91Ó°ÊÓ

Partial derivatives are a fundamental concept in multivariable calculus. When we deal with functions of multiple variables, like \( u = f(x, y) \), we want to understand how the function changes as each individual variable changes.
The partial derivative of \( u \) with respect to \( x \), denoted as \( \frac{\partial u}{\partial x} \), measures how \( u \) changes as \( x \) changes, while keeping \( y \) constant. Similarly, the partial derivative with respect to \( y \), \( \frac{\partial u}{\partial y} \), looks at the change in \( u \) as \( y \) changes, with \( x \) held constant.
In practical terms:

• To find \( \frac{\partial u}{\partial x} \), differentiate \( f(x, y) \) treating \( y \) as a constant.
• To find \( \frac{\partial u}{\partial y} \), differentiate \( f(x, y) \) while treating \( x \) as a constant.

This exercise uses these partial derivatives to transform an equation from a different coordinate system, highlighting their role in understanding changes across various dimensions.

The chain rule in calculus is a powerful tool that allows us to differentiate composite functions. In the context of multivariable calculus, it lets us express derivatives of complex functions through their simpler parts. Consider a scenario where \( u = f(x, y) \), and both \( x \) and \( y \) are themselves functions of \( s \) and \( t \). Here, you can see how interlinked changes in \( s \) and \( t \) impact \( u \).
We use the chain rule to find partial derivatives like \( \frac{\partial u}{\partial s} \) and \( \frac{\partial u}{\partial t} \). The extended chain rule expression is:

• \( \frac{\partial u}{\partial x} = \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x} \)
• \( \frac{\partial u}{\partial y} = \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial y} + \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial y} \)

With these expressions, you can compute how \( u \) changes with respect to \( s \) and \( t \) by considering the interdependencies of the variables. It simplifies complex relationships into manageable calculations, ensuring deeper insights into multi-layered functions.

Coordinate transformation involves changing from one system of coordinates to another, which is crucial for simplifying complex calculus problems. In this exercise, we switch from \( (x, y) \) coordinates to \( (s, t) \) coordinates.
Why is this helpful?

• It can simplify the mathematics involved in solving calculus problems.
• It provides a clearer geometrical interpretation of the situation.

For transformations, knowing how the variables interact is essential. Here, we start with \( x = e^s \cos t \) and \( y = e^s \sin t \), creating a polar to Cartesian conversion scenario. To reverse-engineer back to \( s \) and \( t \), we use:

• \( s = \frac{1}{2} \ln(x^2 + y^2) \)
• \( t = \arctan(\frac{y}{x}) \)

Understanding these transformations is key to moving smoothly between different perspectives on a problem, helping to find solutions that might not be immediately obvious in the initial frame.