Problem 220 - Calculus 3

Question

For the following exercises, use the information provided to solve the problem. Suppose \(f(x, y)=x+y, u=e^{x} \sin y, x=t^{2},\) and \(y=\pi t, \quad\) where \(x=r \cos \theta\) and \(y=r \sin \theta .\) Find \(\frac{\partial f}{\partial \theta}\)

Step-by-Step Solution

Answer

The partial derivative is \(-r\sin\theta + r\cos\theta\).

Step 1: Express Variables in Terms of \(\theta\)

We have \(x = r \cos \theta\) and \(y = r \sin \theta\). Thus, the function \(f(x, y) = x + y\) becomes \(f(r\cos\theta, r\sin\theta) = r\cos\theta + r\sin\theta\).

Step 2: Find \(\frac{\partial f}{\partial \theta}\)

To find \(\frac{\partial f}{\partial \theta}\), return to the expression \(f(x, y) = f(r\cos\theta, r\sin\theta) = r\cos\theta + r\sin\theta\). Compute its derivative with respect to \(\theta\):\[\frac{\partial f}{\partial \theta} = \frac{\partial}{\partial \theta}(r\cos\theta + r\sin\theta) = -r\sin\theta + r\cos\theta.\]

Key Concepts

Partial Derivatives

In multivariable calculus, understanding the concept of partial derivatives is crucial. The partial derivative of a function represents the rate of change of the function with respect to one of its variables, while keeping the other variables constant. This is akin to analyzing how a function reacts when one specific variable undergoes a slight change.
To compute a partial derivative, \(\frac{\partial f}{\partial x}\), fundamentally means taking the derivative of the function \(f(x, y)\) with respect to \(x\) while treating \(y\) as a constant.

• The notation \(\frac{\partial}{\partial \theta}\) in our problem signifies that we are examining changes in \(f\) as \(\theta\) changes.
• For \(f(x, y) = x + y\), we represent it in terms of \(r\) and \(\theta\) to find \(\frac{\partial f}{\partial \theta}\), focusing solely on how \(f(x, y)\) changes with respect to \(\theta\).

Polar Coordinates

Polar coordinates provide a different framework from the Cartesian system to describe the location of a point in a plane. Instead of using \((x, y)\) coordinates, polar coordinates use \((r, \theta)\), where \(r\) denotes the distance from the origin and \(\theta\) represents the angle with the positive x-axis.
This system is especially useful for dealing with problems where functions naturally incorporate angles and radii, such as circular motions.

• In the exercise, the conversion to polar coordinates is crucial for finding \(\frac{\partial f}{\partial \theta}\).
• The conversions \(x = r \cos \theta\) and \(y = r \sin \theta\) enable the expression of \(f(x, y)\) in terms of \((r, \theta)\).
• This representation simplifies taking partial derivatives with respect to \(\theta\).

Understanding polar coordinates allows for straightforward computation and analysis of problems involving circular or rotational symmetry.

Chain Rule

The chain rule is an essential tool in calculus used to differentiate complex functions. When dealing with compositions of functions where one variable is dependent on others, the chain rule helps in finding derivatives efficiently.
In multivariable settings, the chain rule becomes indispensable when differentiating a function that depends on intermediary variables that are functions of other variables.

• In our problem, while finding \(\frac{\partial f}{\partial \theta}\), the chain rule helps manage the dependency between \(x = r\cos\theta\) and \(y = r\sin\theta\).
• This allows us to compute how \(f(x, y)\) changes as \(\theta\) varies indirectly via \(x\) and \(y\).
• By focusing on the composition of these functions, the chain rule streamlines the calculation, ensuring accuracy and efficiency.

Mastering the chain rule enhances one’s ability to deal with the interactions between multiple changing variables.