Problem 219 - Calculus 3

Question

For the following exercises, use the information provided to solve the problem. If \(f(x, y)=x y, x=r \cos \theta,\) and \(y=r \sin \theta,\) find \(\frac{\partial f}{\partial r}\) and express the answer in terms of \(r\) and \(\theta\)

Step-by-Step Solution

Answer

\(\frac{\partial f}{\partial r} = 2r \cos \theta \sin \theta\)

Step 1: Substitute Polar Coordinates into f

The function given is \(f(x, y) = xy\). Using the substitutions for polar coordinates, \(x = r \cos \theta\) and \(y = r \sin \theta\), we rewrite \(f(x, y)\) as \(f(r, \theta) = (r \cos \theta)(r \sin \theta)\).

Step 2: Simplify the Expression

Multiply the expressions from the substitution: \((r \cos \theta)(r \sin \theta) = r^2 \cos \theta \sin \theta\). So, the function in polar coordinates is \(f(r, \theta) = r^2 \cos \theta \sin \theta\).

Step 3: Differentiate w.r.t r

Find the partial derivative of \(f(r, \theta) = r^2 \cos \theta \sin \theta\) with respect to \(r\).\[\frac{\partial f}{\partial r} = \frac{\partial}{\partial r} (r^2 \cos \theta \sin \theta)\]This is a simple power rule differentiation:\[\frac{\partial f}{\partial r} = 2r \cos \theta \sin \theta\]

Step 4: Express the Solution

We now have the solution for \(\frac{\partial f}{\partial r}\) expressed in terms of \(r\) and \(\theta\):\[\frac{\partial f}{\partial r} = 2r \cos \theta \sin \theta\]

Key Concepts

Polar Coordinates

Polar coordinates are an alternative way to describe points in a plane. While rectangular coordinates use an ordered pair \((x, y)\), polar coordinates define any point using a distance \(r\) from the origin and an angle \(\theta\) measured from the positive x-axis.

• \(r\) represents the radial distance from the origin to the point.
• \(\theta\) is the angle between the positive x-axis and the line connecting the origin to the point.

To convert from rectangular coordinates to polar coordinates:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These conversions allow a function to be expressed in terms of \(r\) and \(\theta\), which can be particularly useful when dealing with circular or rotational symmetry in a problem. For instance, converting the function \(f(x, y) = xy\) to polar form helps us to leverage symmetries not apparent in the rectangular form.

Multivariable Functions

A multivariable function is a function that has more than one variable. For example, \(f(x, y)\) is a function depending on two variables, \(x\) and \(y\). These functions are essential in calculus when dealing with geometrical and physical contexts involving multiple dimensions.

• Partial derivatives are used to understand how these functions change as each variable changes independently.
• In our example, we have \(f(x, y) = xy\), where \(x\) and \(y\) are interdependent through the polar transformations.

Manipulating these functions involves understanding their behavior in respect to each variable. This understanding is crucial for fields like physics and engineering, where multivariable calculus models systems with multiple degrees of freedom.

Differentiation in Polar Form

Differentiation in polar form comes into play when expressing a multivariable function in polar coordinates and then finding its derivative. After converting a function to polar coordinates, differentiation is performed with respect to \(r\) or \(\theta\), treating the other variable as a constant. This is referred to as partial differentiation.

• In our example, the function \(f(r, \theta) = r^2 \cos \theta \sin \theta\) is differentiated with respect to \(r\), treating \(\theta\) as a constant.
• The calculated partial derivative, \(\frac{\partial f}{\partial r} = 2r \cos \theta \sin \theta\), expresses how the function changes as \(r\) changes.

This method simplifies solving problems with rotational symmetry, as identifying symmetrical patterns can often make solving more efficient and intuitive.