Problem 216 - Calculus 3

Question

For the following exercises, use the information provided to solve the problem. Let \(w(t, v)=e^{t v}\) where \(t=r+s\) and \(v=r s .\) Find \(\frac{\partial w}{\partial r}\) and \(\frac{\partial w}{\partial s}\)

Step-by-Step Solution

Answer

\( \frac{\partial w}{\partial r} = e^{r^2s + rs^2}(2rs + s^2) \), \( \frac{\partial w}{\partial s} = e^{r^2s + rs^2}(r^2 + 2rs) \).

Step 1: Understand the function

We start with the function given as \( w(t, v) = e^{tv} \), where \( t = r + s \) and \( v = rs \). We need to find the partial derivatives of \( w \) with respect to \( r \) and \( s \).

Step 2: Express function in terms of r and s

Substitute \( t = r + s \) and \( v = rs \) into \( w(t, v) \). So, \( w(r, s) = e^{(r + s)(rs)} = e^{r^2s + rs^2} \).

Step 3: Find partial derivative with respect to r

To find \( \frac{\partial w}{\partial r} \), use the chain rule:\[ \frac{\partial w}{\partial r} = \frac{\partial}{\partial r} \left( e^{r^2s + rs^2} \right) = e^{r^2s + rs^2} \cdot \frac{\partial}{\partial r} (r^2s + rs^2) \]Calculate the derivative of the exponent:\[ \frac{\partial}{\partial r}(r^2s + rs^2) = 2rs + s^2 \]Thus, \[ \frac{\partial w}{\partial r} = e^{r^2s + rs^2}(2rs + s^2) \].

Step 4: Find partial derivative with respect to s

To find \( \frac{\partial w}{\partial s} \), use the chain rule similarly:\[ \frac{\partial w}{\partial s} = \frac{\partial}{\partial s} \left( e^{r^2s + rs^2} \right) = e^{r^2s + rs^2} \cdot \frac{\partial}{\partial s} (r^2s + rs^2) \]Calculate the derivative of the exponent:\[ \frac{\partial}{\partial s}(r^2s + rs^2) = r^2 + 2rs \]Therefore, \[ \frac{\partial w}{\partial s} = e^{r^2s + rs^2}(r^2 + 2rs) \].

Key Concepts

Chain Rule

The chain rule is crucial when dealing with functions of multiple variables, especially when taking derivatives. It allows us to manage the differentiation of composite functions.
In basic calculus, the chain rule finds the derivative of a function formed by the composition of two functions. But, in multivariable calculus, it helps find the partial derivatives of functions that depend indirectly on certain variables.
Here's a simplified approach to remember:

• Identify the outer and inner functions.
• Differentiate the outer function first, with respect to the inner variable.
• Then, multiply by the derivative of the inner function, with respect to the variable of interest.

This technique is applied repeatedly in the exercise. When finding \( \frac{\partial w}{\partial r} \), we first take the derivative of the exponential function, and then multiply it by the derivative of its exponent with respect to \( r \). Similarly, the same steps are applied to ß\( s \). This systematic approach makes complex differentiations more manageable.

Exponential Function

The exponential function \( e^x \) is a fundamental mathematical function with special properties. It is defined as a constant base \( e \) raised to the power of a given variable. The unique characteristic of exponential functions is how they change:

• The derivative of \( e^x \) is itself, \( e^x \). This makes them exceptional for calculus because finding the derivative does not alter the base function.
• In the problem, we have an exponent that's a function itself, \( r^2s + rs^2 \). Using the chain rule, we derive the expression with respect to \( r \) and \( s \) accordingly, leveraging the simplicity that exponential functions provide.

The use of exponential functions in multivariable calculus highlights their analytical power and simplicity, as seen in the ease with which we calculate derivatives using them.

Multivariable Calculus

Multivariable calculus is the extension of calculus into functions of more than one variable. It opens up a world of understanding changes and behavior in multi-dimensional space. Here are key points about it:

• Functions can depend on several variables. In the exercise, \( w \) depends on both \( t \) and \( v \), but those themselves depend on \( r \) and \( s \).
• Partial derivatives are used to find rates of change of functions concerning one variable at a time, holding other variables constant.
• Techniques such as the chain rule allow for manipulation of derivatives when functions are nested or composed within each other, like \( e^{(r^2s + rs^2)} \).

In this context, multivariable calculus allows for the exploration of more complex mathematical models, paving the way for better real-world applications across physics, engineering, and beyond. This makes mastering it not just an academic requirement but a crucial skill for various scientific endeavors.