Question

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

Short answer

For \(z=x^2+2y^2\): \(2r^2\text{sin}(\theta)\text{cos}(\theta)\). For \(z=r^2\tan^2(\theta)\): \(2r^2\tan(\theta)\text{sec}^2(\theta)\).

Step-by-step solution

Understand the Problem

We need to find the partial derivative of two different functions, given transformations for the variables.

Express the First Function

For the first function, express \(z = x^2 + 2y^2\) in terms of \(r\) and \(\theta\) using the coordinate transformations \(x = r\text{cos}(\theta)\) and \(y = r\text{sin}(\theta)\). Substitute these into the equation: \(z = (r\text{cos}(\theta))^2 + 2(r\text{sin}(\theta))^2\).

Simplify the First Function

Simplify the expression: \(z = r^2\text{cos}^2(\theta) + 2r^2\text{sin}^2(\theta)\). Using the Pythagorean identity, this simplifies further to: \(z = r^2(\text{cos}^2(\theta) + 2\text{sin}^2(\theta))\).

Calculate the Partial Derivative of the First Function

Find the partial derivative of \(z\) with respect to \(\theta\), treating \(r\) as a constant: \( \frac{\text{d}}{\text{d}\theta} \big[r^2(\text{cos}^2(\theta) + 2\text{sin}^2(\theta)) \big] = r^2 \big[ -2\text{cos}(\theta)\text{sin}(\theta) + 4\text{sin}(\theta)\text{cos}(\theta)\big] = 2r^2\text{sin}(\theta)\text{cos}(\theta)\). Thus, \( \bigg( \frac{\text{d}z}{\text{d}\theta} \bigg)_r = 2r^2\text{sin}(\theta)\text{cos}(\theta)\).

Express the Second Function

For the second function, express \(z = r^2 \tan^2(\theta)\). Since the function is already in terms of \(r\) and \(\theta\), no further transformations are necessary.

Calculate the Partial Derivative of the Second Function

Find the partial derivative of \(z\) with respect to \(\theta\), treating \(r\) as a constant: \( \frac{\text{d}}{\text{d}\theta} \big[r^2 \tan^2(\theta)\big]\) Using the chain rule, this becomes: \( r^2 \big[ \frac{\text{d}}{\text{d}\theta} (\tan (\theta))^2 \big] = r^2 \big[ 2 \tan (\theta) \times \text{sec}^2 (\theta) \big]\). Thus, \( \bigg( \frac{\text{d}z}{\text{d}\theta} \bigg)_r = 2r^2 \tan (\theta) \text{sec}^2 (\theta) \).

Key concepts

Partial Derivatives

When we talk about partial derivatives, we're looking at how a function changes as one of its variables changes, with others held constant. For example, if we have a function of two variables, like \(z = x^2 + 2y^2\), the partial derivative with respect to \(x\) tells us how \(z\) changes as \(x\) changes, while keeping \(y\) constant. This concept is crucial in understanding the behavior of multivariable functions.
In the provided exercise, we found partial derivatives for two different functions: \(z = x^2 + 2y^2\) and \(z = r^2 \tan^2(\theta)\). This involved taking derivatives with respect to \(\theta\), treating other variables like \(r\) as constants. Through step-by-step calculations, we derived these partial derivatives using established rules, which leads us to our next core concept.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by an angle and a distance from a fixed point called the pole (commonly the origin). The angle is denoted by \(\theta\) and the distance by \(r\). Converting between Cartesian coordinates \((x, y)\) and polar coordinates \((r, \theta)\) is done using the formulas:

• \(x = r\cos(\theta)\)
• \(y = r\sin(\theta)\)

In the exercise, we used these conversions to transform the function \(z = x^2 + 2y^2\) into polar coordinates. By substituting the polar coordinate expressions into the Cartesian function, we could rewrite \(z\) in terms of \(r\) and \(\theta\). This transformation is essential because it simplifies the process of taking partial derivatives in the new coordinate system.

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composition of functions. When applying it to partial derivatives, it helps us determine how changes in one variable propagate through dependent variables. For a function expressed as \(z = f(x, y)\) where \(x\) and \(y\) are further dependent on \(r\) and \(\theta\), the chain rule looks like:
\[ \frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial \theta} \]
In our exercise, we applied the chain rule explicitly for the second function \(z = r^2 \tan^2(\theta)\). By recognizing that \(\tan(\theta)^2\) is a composition involving \(\theta\), we used the chain rule to compute the derivative, which included terms involving the basic derivative of the tangent function \(\frac{d}{d\theta} \tan(\theta) = \sec^2(\theta)\).

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the included variables. These identities are crucial tools for simplifying and solving trigonometric expressions. Some key identities include:

• Pythagorean identity: \( \text{cos}^2(\theta) + \text{sin}^2(\theta) = 1 \)
• Double-angle identities: \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)

In our exercise, we used the Pythagorean identity to simplify the expression \(z = r^2 ( \text{cos}^2(\theta) + 2\text{sin}^2(\theta) )\) into a more manageable form. This simplification helped in taking partial derivatives easily. By recognizing and applying these identities, complex trigonometric expressions become much simpler to handle, facilitating the process of differentiation and integration in calculus.
Understanding these core concepts lays the foundation for tackling more advanced problems involving partial derivatives in polar coordinates. By mastering these fundamentals, one can approach similar problems with greater confidence and efficiency.