Problem 14 - Applied Calculus

Question

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(P=e^{-0.2 t}\)

Step-by-Step Solution

Answer

The derivative of \( P = e^{-0.2t} \) is \( \frac{dP}{dt} = -0.2 e^{-0.2t} \).

Step 1: Identify the Function Type

The given function is an exponential function of the form \( P = e^{kt} \), where \( k = -0.2 \). We need to differentiate this function with respect to \( t \).

Step 2: Recall the Derivative Rule for Exponential Functions

The derivative of \( e^{kt} \) with respect to \( t \) is given by \( \frac{d}{dt}[e^{kt}] = k e^{kt} \). Apply this rule to find the derivative of \( P \).

Step 3: Differentiate the Given Function

Using the rule for exponential functions, differentiate \( P = e^{-0.2t} \). Compute \( \frac{dP}{dt} = -0.2 e^{-0.2t} \).

Step 4: Simplify the Result

The expression from the differentiation step already represents the derivative in its simplest form: \( \frac{dP}{dt} = -0.2 e^{-0.2t} \).

Key Concepts

Exponential Functions

Exponential functions are a special type of mathematical function that have a constant base raised to a variable exponent. One of the most common bases used in exponential functions is the number 'e', approximately 2.718. This leads to the function format like \[ f(x) = e^{kx} \] where 'k' is a constant. These functions are unique because they grow (or decay) at a rate proportional to their current value, providing a wide variety of applications in fields such as biology, finance, and physics.

In this particular exercise, we are dealing with the exponential function \[ P = e^{-0.2t} \]. Here, the base 'e' is raised to the power of \(-0.2t\), implying an exponential decay. Understanding how these functions behave in terms of growth or decay is crucial for differentiating them correctly, which is our next focus in this exercise.

Derivative Rules

Differentiation involves finding the derivative of a function, which is essentially the rate at which a function changes with respect to a variable. For exponential functions, specific derivative rules simplify this process significantly. The rule for differentiating an exponential function of the form \[ e^{kt} \]states that: \[ \frac{d}{dt}[e^{kt}] = k e^{kt} \]. This formula tells us that the derivative of an exponential function is the product of the constant 'k' and the original function itself.

This simplifies differentiation immensely, allowing us to quickly figure out how the function's value changes. In the case of the function \[ P = e^{-0.2t} \], using the rule gives us:

• Calculate the derivative: \(\frac{dP}{dt} = -0.2 e^{-0.2t}\)
• Makes use of 'k = -0.2'
• Preserves the form of the original function

rhe simplicity of this rule is one reason why exponential functions are favored in modeling scenarios involving growth or decay.

Constants in Differentiation

When differentiating functions, constants play a valuable role. In many mathematical functions, constants are numbers that do not change when the function’s primary variable changes. They can be coefficients (like in our exercise) or added terms within the function.

In differentiation, a key principle to remember is that the derivative of a constant by itself is zero. However, when a constant is multiplied by a function, the differentiation can proceed directly on the function, leaving the constant as a multiplier. This leads to formulas like:

• Multiply the constant with the derivative of the other component.
• E.g., for \(k e^{kt}\), the constant 'k' remains constant.

This principle is clearly seen in our example: \( P = e^{-0.2t} \). By differentiating, the constant \(-0.2\) is simply brought out of the derivative operation and multiplied to the result of differentiating the exponential part. Thus we directly arrive at:\(\frac{dP}{dt} = -0.2 e^{-0.2t} \) without altering the nature of the function’s exponential component.