Problem 31 - Applied Calculus

Question

Calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$ f=100 e^{r t} $$

Step-by-Step Solution

Answer

The mixed partial derivatives are equal: \( \frac{\partial^2 f}{\partial r \partial t} = \frac{\partial^2 f}{\partial t \partial r} = 100 e^{r t} + 100 r t e^{r t} \).

Step 1: Identify Variables and Function

We need to first identify the variables involved in the function. The function given is \( f = 100 e^{r t} \), where \( r \) and \( t \) are the variables.

Step 2: Calculate First Partial Derivative with Respect to \( r \)

To find the first partial derivative of \( f \) with respect to \( r \), we treat \( t \) as a constant: \( \frac{\partial f}{\partial r} = \frac{\partial}{\partial r} (100 e^{r t}) = 100 t e^{r t} \).

Step 3: Calculate First Partial Derivative with Respect to \( t \)

Similarly, to find the first partial derivative with respect to \( t \), we treat \( r \) as a constant: \( \frac{\partial f}{\partial t} = \frac{\partial}{\partial t} (100 e^{r t}) = 100 r e^{r t} \).

Step 4: Calculate Second Partial Derivative with Respect to \( r \), Twice

We will differentiate \( \frac{\partial f}{\partial r} = 100 t e^{r t} \) again with respect to \( r \): \( \frac{\partial^2 f}{\partial r^2} = \frac{\partial}{\partial r} (100 t e^{r t}) = 100 t^2 e^{r t} \).

Step 5: Calculate Mixed Partial Derivative, \( \frac{\partial^2 f}{\partial r \partial t} \)

Differentiate \( \frac{\partial f}{\partial r} = 100 t e^{r t} \) with respect to \( t \): \( \frac{\partial^2 f}{\partial t \partial r} = \frac{\partial}{\partial t} (100 t e^{r t}) = 100 e^{r t} + 100 t r e^{r t} \).

Step 6: Calculate Second Partial Derivative with Respect to \( t \), Twice

Differentiate \( \frac{\partial f}{\partial t} = 100 r e^{r t} \) again with respect to \( t \): \( \frac{\partial^2 f}{\partial t^2} = \frac{\partial}{\partial t} (100 r e^{r t}) = 100 r^2 e^{r t} \).

Step 7: Calculate Mixed Partial Derivative, \( \frac{\partial^2 f}{\partial t \partial r} \)

Differentiate \( \frac{\partial f}{\partial t} = 100 r e^{r t} \) with respect to \( r \): \( \frac{\partial^2 f}{\partial r \partial t} = \frac{\partial}{\partial r} (100 r e^{r t}) = 100 e^{r t} + 100 r t e^{r t} \).

Step 8: Verify Equality of Mixed Partials

We compare the mixed partial derivatives \( \frac{\partial^2 f}{\partial r \partial t} \) and \( \frac{\partial^2 f}{\partial t \partial r} \):Both are equal, as both result in \( 100 e^{r t} + 100 r t e^{r t} \), which confirms the equality.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They involve taking the derivative of a function with respect to one variable, while keeping the other variables constant. This gives us a way to analyze how a function changes as one specific variable changes, which is particularly useful in functions of several variables.

• For a function such as \( f = 100 e^{r t} \), the partial derivative with respect to \( r \) is found by treating \( t \) as a constant, leading to \( \frac{\partial f}{\partial r} = 100 t e^{r t} \).
• When calculating the derivative with respect to \( t \), treat \( r \) as a constant, resulting in \( \frac{\partial f}{\partial t} = 100 r e^{r t} \).

Identifying and calculating partial derivatives allow us to examine the behavior and rate of change of a multivariable function in the direction of one of its variables.

Mixed Partials

Mixed partials refer to the derivatives obtained by differentiating twice, with respect to different variables. They help assess how a function changes when two variables change simultaneously. For the function \( f = 100 e^{r t} \), mixed partial derivatives are calculated as follows:

• First, differentiate \( \frac{\partial f}{\partial r} = 100 t e^{r t} \) with respect to \( t \): \( \frac{\partial^2 f}{\partial t \partial r} = 100 e^{r t} + 100 t r e^{r t} \).
• Next, differentiate \( \frac{\partial f}{\partial t} = 100 r e^{r t} \) with respect to \( r \): \( \frac{\partial^2 f}{\partial r \partial t} = 100 e^{r t} + 100 r t e^{r t} \).

Mixed partial derivatives are crucial in determining the combined effect of changes in different variables on the function's outcome.

Calculus

Calculus, especially in the context of partial derivatives, allows us to analyze functions of multiple variables in detail. It provides tools for expressing how these functions behave and change in response to variations in their input variables. When dealing with functions like \( f = 100 e^{r t} \), calculus lets us dissect the function's structure:

• By exploring first-order partial derivatives, we understand the influence of each variable individually.
• Calculating second-order partials, including mixed partials, gives us a deeper insight into how these variables interact.
• Determining mixed partial derivatives furthers our comprehension of the intricate relationships within multivariable functions.

Understanding calculus in this context fosters a comprehensive approach to handling complex mathematical relations and interpreting their real-world implications.

Equality of Mixed Partials

The principle of the equality of mixed partials, also known as Clairaut's theorem, states that if a function \( f \) is continuously differentiable, the order of differentiation for mixed partial derivatives does not matter. Put simply, \( \frac{\partial^2 f}{\partial r \partial t} = \frac{\partial^2 f}{\partial t \partial r} \). For the function \( f = 100 e^{r t} \), we confirmed this with:

• \( \frac{\partial^2 f}{\partial r \partial t} = 100 e^{r t} + 100 r t e^{r t} \)
• \( \frac{\partial^2 f}{\partial t \partial r} = 100 e^{r t} + 100 r t e^{r t} \)

Both mixed partial derivatives are equal, thus verifying the theorem's applicability. The equality of mixed partials is foundational in ensuring consistent results when analyzing multivariable functions, as it ensures the pathway of differentiation does not alter the outcome.