Problem 15 - Calculus Early Transcendentals

Question

Suppose \(f\) is a differentiable function of \(x\) and \(y,\) and \(g(u, v)=f\left(e^{u}+\sin v, e^{u}+\cos v\right) .\) Use the table of values to calculate \(g_{d}(0,0)\) and \(g_{c}(0,0)\) $$\begin{array}{|c|c|c|c|c|}\hline & {f} & {g} & {f_{x}} & {f_{y}} \\\ \hline(0,0) & {3} & {6} & {4} & {8} \\ \hline(1,2) & {6} & {3} & {2} & {5} \\\ \hline\end{array}$$

Step-by-Step Solution

Answer

\( g_d(0,0) = 12 \) and \( g_c(0,0) = 4 \).

Step 1: Understand the Problem

We need to find the partial derivatives of the function \( g(u, v) = f(e^u + \sin v, e^u + \cos v) \) at the point \((0, 0)\). We will compute \( g_d(0, 0) \) and \( g_c(0, 0) \) by finding \( \frac{\partial g}{\partial u}(0, 0) \) and \( \frac{\partial g}{\partial v}(0, 0) \) respectively using the chain rule.

Step 2: Apply the Chain Rule for Partial Derivative with respect to u

The chain rule for the partial derivative of \( g \) with respect to \( u \) is \[ \frac{\partial g}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u} \] where \( x = e^u + \sin v \) and \( y = e^u + \cos v \). Calculating the components, we find: - \( \frac{\partial x}{\partial u} = e^u \)- \( \frac{\partial y}{\partial u} = e^u \)Therefore, at \((0, 0)\), we have:\[ \left(\frac{\partial g}{\partial u}\right)_{(0,0)} = 4 \cdot e^0 + 8 \cdot e^0 = 12 \]

Step 3: Apply the Chain Rule for Partial Derivative with respect to v

The chain rule for the partial derivative of \( g \) with respect to \( v \) is \[ \frac{\partial g}{\partial v} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial v} \] where \( x = e^u + \sin v \) and \( y = e^u + \cos v \). Calculating the components, we find:- \( \frac{\partial x}{\partial v} = \cos v \)- \( \frac{\partial y}{\partial v} = -\sin v \)Therefore, at \((0, 0)\), we have:\[ \left(\frac{\partial g}{\partial v}\right)_{(0,0)} = 4 \cdot \cos 0 + 8 \cdot (-\sin 0) = 4 \]

Step 4: Conclusion

The calculated derivatives at \((0, 0)\) are: \( g_d(0, 0) = 12 \) and \( g_c(0, 0) = 4 \). These represent the partial derivatives of \( g \) with respect to \( u \) and \( v \), respectively.

Key Concepts

Chain Rule in Multivariable Calculus

In multivariable calculus, the chain rule is an essential tool for differentiating composite functions involving multiple variables. This is especially useful when dealing with functions like \( g(u, v) = f(e^u + \sin v, e^u + \cos v) \), where each variable itself depends on other variables.

When applying the chain rule, you decompose the derivative into simpler parts by expressing the change in the output with respect to the input variables, through immediate dependencies. For example, to find \( \frac{\partial g}{\partial u} \), you take the partial derivatives of \( f \) with respect to \( x \) and \( y \), and multiply each by the respective derivative of \( x \) and \( y \) with regard to \( u \). This is expressed as:

• \( \frac{\partial g}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u} \)

This approach ensures that you account for all the intermediate rates of change that contribute to the overall rate of change in the function \( g \). Similarly, the chain rule is employed to find \( \frac{\partial g}{\partial v} \), highlighting how the outcome changes as \( v \) varies, with calculations deeply rooted in understanding these interdependencies.

Understanding Differentiable Functions

A differentiable function, in the context of multivariable calculus, implies that the function has well-defined partial derivatives at a given point. This property is crucial when we apply the chain rule, as it allows us to make sense of how changes in input variables lead to changes in output values.

For our function \( f \) described in the exercise, it being differentiable means that the rates of change \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are clearly defined at any given point like \((0, 0)\). Thus, \( f \) can be locally approximated by linear functions which reflect its small changes, enabling the precise calculation of partial derivatives of \( g \) by leveraging these linear approximations.

• \( \frac{\partial f}{\partial x} = 4 \) and \( \frac{\partial f}{\partial y} = 8 \) at \((0, 0)\)
• Linear approximations help in predicting how \( f \) changes

The differentiability of \( f \) also assures us that the composite function \( g \) behaves smoothly, allowing us to perform further calculus operations to explore the behavior of \( g \) around the origin without encountering abrupt shifts or undefined expressions.

Introduction to Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It covers topics like partial derivatives, multiple integrals, and vector calculus. In the context of the exercise, understanding multivariable calculus helps in analyzing functions like \( g(u, v) \), where multiple variables influence the behavior and output.

The function \( g \) is created by combining the variables \( u \) and \( v \) in different ways through trigonometric and exponential transformations. This introduces the necessity of using tools like the chain rule to unravel the complexity and determine the function's partial derivatives. Multivariable calculus also demands a keen understanding of how variations in each variable (like \( u \) and \( v \)) independently and collectively influence the outcome.

• Explore how variables interact in functional forms
• Utilize calculus tools to manage multiple dependencies

With functions of multiple variables, graphical representations become more dimensional. Rather than lines or curves, you deal with surfaces and volumes, navigating through gradients and contours to visualize outcomes. Thus, mastery in multivariable calculus allows you to perform more expansive analyses and solve more complex mathematical challenges involved in real-world systems.