Problem 22

Question

Find $\partial w / \partial u$ and $\partial w / \partial v$. $$ \begin{aligned} &w=x^{2}-2 y-7 z ; x=v \cos (\pi-u) \\ &y=u \sin (\pi-v), z=u v \end{aligned} $$

Step-by-Step Solution

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Answer

$ \frac{\partial w}{\partial u} = 2x(v \sin(\pi-u)) - 2 \sin(\pi-v) - 7v $, $ \frac{\partial w}{\partial v} = 2x \cos(\pi-u) + 2u \cos(\pi-v) - 7u $.

1Step 1: Understand the Function w

From the given equations, we have a function $ w = x^2 - 2y - 7z $. Here, $ x = v \cos(\pi-u) $, $ y = u \sin(\pi-v) $, and $ z = uv $. Our task is to find the partial derivatives of $ w $ with respect to $ u $ and $ v $.

2Step 2: Partial Derivative of w with respect to u

Firstly, differentiate each component of $ w $ with respect to $ u $ while using the chain rule:- For $ x^2 $, note that $ x = v \cos(\pi-u) $. Therefore, $ \frac{\partial x}{\partial u} = v \sin(\pi-u) $. Then, $ \frac{\partial}{\partial u}(x^2) = 2x\frac{\partial x}{\partial u} = 2x(v \sin(\pi-u)) $.- For $ -2y $, we have $ y = u \sin(\pi-v) $. Thus, $ \frac{\partial y}{\partial u} = \sin(\pi-v) $, and so $ \frac{\partial}{\partial u}(-2y) = -2 \sin(\pi-v) $.- For $ -7z $, $ z = uv $ gives $ \frac{\partial z}{\partial u} = v $, therefore $ \frac{\partial}{\partial u}(-7z) = -7v $.Combine these to get: \[ \frac{\partial w}{\partial u} = 2x(v \sin(\pi-u)) - 2 \sin(\pi-v) - 7v \]

3Step 3: Partial Derivative of w with respect to v

Now differentiate $ w $ with respect to $ v $:- For $ x^2 $, remember $ \frac{\partial x}{\partial v} = \cos(\pi-u) $. So, $ \frac{\partial}{\partial v}(x^2) = 2x \cos(\pi-u) $.- For $ -2y $, we have $ \frac{\partial y}{\partial v} = -u \cos(\pi-v) $. Therefore, $ \frac{\partial}{\partial v}(-2y) = 2u \cos(\pi-v) $.- For $ -7z $, since $ \frac{\partial z}{\partial v} = u $, then $ \frac{\partial}{\partial v}(-7z) = -7u $.Combine these to find: \[ \frac{\partial w}{\partial v} = 2x \cos(\pi-u) + 2u \cos(\pi-v) - 7u \]

4Step 4: Conclusion

The partial derivative of $ w $ with respect to $ u $ is $ \frac{\partial w}{\partial u} = 2x(v \sin(\pi-u)) - 2 \sin(\pi-v) - 7v $. The partial derivative with respect to $ v $ is $ \frac{\partial w}{\partial v} = 2x \cos(\pi-u) + 2u \cos(\pi-v) - 7u $.

Key Concepts

chain rulemultivariable calculusdifferentiation

chain rule

The chain rule is a fundamental concept in calculus, crucial when dealing with composite functions. It helps us find the derivative of a function with respect to a variable that indirectly affects it through other functions.
In the context of partial derivatives, the chain rule allows us to break down complex dependencies between variables into manageable steps.
Consider the function given, where each term in the expression for $ w $ depends on $ x, y, $ and $ z $. Each of these variables is expressed in terms of $ u $ and $ v $. Using the chain rule, we determine how changes in $ u $ or $ v $ influence $ w $ through $ x, y, $ and $ z $.

For $ x^2 $, start with differentiating $ x = v \cos(\pi - u) $ with respect to $ u $ to obtain $ v \sin(\pi - u) $.
This derivative is then multiplied by the derivative of $ x^2 $ with respect to $ x $, which is $ 2x $.

By following such steps, we use the chain rule to handle composite dependencies systematically.

multivariable calculus

Multivariable calculus extends the principles of single-variable calculus to functions of several variables. It is essential in various fields, from physics to engineering, where systems often involve multiple interdependent variables.
In this exercise, we encountered a function $ w $ that depends on multiple variables: $ x, y, $ and $ z $. These, in turn, depend on additional variables $ u $ and $ v $. This hierarchical structure is typical in multivariable calculus problems.

We use operations like partial differentiation to analyze each variable's impact on the function independently.
Understanding these relationships helps us predict how small changes in some variables affect others within a system.

The complexity increases as we incorporate more variables, but the core concept remains rooted in breaking down functions into their constituent parts and examining their direct relationships.

differentiation

Differentiation is the process of finding the derivative, which measures how a function changes as its input changes.
In multivariable calculus, differentiation extends to include partial derivatives, which measure the rate of change of a function with respect to one variable while keeping others constant. This is particularly useful in applications where variables are interrelated, like in the given function $ w $.

Partial derivatives provide insight into how the function changes concerning each independent variable. For $ \frac{\partial w}{\partial u} $, we determine how changes in $ u $ affect $ w $, treating $ v $ as a constant.
Similarly, finding $ \frac{\partial w}{\partial v} $ involves analyzing how $ w $ changes with respect to $ v $, with $ u $ held steady.

This approach allows us to explore each dimension of change individually, making it invaluable for understanding multidimensional systems. Differentiation, thus, provides powerful tools for evaluating intricate dependencies in complex functions.

Problem 22

Other exercises in this chapter

Problem 21

Sketch the graph of $f$. $f(x, y)=\sqrt{4-x^{2}-y^{2}}$

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Problem 22

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ k(u, v)=(u+v)^{

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Problem 22

Determine $d f$. $$ f(x, y, z)=\frac{x}{x^{2}+y^{2}+z^{2}} $$

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Problem 22

From (1) it follows that the directional derivative of a function $f$ at a point is smallest in the direction opposite to the gradient of $f$ at that point.

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