Question

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

Short answer

\(\frac{\partial z}{\partial y} \) for 1st case: \(2 \sin \theta ( \cos^2 \theta + 2 \sin^2 \theta )\), for 2nd case: \(2 \sin \theta \tan^2 \theta\).

Step-by-step solution

Given Functions and Substitutions

Identify the given function and the provided substitutions. For the first scenario: Function: \[z = x^2 + 2y^2\] Substitutions: \[x = r \cos \theta\]\[y = r \sin \theta\]

Substitute Variables in Function

Substitute the provided expressions for \(x\) and \(y\) into the function \[z = (r \cos \theta)^2 + 2(r \sin \theta)^2\]This simplifies to: \[z = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\]

Simplify the Function

Factor out \(r^2\) from the expression: \[z = r^2 (\cos^2 \theta + 2 \sin^2 \theta)\]

Differentiation with Respect to \(y\) (\frac{\partial z}{\partial y}\)

Use the chain rule for partial differentiation: \[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial z}{\partial \theta} \cdot \frac{\partial \theta}{\partial y}\]Since \(\theta\) is independent of \(y\), \[\frac{\partial \theta}{\partial y} = 0\] and since \[r^2 = x^2 + y^2\], \[r = \sqrt{x^2 + y^2}\], \[\frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{\sin \theta}{r}\]

Compute Partial Derivatives

Calculate the partial derivative of \(z\) with respect to \(r\): \[\frac{\partial z}{\partial r} = 2r (\cos^2 \theta + 2 \sin^2 \theta)\]Hence, \[\frac{\partial z}{\partial y} = \frac{\sin \theta}{r} \cdot 2r( \cos^2 \theta + 2 \sin^2 \theta) = 2 \sin \theta ( \cos^2 \theta + 2 \sin^2 \theta)\]

Repeat for Second Function

Identify the new given function: \[z = r^2 \tan^2 \theta\]Compute partial derivative with respect to \(y\): \[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial z}{\partial \theta} \cdot \frac{\partial \theta}{\partial y}\]

Compute \(\frac{d z}{d y}\) for the New Function

Use similar rules as before: \[\frac{\partial z}{\partial r} = 2r \tan^2 \theta\] \[\frac{\partial z}{\partial \theta} = r^2 \cdot 2 \tan \theta \cdot \sec^2 \theta = 2r^2 \tan \theta \sec^2 \theta\] Hence, \[\frac{\partial z}{\partial r} \cdot \frac{\partial r}{\partial y}\] \[= 2r \tan^2 \theta \cdot \frac{\sin \theta}{r} = 2 \sin \theta \tan^2 \theta\]

Key concepts

Chain Rule in Partial Differentiation

When working with functions of multiple variables, the chain rule helps us find the derivative of a function with respect to a given variable by considering its dependency on other intermediate variables. In this exercise, we deal with the function z and its dependencies on variables x and y, which are further dependent on r and \(\theta\).
For example, to find \(\frac{\partial z}{\partial y}\), we employ the chain rule:
\[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial z}{\partial \theta} \cdot \frac{\partial \theta}{\partial y}\]
Here, \(\theta\) is independent of y, making \(\frac{\partial \theta}{\partial y} = 0\), and simplifying our calculations. Understanding this step is crucial to applying the chain rule effectively in partial differentiation. This method ensures all indirect dependencies through intermediate variables are accurately captured.

Trigonometric Substitution

Trigonometric substitution simplifies functions containing trigonometric expressions by substituting variable combinations. For instance, in our exercise, we substitute x and y with trigonometric expressions involving r and \(\theta\).
Given:
\[x = r \cos \theta \]
\[y = r \sin \theta\]
Substituting these into the function \(z = x^2 + 2y^2\), we get:
\[z = (r \cos \theta)^2 + 2(r \sin \theta)^2 = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\]
This approach not only simplifies differentiation but also makes integration and solving differential equations involving trigonometric identities more manageable. The goal is to convert complicated algebraic expressions into more approachable trigonometric forms, easing the application of differentiation rules.

Differentiation and Simplification

Differentiation involves finding the rate at which one quantity changes with respect to another. In our function, partial derivatives measure how z changes as x or y changes. By substituting variables, we simplify and differentiate the function step-by-step.
To differentiate \( z = r^2 (\cos^2 \theta + 2 \sin^2 \theta) \) with respect to y, we first found \(\frac{\partial r}{\partial y} = \frac{\sin \theta}{r}\). Then we computed \(\frac{\partial z}{\partial r} = 2r (\cos^2 \theta + 2 \sin^2 \theta)\), multiplying these together:
\( \frac{\partial z}{\partial y} = 2 \sin \theta (\cos^2 \theta + 2 \sin^2 \theta) \)
Repeating this process for the second function \(z = r^2 \tan^2 \theta\):
\(\frac{\partial z}{\partial r} = 2r \tan^2 \theta\)
\(\frac{\partial z}{\partial \theta} = 2r^2 \tan \theta \sec^2 \theta\)
These steps highlight the importance of following rules methodically to obtain accurate derivatives. Differentiation reveals how changes in one variable influence another, essential in many scientific and engineering applications.