Problem 7 - The Calculus with Analytic Geometry

Question

In Exercises 5 through 10, find the indicated partial derivative by using the chain rule. $$ u=\cosh \frac{y}{x^{\prime}} x=3 r^{2} s ; y=6 s e^{r} ; \frac{\partial u}{\partial r} ; \frac{\partial u}{\partial s} $$

Step-by-Step Solution

Answer

Use the chain rule to find: \( \frac{\partial u}{\partial r} = \text{sinh} \frac{y}{x} \cdot \frac{6se^r}{x} - \text{sinh} \frac{y}{x} \cdot 6rs \cdot \frac{y}{x^2} \). \( \frac{\partial u}{\partial s} = \text{sinh} \frac{y}{x} \cdot \frac{6e^r}{x} - \text{sinh} \frac{y}{x} \cdot \frac{3r^2 y}{x^2} \).

Step 1 - Identify the given functions

Given functions: \( u = \text{cosh} \frac{y}{x} \) \( x = 3r^2 s \) \( y = 6 se^r \).

Step 2 - Calculate \( \frac{\partial u}{\partial y} \) and \( \frac{\partial y}{\partial r} \)

Find the partial derivative of \( u \) with respect to \( y \): \( \frac{\partial u}{\partial y} = \text{sinh} \frac{y}{x} \cdot \frac{1}{x} \). Calculate the partial derivative of \( y \) with respect to \( r \): \( \frac{\partial y}{\partial r} = 6se^r \).

Step 3 - Calculate \( \frac{\partial x}{\partial r} \)

Find the partial derivative of \( x \) with respect to \( r \): \( \frac{\partial x}{\partial r} = 6rs \).

Step 4 - Apply the chain rule for \( \frac{\partial u}{\partial r} \)

Use the chain rule: \( \frac{\partial u}{\partial r} = \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial r} + \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial r} \).

Step 5 - Substitute the partial derivatives found

Substitute the partial derivatives into the chain rule formula: \( \frac{\partial u}{\partial r} = \text{sinh} \frac{y}{x} \cdot \frac{1}{x} \cdot 6se^r + (-\text{sinh} \frac{y}{x} \cdot \frac{y}{x^2}) \cdot 6rs \).

Step 6 - Simplify the expression

Simplify: \( \frac{\partial u}{\partial r} = \text{sinh} \frac{y}{x} \cdot \frac{6se^r}{x} - \text{sinh} \frac{y}{x} \cdot 6rs \cdot \frac{y}{x^2} \).

Step 7 - Calculate \( \frac{\partial u}{\partial s} \)

Repeat the process to find \( \frac{\partial u}{\partial s} \): \( \frac{\partial y}{\partial s} = 6e^r \) and \( \frac{\partial x}{\partial s} = 3r^2 \). Use the chain rule: \( \frac{\partial u}{\partial s} = \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial s} + \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial s} \).

Step 8 - Substitute and simplify

Substitute the partial derivatives: \( \frac{\partial u}{\partial s} = \text{sinh} \frac{y}{x} \cdot \frac{1}{x} \cdot 6e^r + (-\text{sinh} \frac{y}{x} \cdot \frac{y}{x^2}) \cdot 3r^2 \). Simplify: \( \frac{\partial u}{\partial s} = \text{sinh} \frac{y}{x} \cdot \frac{6e^r}{x} - \text{sinh} \frac{y}{x} \cdot \frac{3r^2 y}{x^2} \).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus. It allows us to find the derivative of a composite function by differentiating the inner and outer functions separately. In the context of multivariable calculus, the chain rule helps us compute partial derivatives of a function with respect to one variable, while considering its dependencies on other variables. To use the chain rule for partial derivatives, we need to identify intermediate variables and their relationships. Then, we can apply the chain rule formula:

• Identify inner and outer functions
• Differentiate each inner function with respect to its variables
• Differentiate the outer function with respect to the intermediate variables
• Combine these using the chain rule

Partial Derivatives

Partial derivatives measure the rate at which a function changes as one of its input variables changes, while keeping other input variables constant. For a function \(u(x, y)\), the partial derivative with respect to \(x\), denoted by \(\frac{\partial u}{\partial x}\), focuses on the change of \(u\) with \(x\), holding \(y\) steady.

In our exercise, we calculate:

• \(\frac{\partial u}{\partial y} = \text{sinh} \frac{y}{x} \cdot \frac{1}{x}\)
• \(\frac{\partial y}{\partial r} = 6se^r\)
• \(\frac{\partial x}{\partial r} = 6rs\)

Understanding each of these derivatives is key to applying the chain rule correctly.

Multivariable Calculus

Multivariable calculus deals with functions of multiple variables. Key operations include finding partial derivatives and using the chain rule to manage dependencies among variables.

In the given problem, we have functions dependent on three variables: \(u\), \(x\), and \(y\). Here, \(x\) and \(y\) are dependent on \(r\) and \(s\). The main task is to find the partial derivatives \(\frac{\partial u}{\partial r}\) and \(\frac{\partial u}{\partial s}\) by considering the interdependencies among these variables.

Using multivariable calculus techniques, we identify intermediate derivatives and apply the chain rule systematically to get the solutions.

Hyperbolic Functions

Hyperbolic functions like \(\text{cosh}(x)\) and \(\text{sinh}(x)\) have similar properties to trigonometric functions but are based on hyperbolas instead of circles. Key identities and derivatives of hyperbolic functions include:

• \(\text{cosh}(x) = \frac{e^x + e^{-x}}{2}\)
• \(\text{sinh}(x) = \frac{e^x - e^{-x}}{2}\)
• Derivatives: \(\frac{d}{dx}\text{cosh}(x) = \text{sinh}(x)\) and \(\frac{d}{dx}\text{sinh}(x) = \text{cosh}(x)\)

In our exercise, we use \(\text{sinh}(\cdot)\) when calculating derivatives, which affects how we apply the chain rule:

• \(\frac{\partial u}{\partial y} = \text{sinh} \frac{y}{x} \cdot \frac{1}{x}\)

Derivatives

Derivatives measure how a function changes as its input changes. In calculus, there are several rules and techniques for finding derivatives, such as the product rule, quotient rule, and chain rule.

For functions of multiple variables, we focus on partial derivatives, which provide useful information about the function's behavior with respect to each of its variables.

In our problem, after calculating individual derivatives:

• \(\frac{\partial y}{\partial r} = 6se^r\)
• \(\frac{\partial x}{\partial r} = 6rs\)
• \(\frac{\partial y}{\partial s} = 6e^r\)
• \(\frac{\partial x}{\partial s} = 3r^2\)

We use these results within the chain rule to find \(\frac{\partial u}{\partial r}\) and \(\frac{\partial u}{\partial s}\). Finally, we simplify these results to get the final expressions.