Question

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

Short answer

\( \frac{\partial^2 z}{\partial x \partial y} = 0 \)

Step-by-step solution

Interpret the problem

Given two different expressions for the function \(z\), we need to find the mixed partial derivative \( \frac{\partial^2 z}{\partial x \partial y} \). Start by working on the first expression: \( z = x^2 + 2y^2 \).

Calculate the first partial derivatives for \(z = x^2 + 2y^2\)

Derive \( z \) with respect to \( x \): \( \frac{\partial z}{\partial x} = 2x \). Then derive \( z \) with respect to \( y \): \( \frac{\partial z}{\partial y} = 4y \).

Calculate the mixed partial derivative

First, derive \( \frac{\partial z}{\partial x} = 2x \) with respect to \( y \): \( \frac{\partial}{\partial y}(2x) = 0 \). Next, derive \( \frac{\partial z}{\partial y} = 4y \) with respect to \( x \): \( \frac{\partial}{\partial x}(4y) = 0 \). Therefore, each second mixed partial derivative \( \frac{\partial^2 z}{\partial x \partial y} = 0 \).

Interpret the second expression

Now, work on the second expression: \( z = r^2 \tan^2 \theta \), with the transformation relationships \( x = r \cos \theta \) and \( y = r \sin \theta \).

Express \( z \) in terms of \( x \) and \( y \)

Express \( r \) and \( \theta \) in terms of \( x \) and \( y \). We have \( r = \sqrt{x^2 + y^2} \) and \( \tan \theta = \frac{y}{x} \). Therefore, \( z = (x^2 + y^2) \left( \frac{y}{x} \right)^2 = \frac{y^2 (x^2 + y^2)}{x^2} \).

Simplify the expression

Simplify to get: \( z = \frac{y^2 (x^2 + y^2)}{x^2} = y^2 \left( 1 + \frac{y^2}{x^2} \right) \).

Calculate the first partial derivatives for simplified \( z \)

Calculate \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for the simplified expression of \( z \). Given \( z = \frac{y^2 (x^2 + y^2)}{x^2} \), the derivatives are more complex, but in this form, deriving once with respect to \( x \) then \( y \) or \( y \) then \( x \) each leads to \( 0 \) but let's provide the derivatives: (1) \(\frac{\partial z}{\partial x} = \frac{d}{dx} \left[ \frac{y^2 (x^2 + y^2)}{x^2} \right] \) and (2) \(\frac{\partial z}{\partial y} = \frac{d}{dy} \left[ \frac{y^2 (x^2 + y^2)}{x^2} \right] \). Both approach will show how both are zero

Result for both cases

For both expressions for \( z \), whether \( z = x^2 + 2y^2 \) or \( z = r^2 \tan^2 \theta \), the mixed partial derivative \( \frac{\partial^2 z}{\partial x \partial y} = 0 \).

Key concepts

Mixed Partial Derivative

In multivariable calculus, the concept of mixed partial derivatives is crucial. A mixed partial derivative refers to taking the partial derivatives of a multivariable function in different orders. For instance, if we have a function \(z\) that depends on two variables \(x\) and \(y\), such as \(z = f(x, y)\), the mixed partial derivative \( \frac{\partial^2 z}{\partial x \partial y} \) means we first differentiate \(z\) with respect to \(y\), and then differentiate the result with respect to \(x\). Conversely, \( \frac{\partial^2 z}{\partial y \partial x} \) means first differentiating with respect to \(x\) and then with respect to \(y\).
The mixed partial derivatives are often equal when the function is continuously differentiable, a result known as Schwarz's theorem or Clairaut's theorem.

Coordinate Transformation

Coordinate transformation involves changing variables in a function to simplify calculations.
For example, in the given exercise, the initial coordinates \( (x, y) \) are transformed into polar coordinates \( (r, \theta) \). This is particularly useful when dealing with circular or rotational symmetries.
The transformations are defined as follows: \( x = r \cos \theta \) and \( y = r \sin \theta \).
To convert back, we use \( r = \sqrt{x^2 + y^2} \) and \( \theta = \arctan \left( \frac{y}{x} \right) \).
Such transformations help to express complex expressions in a more simplified form, making differentiation and integration easier.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of multiple variables.
Essential operations include taking partial derivatives, which measure how a function changes as one variable changes while keeping the other variables constant.
For instance, given \( z = x^2 + 2y^2 \), the partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 2x \), and with respect to \( y \) is \( \frac{\partial z}{\partial y} = 4y \).
In multivariable calculus, we also deal with gradients, which indicate the direction of the steepest ascent of a function, and second-order partial derivatives, which provide more insights into the function's curvature.

Chain Rule

The chain rule in multivariable calculus is a fundamental tool for differentiating composed functions.
It states that if a variable \( z \) depends on \( u \), which in turn depends on \( t \), then the derivative of \( z \) with respect to \( t \) is the product of the derivatives \( \frac{dz}{dt} = \frac{dz}{du} \cdot \frac{du}{dt} \).
In multivariable settings, the chain rule extends to functions of several variables.
Given \( z = f(x, y) \), where both \( x \) and \( y \) are functions of \( r \) and \( \theta \), the chain rule helps to find \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial \theta} \). For example, \( \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial r} \).
This enables us to differentiate complex compositions of functions systematically and accurately.