Problem 17 - Applied Calculus

Question

Estimate \(P^{\prime}(0)\) if \(P(t)=200(1.05)^{t} .\) Explain how you obtained your answer.

Step-by-Step Solution

Answer

The derivative at \( t = 0 \) is approximately 9.758.

Step 1: Identify the Function

The function given is an exponential function: \( P(t) = 200(1.05)^t \). It represents some quantity that starts at 200 and grows by a factor of 1.05 each time period \( t \).

Step 2: Differentiate the Function

To find \( P'(0) \), we first need to find the derivative of \( P(t) \) with respect to \( t \). Using the chain rule for differentiation, the derivative of \( P(t) = 200(1.05)^t \) is \( P'(t) = 200 \times \ln(1.05) \times (1.05)^t \).

Step 3: Substitute to Find P'(0)

To find \( P'(0) \), substitute \( t = 0 \) into the derivative \( P'(t) = 200 \times \ln(1.05) \times (1.05)^t \). This gives us: \[ P'(0) = 200 \times \ln(1.05) \times (1.05)^0 = 200 \times \ln(1.05) \].

Step 4: Calculate the Result

Calculate \( 200 \times \ln(1.05) \). Using the fact that \( \ln(1.05) \approx 0.04879 \), we find that \( P'(0) \approx 200 \times 0.04879 = 9.758 \).

Key Concepts

Exponential Functions

Exponential functions are a special type of mathematical function that represent rapid growth or decay. Their general form is \( f(t) = ab^t \), where \( a \) is the initial value, \( b \) is the base representing a growth (or decay) factor, and \( t \) is the exponent or time variable.
In our exercise, the function \( P(t) = 200(1.05)^t \) is an exponential function used to model a scenario where the initial quantity starts at 200 and grows by 5% per time period. This is because the base, 1.05, indicates the quantity increases by 5%.
Understanding exponential functions is vital as they occur frequently in real-world applications, such as population growth, radioactive decay, and interest calculations in finance.

• They have a constant percentage rate of change.
• They can model situations of consistent proportional growth or decay.
• The base \( b \) must be greater than zero; if \( b > 1 \), the function models growth. If \( 0 < b < 1 \), it models decay.

Chain Rule

The Chain Rule is a fundamental concept in calculus that allows us to differentiate composite functions. Composite functions are functions made up of two or more functions combined.
When you have a function inside another function, like in \( P(t) = 200(1.05)^t \), you need the chain rule. The general formula for the chain rule is: if \( y = f(g(x)) \), then the derivative \( y' = f'(g(x)) \cdot g'(x) \).
In our case, the outer function is \( a^t \) where \( a = 1.05 \), and the inner function is \( t \). The chain rule gives us a way to find the derivative of such exponential functions by handling the inside and outside separately.

• Differentiate the outer function respecting the inner function.
• Multiply by the derivative of the inner function.

This rule is particularly helpful in making differentiation straightforward, even when functions get a bit complex.

Derivative Calculation

Calculating the derivative, especially for exponential functions, helps us determine the rate of change of a function at any point.
The derivative indicates the function's sensitivity to changes in the independent variable, which in this case is \( t \). For our problem, we need to calculate the derivative of \( P(t) = 200(1.05)^t \) with respect to \( t \).
Using the chain rule, the derivative is \( P'(t) = 200 \times \ln(1.05) \times (1.05)^t \).
To find \( P'(0) \), which is the rate of change at \( t = 0 \), substitute \( t = 0 \) into the derivative. This results in
\[ P'(0) = 200 \times \ln(1.05) \times (1.05)^0 = 200 \times \ln(1.05) \].
Then we compute using \( \ln(1.05) \approx 0.04879 \), giving \( P'(0) \approx 9.758 \).

• This indicates that at \( t = 0 \), the function's rate of increase is approximately 9.758 units.
• Derivatives provide a precise understanding of function behavior over time.