Problem 13 - Applied Calculus

Question

Differentiate the functions in Problems 1-28. Assume that \(A\), \(B\), and \(C\) are constants. \(y=e^{-4 t}\) 14\. \(P=e^{-0.2 t}\) 15\. \(P=50 e^{-0.6 t}\) 16\. \(P=200 e^{0.12 t}\) 17\. \(P(t)=3000(1.02)^{t}\) 18\. \(P(t)=12.41(0.94)^{t}\) 19\. \(P(t)=C e^{t}\). 20\. \(y=B+A e^{t}\) 21\. \(f(x)=A e^{x}-B x^{2}+C\) 22\. \(y=10^{x}+\frac{10}{x}\) 23\. \(R=3 \ln q\) 24\. \(D=10-\ln p\) 25\. \(y=t^{2}+5 \ln t\) 26\. \(R(q)=q^{2}-2 \ln q\) 27\. \(y=x^{2}+4 x-3 \ln x\) 28\. \(f(t)=A e^{t}+B \ln t\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Function

We are given the function for problem 14: \( P = e^{-0.2t} \). This is an exponential function.

Step 2: Recall the Exponential Differentiation Rule

The differentiation of an exponential function \( e^{kt} \) with respect to \( t \) is \( ke^{kt} \). For this function, \( k = -0.2 \).

Step 3: Apply the Differentiation Rule

Differentiate \( P = e^{-0.2t} \) using the rule: \( \frac{d}{dt}e^{kt} = ke^{kt} \). Therefore, \( \frac{dP}{dt} = -0.2e^{-0.2t} \).

Step 4: Express the Result

Finally, the derivative of the function \( P = e^{-0.2t} \) with respect to \( t \) is \( \frac{dP}{dt} = -0.2e^{-0.2t} \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. This is a key concept in mathematics and appears frequently in calculus, especially in differentiation problems.
Exponential functions often take the form of:

• \(y = a^{x}\) Where \(a\) is a constant base, and \(x\) is an exponent.
• For natural exponential functions, the base \(e\) (approximately 2.718) is used, such as \(e^x\). This particular form is popular due to its unique properties in calculus.

Exponential functions are known for their continuous and smooth growth or decay, depicted on graphs as a gentle curve that never touches the x-axis. They're pivotal in modeling processes like population growth and radioactive decay.

Derivative Rules

Derivative rules are essential tools in calculus that help us find the rate of change of a function. Understanding these rules allows you to perform differentiation efficiently and accurately.
Here are the fundamental rules for differentiation:

• Power Rule: For any function \(f(x) = x^n\), the derivative \(f'(x) = nx^{n-1}\).
• Constant Rule: The derivative of a constant \(c\) is zero, \(d/dx(c) = 0\).
• Exponential Rule: The derivative of \(e^{x}\) is simply \(e^{x}\), but for \(e^{kt}\), it becomes \(ke^{kt}\) where \(k\) is a constant.
• Sum and Difference Rule: The derivative of a sum/difference of functions is the sum/difference of their derivatives: \((f+g)' = f' + g'\).

By mastering these basic rules, you can handle more complex functions with ease.

Constants in Calculus

Constants in calculus play a crucial role in differentiation and integration. They often stand as fixed values that don't change as variables do.
When dealing with differentiation, here's how constants interact:

• A constant multiplied by a variable function is taken out of the derivative. For example, for \(y = A e^x\), \(A\) doesn't change the derivative other than as a multiplier: \( y' = A e^x\).
• In calculus, if a function is solely a constant, its derivative is zero, because there is no change: \(d/dx(C) = 0\).

Constant coefficients are particularly important in defining linear functions and exponential functions with coefficients, influencing the rate without participating directly in derivative calculations.

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique that simplifies the process of differentiating functions that are products, quotients, or have exponents.
This technique makes use of the natural logarithm \(\ln\) to transform complex functions into simpler ones. Here's how it works:

• Take the natural logarithm of both sides of the equation: \(y = f(x)\) becomes \(\ln(y) = \ln(f(x))\).
• Differentiating both sides with respect to \(x\) involves using the chain rule on \(\ln(y)\), yielding \(1/y \cdot dy/dx\) on the left-hand side.
• This method is especially useful for functions like \(y = x^x\) or \(y = a^x\cdot x^b\), which can be cumbersome to differentiate using usual rules.

Logarithmic differentiation can significantly streamline the process, offering a clearer path to finding derivatives.