Problem 27 - Applied Calculus

Question

For Problems calculate all four second-order partial derivatives and confirm that the mixed partials are equal. $$B=5 x e^{-2 t}$$

Step-by-Step Solution

Answer

The second-order partial derivatives confirm that mixed partial derivatives are equal: \(-10 e^{-2t}\).

Step 1: Identify the variables

The given function is \( B = 5x e^{-2t} \), where \( B \) is dependent on the variables \( x \) and \( t \). We need to find the second-order partial derivatives with respect to these variables.

Step 2: First-order partial derivatives

Calculate the first partial derivatives. For the partial derivative with respect to \( x \), differentiate treating \( t \) as a constant: \[ \frac{\partial B}{\partial x} = 5 e^{-2t} \] For the partial derivative with respect to \( t \), differentiate treating \( x \) as a constant:\[ \frac{\partial B}{\partial t} = 5x \cdot (-2)e^{-2t} = -10x e^{-2t} \]

Step 3: Second-order partial derivatives with respect to x

Calculate the second partial derivative with respect to \( x \):\[ \frac{\partial^2 B}{\partial x^2} = \frac{\partial}{\partial x} \left( 5 e^{-2t} \right) = 0 \]Here, differentiating a constant term, \( 5 e^{-2t} \), with respect to \( x \) gives zero.

Step 4: Mixed partial derivative \( \frac{\partial^2 B}{\partial x \partial t} \)

Find the mixed partial derivative first with respect to \( x \), then \( t \):\[ \frac{\partial^2 B}{\partial t \partial x} = \frac{\partial}{\partial t} \left( 5 e^{-2t} \right) = 5 \cdot (-2) e^{-2t} = -10 e^{-2t} \]

Step 5: Second-order partial derivatives with respect to t

Calculate the second partial derivative with respect to \( t \):\[ \frac{\partial^2 B}{\partial t^2} = \frac{\partial}{\partial t} \left( -10x e^{-2t} \right) = -10x \cdot (-2) e^{-2t} = 20x e^{-2t} \]

Step 6: Mixed partial derivative \( \frac{\partial^2 B}{\partial t \partial x} \)

Find the mixed partial derivative first with respect to \( t \), then \( x \):\[ \frac{\partial^2 B}{\partial x \partial t} = \frac{\partial}{\partial x} \left( -10x e^{-2t} \right) = -10 e^{-2t} \]

Step 7: Verify equality of mixed partial derivatives

Check that \( \frac{\partial^2 B}{\partial x \partial t} = \frac{\partial^2 B}{\partial t \partial x} \):Both calculated values are \(-10 e^{-2t}\), confirming the equality of mixed partial derivatives.

Key Concepts

Mixed Partials

In the world of calculus and in particular partial derivatives, mixed partials play a crucial role. When dealing with a function of several variables, mixed partial derivatives are the derivatives taken with respect to two different variables. For instance, if you have a function like \( B = 5x e^{-2t} \), you could take the derivative with respect to \( x \) and then \( t \), or vice versa. This is called mixed partial differentiation.

Here are the key steps:

• First Partial Derivative: Take the derivative with respect to one variable, treating others as constants.
• Second Mixed Partial Derivative: Differentiate again, now with respect to another variable.

The interesting and quite magical part is, under most conditions (continuity being one), the order of differentiation does not matter. This means \( \frac{\partial^2 B}{\partial x \partial t} = \frac{\partial^2 B}{\partial t \partial x} \). In our example function, both yield \(-10 e^{-2t}\), showing the equality of mixed partials. This principle is a foundational element of multivariable calculus, making complex functions more manageable.

Function Derivatives

Function derivatives are at the heart of understanding how functions change. When you compute a derivative, you are essentially finding the rate at which a function changes at any given point. In single-variable calculus, this involves finding the slope of the tangent line to the curve at a point. However, with functions of multiple variables, like \( B = 5x e^{-2t} \), the notion of derivatives extends to partial derivatives.

Here's what happens:

• Partial Derivative with respect to \( x \): You treat \( t \) as a constant and differentiate with respect to \( x \). This results in \( \frac{\partial B}{\partial x} = 5 e^{-2t} \).
• Partial Derivative with respect to \( t \): Similarly, treating \( x \) as a constant, differentiating with respect to \( t \) gives \( \frac{\partial B}{\partial t} = -10x e^{-2t} \).

These derivatives show how the function \( B \) changes if you vary just one variable while keeping the other fixed.

Differentiation

Differentiation, a cornerstone concept in calculus, involves computing the derivative of a function to understand its behavior and rate of change. When you differentiate a function like \( B = 5x e^{-2t} \), you apply the principles of finding derivatives to discover how the function's output changes as its inputs vary.

In a step-by-step process:

• Start by identifying the variables involved. Here, \( B \) depends on \( x \) and \( t \).
• Apply the rules of differentiation to find first-order partial derivatives with respect to each variable.
• Continue the process to find second-order derivatives, which include second partial derivatives like \( \frac{\partial^2 B}{\partial x^2} \) and mixed partials like \( \frac{\partial^2 B}{\partial x \partial t} \).

This technique helps uncover complex relationships within multivariable functions. Moreover, the equality of mixed partials, as seen in \( \frac{\partial^2 B}{\partial x \partial t} = \frac{\partial^2 B}{\partial t \partial x} = -10 e^{-2t} \), provides a clearer picture of the function's behavior. Differentiation is not just a tool; it's a gateway to understanding dynamic changes in systems and functions.