Problem 21 - APEX Calculus

Question

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. $$ z=x^{2}+y^{2}, \quad x=s \cos t, \quad y=s \sin t ; \quad s=2, t=\pi / 4 $$

Step-by-Step Solution

Answer

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

Step 1: 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 \( \frac{\partial z}{\partial t} \).

Step 2: 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:\[ \frac{\partial z}{\partial s} = 2x \cdot \cos t + 2y \cdot \sin t \]

Step 3: 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:\[ \frac{\partial z}{\partial t} = 2x \cdot (-s \sin t) + 2y \cdot (s \cos t) \]

Step 4: 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 \( \frac{\partial z}{\partial t} \):\[ \frac{\partial z}{\partial t} = 2(\sqrt{2})(-2\cdot\frac{1}{\sqrt{2}}) + 2(\sqrt{2})(2\cdot\frac{1}{\sqrt{2}}) = 0 \]

Step 5: Final Result

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

Key Concepts

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.