Problem 23 - Calculus Early Transcendentals

Question

\(21-26\) Use the Chain Rule to find the indicated partial derivatives. $$\begin{array}{l}{R=\ln \left(u^{2}+v^{2}+w^{2}\right)} \\ {u=x+2 y, \quad v=2 x-y, \quad w=2 x y} \\ {\frac{\partial R}{\partial x}, \frac{\partial R}{\partial y} \text { when } x=y=1}\end{array}$$

Step-by-Step Solution

Answer

\(\frac{\partial R}{\partial x} = \frac{9}{7}\), \(\frac{\partial R}{\partial y} = \frac{9}{7}\).

Step 1: Identify the components

First, identify the components involved in the expression. We have \(R\) as a function of \(u, v, w\), which are functions of \(x\) and \(y\). The expressions are \(u = x + 2y\), \(v = 2x - y\), and \(w = 2xy\). Our goal is to find the partial derivatives \(\frac{\partial R}{\partial x}\) and \(\frac{\partial R}{\partial y}\).

Step 2: Use the Chain Rule

Apply the chain rule for partial derivatives. The chain rule states that to find \(\frac{\partial R}{\partial x}\) and \(\frac{\partial R}{\partial y}\), we need to differentiate \(R\) with respect to each variable individually while considering changes in \(u, v,\) and \(w\):\[\frac{\partial R}{\partial x} = \frac{\partial R}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial x}\]\[\frac{\partial R}{\partial y} = \frac{\partial R}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial R}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial R}{\partial w}\frac{\partial w}{\partial y}\]

Step 3: Compute Partial Derivatives of Function R with respect to u, v, w

Derive each component of \(R\) with respect to \(u\), \(v\), and \(w\):\[\frac{\partial R}{\partial u} = \frac{2u}{u^2 + v^2 + w^2}, \quad \frac{\partial R}{\partial v} = \frac{2v}{u^2 + v^2 + w^2}, \quad \frac{\partial R}{\partial w} = \frac{2w}{u^2 + v^2 + w^2}\]

Step 4: Compute Partial Derivatives of u, v, w with respect to x and y

Calculate the derivatives of \(u, v, w\) separately with respect to \(x\) and \(y\):\[\frac{\partial u}{\partial x} = 1, \quad \frac{\partial u}{\partial y} = 2\]\[\frac{\partial v}{\partial x} = 2, \quad \frac{\partial v}{\partial y} = -1\]\[\frac{\partial w}{\partial x} = 2y, \quad \frac{\partial w}{\partial y} = 2x\]

Step 5: Substitute Values for x and y

Substitute \(x = 1\) and \(y = 1\) into the expressions for \(u, v,\) and \(w\):\[u = 1 + 2 \times 1 = 3, \quad v = 2 \times 1 - 1 = 1, \quad w = 2 \times 1 \times 1 = 2\]

Step 6: Evaluate the partial derivatives for R

Substitute the values of \(u, v, w\) from Step 5 and combine with derivatives from Steps 3 and 4:Find \(\frac{\partial R}{\partial x}\):\[\frac{\partial R}{\partial x} = \frac{3}{7} \times 1 + \frac{1}{7} \times 2 + \frac{2}{7} \times 2 = \frac{3}{7} + \frac{2}{7} + \frac{4}{7} = \frac{9}{7}\]Find \(\frac{\partial R}{\partial y}\):\[\frac{\partial R}{\partial y} = \frac{3}{7} \times 2 + \frac{1}{7} \times (-1) + \frac{2}{7} \times 2 = \frac{6}{7} - \frac{1}{7} + \frac{4}{7} = \frac{9}{7}\]

Step 7: Conclusion: Compilation of Results

The final results for the partial derivatives at \(x = 1\) and \(y = 1\) are:\(\frac{\partial R}{\partial x} = \frac{9}{7}\) and \(\frac{\partial R}{\partial y} = \frac{9}{7}\).

Key Concepts

Partial Derivatives

Partial derivatives are an essential tool in multivariable calculus, allowing us to understand how a function changes as we vary one of several input variables. Imagine you have a surface or a landscape, and you want to know how steep it is in certain directions. That's where partial derivatives come in handy. They help you find the rate of change of the function with respect to one variable, keeping the others constant.

For example, if you have a function like a temperature over a geographical area, the partial derivative with respect to the x-direction would tell you how temperature changes if you move east or west, without moving north or south.

• To calculate a partial derivative, we differentiate the multi-variable function with respect to one of its variables while treating the other variables as constants.
• This concept is useful in fields such as physics, engineering, and economics, where understanding the sensitivity of a function to its input variables is crucial for analysis and decision-making.

In our exercise, finding \(\frac{\partial R}{\partial x}\) means we assess how changes in x affect the function R, while \(\frac{\partial R}{\partial y}\) tells us the impact of changing y.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of two or more variables. It allows for a richer analysis and understanding of systems where multiple factors interact. Instead of dealing with just lines or curves, you deal with surfaces and volumes.

This field includes topics such as partial derivatives, multiple integrals, and vector calculus. These tools are valuable in various scientific and engineering disciplines, from predicting the weather to designing complex systems.

• In multivariable calculus, understanding functions in 3D or higher dimensions is crucial as it helps model real-world scenarios.
• We look at functions that have several variables, and we often need to understand how changes in these variables affect the system.

In the problem we're solving, the function R is dependent on three variables, u, v, and w, which themselves are functions of two variables, x and y. Multivariable calculus helps us navigate and simplify these complex relationships.

Function Differentiation

Function differentiation in multivariable contexts involves using rules like the Chain Rule to systematically differentiate complicated functions. The Chain Rule is a powerful tool that helps relate the derivative of a function to its individual components, even if those components are dependent on other variables.

Picture a scenario where you have a nested function: differentiating directly might be complex, but breaking it down using the Chain Rule simplifies the task. The Chain Rule states that if a variable depends on one or more other variables, and you know how to differentiate each of them, you can find the derivative of the original function.

• In mathematical notation, for a function R that depends on intermediate variables u, v, and w, which in turn depend on x and y, the partial derivatives with respect to x and y involve differentiating R with respect to u, v, and w first, then multiplying by the derivatives of u, v, and w with respect to x or y.
• This step-by-step differentiation ensures that all interdependencies are considered, leading to accurate derivative calculations.

In our problem, function differentiation, through the Chain Rule, allows us to handle the complexity efficiently and obtain the required \(\frac{\partial R}{\partial x}\) and \(\frac{\partial R}{\partial y}\).