Problem 14 - Calculus with Concepts in Calculus

Question

Compute \(\partial z / \partial r\) and \(\partial z / \partial s\). $$ z=2^{u-v} ; u=r \cos s, v=r \sin s $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial r} = \ln(2) \cdot 2^{u-v} (\cos s - \sin s) \) and \( \frac{\partial z}{\partial s} = \ln(2) \cdot r \cdot 2^{u-v} (\cos s - \sin s) \).

Step 1: Express z in terms of u and v

The given expression for \( z \) is \( z = 2^{u-v} \). This implies that \( z \) is a function of two variables \( u \) and \( v \).

Step 2: Determine the partial derivatives of z with respect to u and v

First, compute \( \frac{\partial z}{\partial u} \). Since \( z = 2^{u-v} \), we apply the chain rule for \( z = a^x \), which gives us:\[ \frac{\partial z}{\partial u} = \ln(2) \cdot 2^{u-v} \]Similarly, for \( \frac{\partial z}{\partial v} \), we have:\[ \frac{\partial z}{\partial v} = -\ln(2) \cdot 2^{u-v} \]

Step 3: Find partial derivatives of u and v with respect to r

Given \( u = r \cos s \) and \( v = r \sin s \), calculate the partial derivatives:\[ \frac{\partial u}{\partial r} = \cos s \ \frac{\partial v}{\partial r} = \sin s \]

Step 4: Apply the chain rule for \( \partial z / \partial r \)

Use the chain rule, \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial r} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial r}\):\[ \frac{\partial z}{\partial r} = \ln(2) \cdot 2^{u-v} \cdot \cos s - \ln(2) \cdot 2^{u-v} \cdot \sin s \]

Step 5: Simplify \( \partial z / \partial r \)

Combine like terms to simplify the expression:\[ \frac{\partial z}{\partial r} = \ln(2) \cdot 2^{u-v} (\cos s - \sin s) \]

Step 6: Find partial derivatives of u and v with respect to s

For \( u = r \cos s \) and \( v = r \sin s \), compute the partial derivatives:\[ \frac{\partial u}{\partial s} = -r \sin s \ \frac{\partial v}{\partial s} = r \cos s \]

Step 7: Apply the chain rule for \( \partial z / \partial s \)

Apply the chain rule, \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial s} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial s}\):\[ \frac{\partial z}{\partial s} = \ln(2) \cdot 2^{u-v} \cdot (-r \sin s) + \ln(2) \cdot 2^{u-v} \cdot r \cos s \]

Step 8: Simplify \( \partial z / \partial s \)

Simplify the expression by combining like terms:\[ \frac{\partial z}{\partial s} = \ln(2) \cdot r \cdot 2^{u-v} (\cos s - \sin s) \]

Key Concepts

chain rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. In the context of multivariable functions, this rule helps us find how a function changes with respect to one variable by considering its dependence on other variables.

Consider functions like our given example, where you have a function, say, \[ z = f(u, v) \] with \[ u = g(r, s) \] and\[ v = h(r, s) \].To find \( \frac{\partial z}{\partial r} \), the chain rule in partial differentiation involves the formula:

• \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial r} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial r} \)

This method allows us to track how changes in the intermediate variables \( u \) and \( v \) contribute to changes in \( z \) with respect to \( r \).

In simpler terms, you can imagine each step in the chain as a tiny shift, where every function's change contributes cumulatively to the whole. Breaking down complex dependencies into smaller pieces using the chain rule makes tackling multivariable problems straightforward and structured.

multivariable functions

Multivariable functions are functions that depend on more than one input variable. In our case, the function \( z = 2^{u-v} \) depends on the variables \( u \) and \( v \). A multivariable function's output depends on the different interactions between its variables.

These functions are essential in modeling real-world phenomena, where outcomes are influenced by multiple factors.

• For instance, weather predictions involve variables like temperature, humidity, and pressure.
• Economics might use variables such as interest rates, inflation, and GDP.

Understanding multivariable functions necessitates studying how each variable influences the output individually and collectively. This exploration often involves tools like **partial derivatives** that help isolate the effect of one variable at a time.

Moreover, equations where more than one variable affects the result are complex but also incredibly insightful, offering detailed perspectives on systemic behavior. Thorough grasp of concepts like our function \( z = 2^{u-v} \) makes it easier for learners to handle intricacies in advanced calculus and applied fields.

partial derivatives

Partial derivatives are an extension of derivatives to multivariable functions. They measure the rate of change of a function concerning one variable, keeping others constant. In our example, we find partial derivatives like \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \) that tell us how \( z \) changes as \( u \) or \( v \) changes, with the existence of multiple variables.

Calculating partial derivatives is crucial when you want to understand how individual components of a multivariable function interact.

• For \( z = 2^{u-v} \), utilizing the formula for exponential functions, we find \( \frac{\partial z}{\partial u} = \ln(2) \cdot 2^{u-v} \).
• Meanwhile, \( \frac{\partial z}{\partial v} = -\ln(2) \cdot 2^{u-v} \).

These expressions highlight the nature of influence: \( u \) contributes positively, while \( v \) contributes negatively.

Overall, understanding partial derivatives involves recognizing these subtle impacts within a bigger picture, facilitating more controlled and predictable outcomes in both academic and practical applications.