Problem 32 - Applied Calculus

Question

For Problems calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$P=2 K L^{2}$$

Step-by-Step Solution

Answer

The second-order partial derivatives are \(0, 4K, 4L, 4L\). Mixed partials are equal.

Step 1: Identify Variables and Function

Given the function \( P = 2KL^2 \), we need to find partial derivatives with respect to the variables \( K \) and \( L \).

Step 2: Find First-Order Partial Derivatives

Compute the first-order partial derivative of \( P \) with respect to \( K \):\[ \frac{\partial P}{\partial K} = \frac{\partial}{\partial K}(2KL^2) = 2L^2 \]Now, find the first-order partial derivative of \( P \) with respect to \( L \):\[ \frac{\partial P}{\partial L} = \frac{\partial}{\partial L}(2KL^2) = 4KL \]

Step 3: Calculate Second-Order Partial Derivative (w.r.t. K then L)

Now take the partial derivative of \( \frac{\partial P}{\partial K} = 2L^2 \) with respect to \( L \):\[ \frac{\partial^2 P}{\partial L \partial K} = \frac{\partial}{\partial L}(2L^2) = 4L \]

Step 4: Calculate Second-Order Partial Derivative (w.r.t. L then K)

Now take the partial derivative of \( \frac{\partial P}{\partial L} = 4KL \) with respect to \( K \):\[ \frac{\partial^2 P}{\partial K \partial L} = \frac{\partial}{\partial K}(4KL) = 4L \]

Step 5: Confirm Equality of Mixed Partial Derivatives

Compare the mixed partial derivatives calculated in Steps 3 and 4:\[ \frac{\partial^2 P}{\partial L \partial K} = 4L \quad \text{and} \quad \frac{\partial^2 P}{\partial K \partial L} = 4L \]The mixed partial derivatives are equal, confirming Clairaut's Theorem.

Step 6: Calculate Second-Order Partial Derivative (w.r.t. K twice)

Find the second-order partial derivative by differentiating \( \frac{\partial P}{\partial K} = 2L^2 \) with respect to \( K \) again:\[ \frac{\partial^2 P}{\partial K^2} = \frac{\partial}{\partial K}(2L^2) = 0 \]

Step 7: Calculate Second-Order Partial Derivative (w.r.t. L twice)

Find the second-order partial derivative by differentiating \( \frac{\partial P}{\partial L} = 4KL \) with respect to \( L \) again:\[ \frac{\partial^2 P}{\partial L^2} = \frac{\partial}{\partial L}(4KL) = 4K \]

Key Concepts

Clairaut's Theorem

Clairaut's Theorem is a fascinating result in calculus, specifically in the field of partial derivatives. This theorem deals with the equality of mixed partial derivatives, meaning it checks whether differentiating in different orders produces the same result.
When we talk about the partial derivatives of a function of two or more variables, we must consider the impact of changing one variable while keeping others constant. The theorem provides us with a simple condition: for functions that meet certain criteria, the order of differentiation does not matter, as long as the second partial derivatives are continuous at the point of interest.
In simpler terms, if you first take the derivative of a function with respect to one variable and then with respect to another, the result should be the same as if you take it in the reverse order. A practical illustration of Clairaut's Theorem can be seen in the solution: For the function \( P = 2KL^2 \), we calculated:

• \( \frac{\partial^2 P}{\partial L \partial K} = 4L \)
• \( \frac{\partial^2 P}{\partial K \partial L} = 4L \)

Both computations lead to the same result, showcasing the power of this theorem. It assures us that under suitable conditions, the computations of mixed partial derivatives will be equal, simplifying the analysis of complex functions.

First-Order Partial Derivatives

First-order partial derivatives are a cornerstone in understanding how functions of several variables behave. When we compute a first-order partial derivative, we are examining how the function changes when we vary one variable slightly, while keeping the others fixed.
In the given exercise, the function is \( P = 2KL^2 \).
We start by finding the first-order partial derivatives:

• For \( K \), we find \( \frac{\partial P}{\partial K} = 2L^2 \). Here, we treat \( L \) as a constant and differentiate accordingly.
• For \( L \), we find \( \frac{\partial P}{\partial L} = 4KL \). Here, we treat \( K \) as a constant and differentiate accordingly.

These derivatives tell us the rate at which \( P \) changes as only \( K \) or \( L \) changes, which is extremely useful in applications like physics or economics. They help to identify which variable has a more significant impact on the function at a particular point.

Mixed Partial Derivatives

Mixed partial derivatives take a deeper look into how a function responds to changes in more than one variable. They involve taking derivatives with respect to different variables sequentially.
In mathematical terms, mixed partial derivatives of a function \( f(x, y) \) would involve taking the partial derivative of the function first with respect to \( x \) and then with respect to \( y \) (or vice versa).

In our problem, the mixed partial derivatives are calculated as follows:

• \( \frac{\partial^2 P}{\partial L \partial K} = 4L \), where we first differentiated \( f \) with respect to \( K \) and then with respect to \( L \).
• \( \frac{\partial^2 P}{\partial K \partial L} = 4L \), where the order of differentiation was reversed.

As per Clairaut's Theorem, these should be equal if computed correctly, and indeed, here they are both \( 4L \).
Mixed partial derivatives are particularly important in advanced fields like optimization and multivariable calculus, where understanding the intricate relationship between variables is key.