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 the first function: \( \frac{\partial^2 z}{\partial r \partial \theta} = 2r \sin(2\theta) \). For the second function: \( \frac{\partial^2 z}{\partial r \partial \theta} = 4r \tan \theta \sec^2 \theta \).

Step-by-step solution

Convert to polar coordinates

First, convert the given function into polar coordinates. For the first function: \( z = x^2 + 2y^2 \). Using \( x = r \cos \theta \) and \( y = r \sin \theta \), we get: \[ z = (r \cos \theta)^2 + 2(r \sin \theta)^2 = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta \]. Simplify this to obtain: \[ z = r^2 ( \cos^2 \theta + 2 \sin^2 \theta ) = r^2 (1 + \sin^2 \theta) \]

Calculate partial derivatives for first function

Now, find the partial derivatives: \( \frac{\partial z}{\partial r} = \frac{\partial}{\partial r} \left[ r^2 (1 + \sin^2 \theta) \right] = 2r (1 + \sin^2 \theta) \)\( \frac{\partial z}{\partial \theta} = \frac{\partial}{\partial \theta} \left[ r^2 (1 + \sin^2 \theta) \right] = r^2 \frac{\partial}{\partial \theta} [1 + \sin^2 \theta] = r^2 \cdot 2\sin \theta \cos \theta = r^2 \sin(2\theta) \)

Calculate mixed partial derivative for first function

To find the mixed partial derivative, we continue from previous results:\( \frac{\partial^2 z}{\partial r \partial \theta} = \frac{\partial}{\partial \theta} \left( 2r (1 + \sin^2 \theta) \right) = 2r \cdot 2 \sin \theta \cos \theta = 2r \sin(2\theta) \)

Convert second function to polar coordinates

Now, consider the second function: \( z = r^2 \tan^2 \theta \). This function is already in polar coordinates, so no further conversion is necessary.

Calculate partial derivatives for second function

Find the partial derivatives for the second function:\( \frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 \tan^2 \theta] = 2r \tan^2 \theta \)\( \frac{\partial z}{\partial \theta} = \frac{\partial}{\partial \theta} [r^2 \tan^2 \theta] = r^2 \cdot 2 \tan \theta (\sec^2 \theta) = 2r^2 \tan \theta \sec^2 \theta \)

Calculate mixed partial derivative for second function

To find the mixed partial derivative for the second function, differentiate \( \frac{\partial z}{\partial r} \) with respect to \(\theta\):\( \frac{\partial^2 z}{\partial r \partial \theta} = \frac{\partial}{\partial \theta} [2r \tan^2 \theta] = 2r \cdot 2 \tan \theta \sec^2 \theta = 4r \tan \theta \sec^2 \theta \)

Key concepts

polar coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is defined by a distance from a reference point (called the origin) and an angle from a reference direction. The distance is denoted by \( r \) and the angle by \( \theta \). Converting Cartesian coordinates \( (x, y) \) to polar coordinates involves the equations: \( x = r \cos \theta \) and \( y = r \sin \theta \). This transformation is particularly useful in problems with symmetry about a point, as it simplifies the mathematics involved. For example, transforming the function \( z = x^2 + 2y^2 \) using \( x = r \cos \theta \) and \( y = r \sin \theta \) results in \( z = r^2 (1 + \sin^2 \theta) \), which can be easier to differentiate in specific cases.

mixed partial derivatives

Mixed partial derivatives involve taking partial derivatives of a function with respect to two different variables. For example, if we have a function \( z = f(r, \theta) \), the mixed partial derivative \( \frac{\partial^2 z}{\partial r \partial \theta} \) involves first differentiating \( z \) with respect to \( r \) and then with respect to \( \theta \). These mixed derivatives are crucial when analyzing the behavior of multivariable functions, as they provide insights into how the function changes as we move along different variable axes. It's important to note that under certain conditions (specifically, if the function is continuously differentiable), the order of differentiation does not matter, i.e., \( \frac{\partial^2 z}{\partial r \partial \theta} = \frac{\partial^2 z}{\partial \theta \partial r} \).

Jacobian transformation

The Jacobian transformation is a method used to change variables in multiple integrals. It involves the determinant of the Jacobian matrix, which is composed of the partial derivatives of the new variables with respect to the old ones. In transforming to polar coordinates, the Jacobian determinant, given by \( \frac{\partial (x, y)}{\partial (r, \theta)} = r \), helps convert integrals from Cartesian coordinates to polar coordinates. This transformation is essential in various fields such as physics and engineering, where working with radial symmetry simplifies the problem. When dealing with the mixed partial derivatives of functions expressed in polar coordinates, the Jacobian determinant can ensure that the transformation is properly accounted for.

chain rule in multivariable calculus

The chain rule in multivariable calculus extends the chain rule from single-variable calculus. It is used to differentiate composite functions. If a function \( z = f(x, y) \) depends on variables \( x \) and \( y \), which in turn depend on other variables \( r \) and \( \theta \), then the partial derivative of \( z \) with respect to \( r \) or \( \theta \) involves using the chain rule. Specifically, when converting to polar coordinates, we have: \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial r} \) and \( \frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial \theta} \). This allows for systematic calculation of derivatives when dealing with composite functions, ensuring each dependency is properly accounted for. By applying the chain rule, we can accurately find partial and mixed derivatives for functions expressed in different coordinate systems.