Problem 21 In Exercises $19-22,$ function... [FREE SOLUTION]

Chapter 12: Problem 21

In Exercises $19-22,$ functions $z=f(x, y), x=g(s, t)$ and $y=h(s, t)$ are given. (a) Use the Multivariable Chain Rule to compute $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ (b) Evaluate $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ at the indicated $s$ and $t$ values.$ 12. $$ z=x^{2}+y^{2}, \quad x=s \cos t, \quad y=s \sin t ; \quad s=2, t=\pi / 4 $$

Short Answer

Expert verified

$\frac{\partial z}{\partial s} = 4$ and $\frac{\partial z}{\partial t} = 0$.

Step by step solution

Identify the Given Functions

We are given the function $ z = f(x, y) = x^2 + y^2 $ and the parametric functions $ x = g(s, t) = s \cos t $ and $ y = h(s, t) = s \sin t $. We need to use these to find the partial derivatives $ \frac{\partial z}{\partial s} $ and $ \frac{\partial z}{\partial t} $.

Apply the Multivariable Chain Rule for $\frac{\partial z}{\partial s}$

Using the multivariable chain rule, we find:\[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \]Compute each partial derivative:- $ \frac{\partial z}{\partial x} = 2x $- $ \frac{\partial z}{\partial y} = 2y $- $ \frac{\partial x}{\partial s} = \cos t $- $ \frac{\partial y}{\partial s} = \sin t $Substitute these into the chain rule expression:\[ \frac{\partial z}{\partial s} = 2x \cdot \cos t + 2y \cdot \sin t \]

Apply the Multivariable Chain Rule for $\frac{\partial z}{\partial t}$

Using the multivariable chain rule, we find:\[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \]Compute each partial derivative:- $ \frac{\partial x}{\partial t} = -s \sin t $- $ \frac{\partial y}{\partial t} = s \cos t $Substitute these into the chain rule expression:\[ \frac{\partial z}{\partial t} = 2x \cdot (-s \sin t) + 2y \cdot (s \cos t) \]

Substitute Calculated Values into Derivative Formulas

First, evaluate $ x $ and $ y $ at $ s = 2 $ and $ t = \frac{\pi}{4} $:- $ x = 2 \cos \frac{\pi}{4} = \sqrt{2} $- $ y = 2 \sin \frac{\pi}{4} = \sqrt{2} $Substitute $ x = \sqrt{2} $ and $ y = \sqrt{2} $ into the expression for $ \frac{\partial z}{\partial s} $:\[ \frac{\partial z}{\partial s} = 2(\sqrt{2}) \cdot \frac{1}{\sqrt{2}} + 2(\sqrt{2}) \cdot \frac{1}{\sqrt{2}} = 4 \]Do the same for $ \frac{\partial z}{\partial t} $:\[ \frac{\partial z}{\partial t} = 2(\sqrt{2})(-2\cdot\frac{1}{\sqrt{2}}) + 2(\sqrt{2})(2\cdot\frac{1}{\sqrt{2}}) = 0 \]

Final Result

The partial derivatives, evaluated at $ s = 2 $ and $ t = \frac{\pi}{4} $, are:- $ \frac{\partial z}{\partial s} = 4 $- $ \frac{\partial z}{\partial t} = 0 $

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule

The Chain Rule in multivariable calculus is a method to find derivatives of functions that are composed of other functions. When a function, say $ z = f(x, y) $, is dependent on variables that are themselves functions of other variables, such as $ x = g(s, t) $ and $ y = h(s, t) $, the Chain Rule helps us find how a change in $ s $ or $ t $ affects $ z $.

For the function $ z = x^2 + y^2 $ with $ x = s \cos t $ and $ y = s \sin t $, we apply the multivariable Chain Rule to differentiate $ z $ with respect to $ s $ and $ t $.

To find $ \frac{\partial z}{\partial s} $, use:
$ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} $
To find $ \frac{\partial z}{\partial t} $, use:
$ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} $

These formulas break down the effects of changes in $ s $ and $ t $ into their contributions through $ x $ and $ y $.

Partial Derivatives

Partial derivatives are a way to study the change in multivariable functions concerning one variable at a time, keeping other variables constant. In our example, the function $ z = x^2 + y^2 $ includes variables $ x $ and $ y $ which each depend on $ s $ and $ t $.

To calculate partial derivatives for $ z $:

$ \frac{\partial z}{\partial x} = 2x $: This derivative follows from the power rule, considering $ y $ is constant.
$ \frac{\partial z}{\partial y} = 2y $: Similarly, by keeping $ x $ constant, this derivative is straightforward.

These derivatives allow us to understand how changes in $ x $ or $ y $ influence $ z $ directly.

Parametric Equations

Parametric equations express a set of related quantities as explicit functions of one or more independent parameters. In our exercise, the equations $ x = s \cos t $ and $ y = s \sin t $ are parametric because they describe $ x $ and $ y $ parameters $ s $ and $ t $.

Using these equations, we can:

Translate trigonometric inputs into spatial coordinates.
Relate the circular motion described by $ \cos t $ and $ \sin t $ with the magnitude controlled by $ s $.

These parametric descriptions help transform the original function $ z $ so that it can be analyzed with respect to $ s $ and $ t $.

Function Evaluation

Function evaluation involves substituting specific values into a function to find the corresponding output. In the solution, after finding the expressions for $ \frac{\partial z}{\partial s} $ and $ \frac{\partial z}{\partial t} $, we evaluate these partial derivatives at $ s = 2 $ and $ t = \frac{\pi}{4} $.

The process includes:

Calculating $ x $ and $ y $ at these values: $ x = \sqrt{2} $ and $ y = \sqrt{2} $.
Substituting $ x $ and $ y $ into our expressions for the derivatives.
Evaluating the expression to get $ \frac{\partial z}{\partial s} = 4 $ and $ \frac{\partial z}{\partial t} = 0 $.

This evaluation demonstrates how a theoretical mathematical solution applies to concrete numbers.

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Short Answer

Step by step solution

Identify the Given Functions

Apply the Multivariable Chain Rule for \(\frac{\partial z}{\partial s}\)

Apply the Multivariable Chain Rule for \(\frac{\partial z}{\partial t}\)

Substitute Calculated Values into Derivative Formulas

Final Result

Key Concepts

Chain Rule

Partial Derivatives

Parametric Equations

Function Evaluation

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