Problem 101 - Advanced Calculus

Question

Let \(F=F(x, y)\) have second partial derivatives in a region of the \(\mathrm{xy}\) -plane and introduce polar coordinates in this region by writing \(\mathrm{x}=\mathrm{r} \cos \theta, \mathrm{y}=\mathrm{r} \sin \theta\). Find \(\partial^{2} \mathrm{~F} / \partial \mathrm{r} \partial \theta\) in terms of derivatives of \(\mathrm{F}\) with respect to \(\mathrm{x}\) and \(\mathrm{y}\).

Step-by-Step Solution

Answer

The second-order mixed partial derivative of \(F(x, y)\) with respect to r and θ is: \(\frac{\partial^2 F}{\partial r\partial \theta}=-\sin\theta\frac{\partial^2 F}{\partial x^2}+r\cos\theta\frac{\partial^2 F}{\partial x\partial y}-\cos\theta\frac{\partial^2 F}{\partial y^2}-\sin\theta\frac{\partial F}{\partial x}\)

Step 1: Find the partial derivatives of x and y with respect to r and θ.

As given, we have x = r cos θ and y = r sin θ. Differentiate both equations with respect to r and θ: \(\frac{\partial x}{\partial r} = \cos \theta\) \(\frac{\partial x}{\partial \theta} = -r \sin \theta\) \(\frac{\partial y}{\partial r} = \sin \theta\) \(\frac{\partial y}{\partial \theta} = r \cos \theta\)

Step 2: Derive the first-order partial derivatives of F with respect to r and θ using the chain rule.

Now, we can find the first-order partial derivatives of F with respect to r and θ using the chain rule: \(\frac{\partial F}{\partial r} = \frac{\partial F}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial r} = (\frac{\partial F}{\partial x}\cos\theta) + (\frac{\partial F}{\partial y}\sin\theta)\) \(\frac{\partial F}{\partial \theta} = \frac{\partial F}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial \theta} = -(\frac{\partial F}{\partial x}r\sin\theta) + (\frac{\partial F}{\partial y}r\cos\theta)\)

Step 3: Use the chain rule again to find the second-order mixed partial derivative of F with respect to r and θ.

Finally, we can use the chain rule again to find ∂²F/∂r∂θ, using the derivatives found in Step 2: \(\frac{\partial^2 F}{\partial r\partial \theta} = \frac{\partial}{\partial r}\frac{\partial F}{\partial \theta} = \frac{\partial}{\partial r}(-(\frac{\partial F}{\partial x}r\sin\theta) + (\frac{\partial F}{\partial y}r\cos\theta))\) Now, differentiate with respect to r: \(\frac{\partial^2 F}{\partial r\partial \theta}=-\sin\theta\frac{\partial^2 F}{\partial x^2}+r\cos\theta\frac{\partial^2 F}{\partial x\partial y}-\cos\theta\frac{\partial^2 F}{\partial y^2}-\sin\theta\frac{\partial F}{\partial x}\) Thus, we have found the second-order mixed partial derivative ∂²F/∂r∂θ in terms of the derivatives of F with respect to x and y.

Key Concepts

Polar Coordinates

Polar coordinates are a way to describe the location of a point in a plane using a radius and an angle. Unlike Cartesian coordinates, which use horizontal and vertical lines for positions, polar coordinates use a circular grid.Key elements of polar coordinates include:

• Radius (r): The distance from the origin (center of the grid) to the point.
• Angle (θ): The angle measured counterclockwise from the positive x-axis to the line segment connecting the origin to the point.

Converting between Cartesian and polar coordinates is essential. You'll often see equations like:

• Using polar coordinates: \( x = r \cos \theta \) and \( y = r \sin \theta \)
• Switching back to Cartesian form involves: \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)

Polar coordinates are particularly useful in solving problems involving circular paths or rotational movements.

Chain Rule

The chain rule is a fundamental rule in calculus for finding the derivative of composite functions. When dealing with functions defined in polar coordinates, the chain rule helps us express derivatives with respect to new variables, such as \( r \) and \( \theta \), in terms of derivatives with respect to Cartesian coordinates, \( x \) and \( y \).Here's how the chain rule works:

• For a function \( F(x, y) \), the partial derivative with respect to \( r \) can be expressed as:\[\frac{\partial F}{\partial r} = \frac{\partial F}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial F}{\partial y} \cdot \frac{\partial y}{\partial r}\]The formula combines derivatives obtained by differentiating \( x \, \text{and} \, y \) with respect to \( r \).
• Similarly, for \( \theta \):\[\frac{\partial F}{\partial \theta} = \frac{\partial F}{\partial x} \cdot \frac{\partial x}{\partial \theta} + \frac{\partial F}{\partial y} \cdot \frac{\partial y}{\partial \theta}\]This helps transition from Cartesian to polar coordinates, thus simplifying integration or differentiation when problems are naturally circular or radial.

By using the chain rule correctly, you can compute necessary derivatives efficiently in a transformed coordinate space.

Mixed Partial Derivatives

Mixed partial derivatives involve second-order derivatives where the function is differentiated with respect to multiple variables. For instance, if \( F \) is a function of \( x \) and \( y \), then a mixed partial derivative could be \( \frac{\partial^2 F}{\partial x \partial y} \).Special attributes of mixed partial derivatives include:

• They often test how a function's rate of change with respect to one variable affects its rate of change with respect to another.
• Under certain conditions (if both are continuous), the order of differentiation does not matter, which means:\[\frac{\partial^2 F}{\partial x \partial y} = \frac{\partial^2 F}{\partial y \partial x}\]
• In the solution provided, mixed partial derivatives are used to compute \( \frac{\partial^2 F}{\partial r \partial \theta} \). Using the chain rule, we understand changes in \( F \) between the radial direction and the angular direction in polar coordinates.

Mixed partial derivatives are valuable in analyzing the behavior of multivariable functions, especially in areas of cross-influence between different directions of change.