Problem 17 - Applied Calculus

Question

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(P(t)=3000(1.02)^{t}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dP}{dt} = 3000 \times (1.02)^{t} \times 0.0198 \).

Step 1: Identify the Function Type

The function given is \( P(t) = 3000(1.02)^{t} \). This is an exponential function of the form \( a imes b^{t} \), where \(a = 3000\) and \(b = 1.02\).

Step 2: Apply the Exponential Differentiation Rule

To differentiate \( a imes b^{t} \) with respect to \(t\), use the rule: \[ \frac{d}{dt} \left( a imes b^{t} \right) = a imes b^{t} imes \ln(b) \]. Here, \(a = 3000\) and \(b = 1.02\).

Step 3: Differentiate the Function

Apply the differentiation rule to the function: \[ \frac{dP}{dt} = 3000 \times (1.02)^{t} \times \ln(1.02) \].

Step 4: Simplify the Derivative

Calculate \(\ln(1.02)\). Using a calculator, \(\ln(1.02) \approx 0.0198\). Then, the derivative becomes: \[ \frac{dP}{dt} = 3000 \times (1.02)^{t} \times 0.0198 \].

Key Concepts

Exponential Function

An exponential function is a special mathematical expression where a constant base is raised to a variable exponent. In our given function, \( P(t) = 3000(1.02)^{t} \), we have a typical form of an exponential function: \( a \times b^{t} \). Here:

• \(a\) represents the initial amount, in this case, 3000.
• \(b\) represents the base of the exponent, which is 1.02.
• \(t\) is the exponent variable, representing time in our function.
• \((1.02)^{t}\) shows how the quantity grows or decays exponentially. Since 1.02 is slightly more than 1, it indicates a growth process over time.

Understanding this form helps in identifying how exponential functions describe real-world phenomena such as population growth, radioactive decay, and financial investments. Each scenario relies on the base value and the rate of growth or decay to predict how the situation evolves over time.

Derivative Calculation

Differentiating an exponential function like \( P(t) = 3000(1.02)^{t} \) involves using specific rules that simplify the process. To calculate its derivative, identify the constants and variable parts first:

• The constant multiplier is 3000, which stays unchanged throughout the differentiation.
• The expression \((1.02)^{t}\) is where differentiation occurs since \(t\) is the variable.

We use the formula: \[ \frac{d}{dt} \left( a \times b^{t} \right) = a \times b^{t} \times \ln(b) \]Substituting the values, we find:

• \(a = 3000\)
• \(b = 1.02\)
• \(\ln(1.02)\) is calculated using a calculator, which provides approximately 0.0198.

The result demonstrates how exponential growth or decay dynamically affects a quantity by introducing the natural log factor into the differentiation process.

Differentiation Rules

Differentiation rules are essential tools in calculus that guide us through finding derivatives efficiently. For exponential functions, the differentiation rules are straightforward yet powerful:

• The function \( a \times b^{t} \) differentiates into itself multiplied by \( \ln(b) \). This uses the natural logarithm, a critical aspect in calculating derivatives for exponential growth or decay.
• Constant factors such as \(3000\) in our example do not change when differentiating the variable part. This is due to the product rule, which implies that constants remain unaltered when multiplied by the derivative of the function.

Applying these rules helps streamline derivative calculations, making it more manageable even for complex functions. It also ensures accuracy in understanding how base changes affect the growth rate described by the exponential function.