Problem 46 - Calculus Early Transcendentals

Question

\(45-48\) Assume that all the given functions are differentiable. If \(u=f(x, y),\) where \(x=e^{s} \cos t\) and \(y=e^{s} \sin t,\) show that $$\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=e^{-2 s}\left[\left(\frac{\partial u}{\partial s}\right)^{2}+\left(\frac{\partial u}{\partial t}\right)^{2}\right]$$

Step-by-Step Solution

Answer

The given identity is confirmed by substituting and simplifying derivative relationships.

Step 1: Calculate Partial Derivatives of x and y

We have transformed coordinates where \( x = e^{s} \cos t \) and \( y = e^{s} \sin t \). First, find the partial derivatives of \(x\) and \(y\) with respect to \(s\) and \(t\): \( \frac{\partial x}{\partial s} = e^s \cos t, \quad \frac{\partial x}{\partial t} = -e^s \sin t \).\( \frac{\partial y}{\partial s} = e^s \sin t, \quad \frac{\partial y}{\partial t} = e^s \cos t \).

Step 2: Express Partial Derivatives of u Using Chain Rule

Using the chain rule, express the partial derivatives \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\) in terms of \(\frac{\partial u}{\partial s}\) and \(\frac{\partial u}{\partial t}\).Since \(u = f(x, y)\), we have:\( \frac{\partial u}{\partial x} = \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial x}\),\( \frac{\partial u}{\partial y} = \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial y} + \frac{\partial u}{\partial t} \cdot \frac{\partial t}{\partial y}\).

Step 3: Apply the Inverse of Coordinate Transformation

Express \(s\) and \(t\) as functions of \(x\) and \(y\). We know that:\( x^2 + y^2 = e^{2s} \) implies that \( e^{2s} = x^2 + y^2 \), thus \( s = \frac{1}{2} \ln(x^2 + y^2) \).The angle \( t \) can be found such that \( \tan t = \frac{y}{x} \), hence \( t = \arctan\left(\frac{y}{x}\right) \).

Step 4: Find Derivatives with respect to x and y

Calculate the derivative relationships:\( \frac{\partial s}{\partial x} = \frac{x}{x^2 + y^2}, \quad \frac{\partial s}{\partial y} = \frac{y}{x^2 + y^2} \).\( \frac{\partial t}{\partial x} = -\frac{y}{x^2 + y^2}, \quad \frac{\partial t}{\partial y} = \frac{x}{x^2 + y^2} \).

Step 5: Substitute and Simplify the Expression

Substitute the relationships from Steps 2 and 4 into the equation:\((\frac{\partial u}{\partial x})^2 = e^{-2s} ((\frac{\partial u}{\partial s})^2 (\frac{x^2}{x^2+y^2}) + (\frac{\partial u}{\partial t})^2 (\frac{y^2}{x^2+y^2}))\).\((\frac{\partial u}{\partial y})^2 = e^{-2s} ((\frac{\partial u}{\partial s})^2 (\frac{y^2}{x^2+y^2}) + (\frac{\partial u}{\partial t})^2 \frac{x^2}{x^2+y^2})\).

Step 6: Verify the Original Expression

Adding the contributions from \(\left(\frac{\partial u}{\partial x}\right)^2\) and \(\left(\frac{\partial u}{\partial y}\right)^2\), we get:\[ (\frac{\partial u}{\partial x})^2 + (\frac{\partial u}{\partial y})^2 = e^{-2s} [(\frac{\partial u}{\partial s})^2 + (\frac{\partial u}{\partial t})^2] \].The left-hand side simplifies to the form stated in the problem, confirming the relation.

Key Concepts

Partial Derivatives

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.

Chain Rule

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

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.