Question

If \(z=x^{2}+2 y^{2}, x=r \cos \theta, y=r \sin \theta,\) find the following partial derivatives. Repeat if \(z=r^{2} \tan ^{2} \theta\). $$\frac{\partial^{2} z}{\partial r \partial \theta}$$

Short answer

For \ z = x^2 + 2y^2: \ \frac{\partial^2 z}{\partial r \partial \theta} = 4r \cos \theta \sin \theta. \ For \ z = r^2 \tan^2 \theta: \ \frac{\partial^2 z}{\partial r \partial \theta} = 4r \tan \theta \sec^2 \theta.

Step-by-step solution

Convert Functions to Polar Coordinates

Given functions are already expressed in polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute these into the function for \(z\).

Rewrite First Function in Polar Coordinates

For the first function, substitute \(x\) and \(y\) into \(z = x^2 + 2y^2\): \[z = (r \cos \theta)^2 + 2(r \sin \theta)^2\]\[z = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\]\[z = r^2 ( \cos^2 \theta + 2 \sin^2 \theta )\].

Calculate First Partial Derivative with Respect to \( r \)

Differentiate \(z\) with respect to \(r\) while treating \(\theta\) as a constant:\[\frac{\partial z}{\partial r} = \frac{\partial}{\partial r}\left[r^2 ( \cos^2 \theta + 2 \sin^2 \theta )\right]\]\[\frac{\partial z}{\partial r} = 2r (\cos^2 \theta + 2 \sin^2 \theta)\].

Calculate Second Partial Derivative with Respect to \(\theta\)

Differentiate the result with respect to \(\theta\):\[\frac{\partial}{\partial \theta} \left[ 2r ( \cos^2 \theta + 2 \sin^2 \theta ) \right]\]Use the chain rule:\[\frac{\partial}{\partial \theta} [ 2r \cos^2 \theta ] + \frac{\partial}{\partial \theta} [ 4r \sin^2 \theta ]\]\[-4r \cos \theta \sin \theta + 8r \cos \theta \sin \theta\]\[= 4r \cos \theta \sin \theta\].

Rewrite Second Function in Polar Coordinates

Given \(z = r^2 \tan^2 \theta\), it is already in polar form.

Calculate First Partial Derivative with Respect to \( r \)

Differentiate \(z\) with respect to \(r\) while treating \(\theta\) as a constant:\[\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 \tan^2 \theta]\]\[\frac{\partial z}{\partial r} = 2r \tan^2 \theta\].

Calculate Second Partial Derivative with Respect to \(\theta\)

Differentiate the result with respect to \(\theta\):\[\frac{\partial}{\partial \theta} [2r \tan^2 \theta]\]Use the chain rule:\[2r \cdot 2\tan \theta \cdot \sec^2 \theta\]\[= 4r \tan \theta \sec^2 \theta\].

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points in the plane using a distance and an angle. Instead of using traditional Cartesian coordinates (\(x, y\)), we use \(r\) and \(\theta\), where \(r\) represents the radius (distance from the origin) and \(\theta\) represents the angle from the positive \(x\)-axis. This system is especially useful when dealing with problems involving circular or rotational symmetry.

Here's how you can convert between Cartesian and polar coordinates:

• \textbf{From Cartesian to Polar}: Given \(x\) and \(y\), find \(r\) and \(\theta\) using:
  \[r = \sqrt{x^2 + y^2} \]
  \[\theta = \arctan\left(\frac{y}{x}\right)\]
• \textbf{From Polar to Cartesian}: Given \(r\) and \(\theta\), find \(x\) and \(y\) using:
  \[x = r \cos \theta \]
  \[y = r \sin \theta \]

For the given exercise, we converted \(x\) and \(y\) to polar coordinates:\
\(x = r \cos \theta\) and \(y = r \sin \theta\). This simplification helps in solving problems involving partial derivatives more efficiently.

Chain Rule

The chain rule is a fundamental calculus technique used to differentiate composite functions. It allows us to split the differentiation process into manageable steps.

To understand the chain rule, suppose you have a composite function \(z = f(g(h(x)))\). The derivative of \(z\) with respect to \(x\) is calculated as follows:
\[\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{du} \cdot \frac{du}{dx} \]
In other words, you differentiate the outer function and multiply it by the derivative of the inner function.

For partial derivatives, the chain rule helps us differentiate a function that depends indirectly on variables. In our exercise, we used the chain rule to differentiate \(z\) with respect to \(r\) and \(\theta\). For example:

• Differentiate \(z\) with respect to \(r\):
  \[\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 ( \cos^2 \theta + 2 \sin^2 \theta )] = 2r (\cos^2 \theta + 2 \sin^2 \theta) \]
• Then, differentiate with respect to \(\theta\):
  \[ \frac{\partial}{\partial \theta} [2r ( \cos^2 \theta + 2 \sin^2 \theta )] = \frac{\partial}{\partial \theta} [ 2r \cos^2 \theta] + \frac{\partial}{\partial \theta} [ 4r \sin^2 \theta ] \]

Combining the results of these steps gives us the desired second partial derivative.

Second Partial Derivatives

Second partial derivatives provide information on how a function curves with respect to its variables. Instead of just finding the first rates of change, they give insight into the rate at which these first rates change.

\[ \frac{\partial^2 z}{\partial x^2}, \frac{\partial^2 z}{\partial y^2}, \text{and} \frac{\partial^2 z}{\partial x \partial y} \]
In our problem, we specifically calculated:
\[ \frac{\partial^2 z}{\partial r \partial \theta} \]
Here's the step-by-step process to compute this without complicating it too much:

Start with the first partial derivative \(\frac{\partial z}{\partial r}\)

Then, differentiate this result with respect to the second variable, \(\theta\)

• Example from our exercise:
  \[ \frac{\partial}{\partial \theta} \[2r ( \cos^2 \theta + 2 \sin^2 \theta ) \] = -4r \cos \theta \sin \theta + 8r \cos \theta \sin \theta \text { which simplifies to } 4r \cos \theta \sin \theta\]

Second partial derivatives can sometimes indicate points of inflection or extremum points of the function. This deeper understanding is crucial while studying the behavior of functions.

Function Transformation

Function transformations help us understand better how a function behaves when its inputs undergo changes. In polar coordinates especially, transformation can align complex problems into simpler scenarios.

For example, consider the transformation from Cartesian coordinates to polar form. This transformation may untangle circular or rotational symmetries much more swiftly than Cartesian representations.
Looking at our specific exercise, we started with:

• Cartesian functions:
  \( z = x^2 + 2y^2\) or \( z = r^2 \tan^2 \theta\)
• By converting these to polar form, we optimized solving their partial derivatives:
  \( z = r^2 ( \cos^2 \theta + 2 \sin^2 \theta)\) or \( z = r^2 \tan^2 \theta\)

This transformation approach not only simplifies the differentiation process but also helps visualize the function’s geometry.

Remember: If you're ever stuck on a problem, try transforming the function into another coordinate system. It could reveal symmetries or simplify calculations.