Problem 21 - Calculus Early Transcendentals

Question

\(21-26\) Use the Chain Rule to find the indicated partial derivatives. $$\begin{array}{ll}{z=x^{2}+x y^{3},} & {x=u v^{2}+w^{3}, \quad y=u+v e^{m}} \\\ {\frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}, \frac{\partial z}{\partial w}} & {\text { when } u=2, v=1, w=0}\end{array}$$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial u} = 80, \frac{\partial z}{\partial v} = 158, \frac{\partial z}{\partial w} = 0 \).

Step 1: Understand the Function Structure

We have a function \( z \) that depends on \( x \) and \( y \). Specifically, \( z = x^2 + xy^3 \). The functions \( x \) and \( y \) themselves depend on \( u \), \( v \), and \( w \): \( x = uv^2 + w^3 \) and \( y = u + ve^m \). Our goal is to find the partial derivatives of \( z \) with respect to \( u \), \( v \), and \( w \) using the chain rule.

Step 2: Apply the Chain Rule for \( \frac{\partial z}{\partial u} \)

Use the chain rule to differentiate \( z \) with respect to \( u \): \[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u} \]Calculate each component:\( \frac{\partial z}{\partial x} = 2x + y^3 \),\( \frac{\partial z}{\partial y} = 3xy^2 \),\( \frac{\partial x}{\partial u} = v^2 \),\( \frac{\partial y}{\partial u} = 1 \).Evaluate at \( u=2, v=1, w=0 \) to find the actual partial derivative.

Step 3: Compute Intermediate Values for \( \frac{\partial z}{\partial u} \)

Substitute \( u=2, v=1, w=0 \) into the functions:\( x = 2 \times 1^2 + 0^3 = 2 \),\( y = 2 + 1 \cdot e^0 = 3 \),\( \frac{\partial z}{\partial x} = 2 \times 2 + 3^3 = 26 \),\( \frac{\partial z}{\partial y} = 3 \times 2 \times 3^2 = 54 \),\( \frac{\partial x}{\partial u} = 1 \),\( \frac{\partial y}{\partial u} = 1 \).Thus, \( \frac{\partial z}{\partial u} = 26 \times 1 + 54 \times 1 = 80 \).

Step 4: Apply the Chain Rule for \( \frac{\partial z}{\partial v} \)

Use the chain rule:\[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v} \]Components:\( \frac{\partial x}{\partial v} = 2uv \),\( \frac{\partial y}{\partial v} = e^m \).We already know \( \frac{\partial z}{\partial x} = 26 \) and \( \frac{\partial z}{\partial y} = 54 \).Evaluate using \( u=2, v=1, w=0 \).

Step 5: Compute Intermediate Values for \( \frac{\partial z}{\partial v} \)

Evaluate the components:\( \frac{\partial x}{\partial v} = 2 \times 2 \times 1 = 4 \),\( \frac{\partial y}{\partial v} = e^0 = 1 \).Thus, \( \frac{\partial z}{\partial v} = 26 \times 4 + 54 \times 1 = 158 \).

Step 6: Apply the Chain Rule for \( \frac{\partial z}{\partial w} \)

Using the chain rule:\[ \frac{\partial z}{\partial w} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial w} \]Components: \( \frac{\partial x}{\partial w} = 3w^2 \).We already calculated \( \frac{\partial z}{\partial x} = 26 \).Evaluate at \( u=2, v=1, w=0 \).

Step 7: Compute Intermediate Values for \( \frac{\partial z}{\partial w} \)

Evaluate \( \frac{\partial x}{\partial w} \):\( \frac{\partial x}{\partial w} = 3 \times 0^2 = 0 \).Thus, \( \frac{\partial z}{\partial w} = 26 \times 0 = 0 \).

Step 8: Compile Final Results

The final partial derivatives are evaluated:\( \frac{\partial z}{\partial u} = 80 \),\( \frac{\partial z}{\partial v} = 158 \),\( \frac{\partial z}{\partial w} = 0 \). These values depend on both direct and indirect dependencies of \( u \), \( v \), and \( w \) in the function \( z \).

Key Concepts

Partial Derivatives

Let's dive into the fascinating world of partial derivatives. In multivariable calculus, you encounter functions of multiple variables. This often involves finding how a change in one variable affects the function while keeping others constant. This is where partial derivatives come into play. Partial derivatives are like regular derivatives, but instead of analyzing a single curve, they examine a surface. Here's how you can recognize them:

• If you have a function \( f(x, y) \), a partial derivative with respect to \( x \) would look like \( \frac{\partial f}{\partial x} \).
• Similarly, the partial derivative concerning \( y \) is \( \frac{\partial f}{\partial y} \).

For example, in the original problem:

• \( \frac{\partial z}{\partial x} = 2x + y^3 \) shows how \( z \) changes as \( x \) tweaks, keeping \( y \) steady.
• This provides insight into the surface’s inclination in the direction of the \( x \)-axis at a given \( (x, y) \).

Partial derivatives are pivotal because they help in understanding how different variables individually influence a multivariable function.

Multivariable Calculus

In multivariable calculus, you expand upon the familiar concepts of single-variable calculus to functions with two or more variables. This involves not just dealing with lines and curves, but also tackling surfaces and multidimensional shapes. The primary goal here is to understand how these functions behave and change, which involves concepts like gradients, level curves, and partial derivatives:

• Gradients give a vector that points in the direction of the greatest rate of increase of the function.
• Level curves are contour lines of constant function value in one plane.

In our specific problem, the function \( z = x^2 + xy^3 \) is a surface affected by two input variables, \( x \) and \( y \). These, in turn, are affected by other variables \( u, v, \) and \( w \), adding layers to the computational graph. The fun and challenge of multivariable calculus lie in unraveling these relationships and understanding how the changing input variables map through complex compositions to influence the function as a whole.

Function Composition

Function composition involves nesting one function inside another. This can make understanding how inputs affect outputs more complex, but also more fascinating. For the problem at hand, you’re tasked with finding partial derivatives of a composed function. Here's how composition figures into our scenario:

• The function \( z = f(x, y) = x^2 + xy^3 \) is composed with functions \( x = g(u, v, w) = uv^2 + w^3 \) and \( y = h(u, v) = u + ve^m \).
• This means any change in \( u, v, \) or \( w \) affects \( x \) and \( y \) first, subsequently altering \( z \).

The key to function composition is understanding how these inner functions impact the outer function. This is managed using the chain rule, which breaks down the derivative of the outer function with respect to the inner ones, offering a clear map of influences. In real-world applications, function composition is everywhere, from physics to computer graphics, making its mastery vital in fields where multivariable calculus is applied.