function \( P = P_{0} e^{t} \). The derivative helps us understand how \( P \) changes with respect to \( t \).

The derivative can be thought of as the "slope" of the function at any given point. For exponential functions like \( e^{t} \), the rate of change is fascinating because the derivative of \( e^{t} \) is \( e^{t} \) itself:

• Derivatives provide critical insight into the behavior of functions, showing how fast they increase or decrease.
• This is integral in forecasting future values or understanding dynamic systems.

The derivative confirms exponential growth by illustrating continuous and rapid change. In practical scenarios, understanding derivatives can vastly improve one's ability to predict outcomes and make informed decisions.

Function Differentiation

Function differentiation involves finding the derivative of a function. It's a method used to determine how a function's output value changes as its input value changes. In our case, we differentiated \( P = P_{0} e^{t} \) with respect to \( t \) to show that \( \frac{dP}{dt} = P_{0} e^{t} \).

Function differentiation requires applying rules like the product rule or the chain rule, depending on the function's complexity. Fortunately, for \( P = P_{0} e^{t} \), it's simpler because:

• The differentiation of \( e^{t} \) is directly \( e^{t} \).
• We simply multiply by the constant \( P_{0} \).

Function differentiation is a powerful tool in mathematics and engineering. It allows us to deeply understand how variables interrelate and predict how systems respond to changes. Once students learn to differentiate functions like \( P = P_{0} e^{t} \), they gain foundational skills used in countless scientific and practical applications.