Question

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

Short answer

\(\frac{\partial z}{\partial r} = 2r(\cos^2 \theta + 2\sin^2 \theta)\) and \(\frac{\partial z}{\partial r} = 2r \tan^2 \theta\).

Step-by-step solution

Identify the expressions for the variables

Given: \(z = x^2 + 2y^2\) with \(x = r \cos \theta\) and \(y = r \sin \theta\). Combine these into a single expression for \(z\) in terms of \(r\) and \(\theta\) first.

Substitute expressions for x and y

Substitute \(x = r \cos \theta\) and \(y = r \sin \theta\) into \(z = x^2 + 2y^2\).\(z = (r \cos \theta)^2 + 2(r \sin \theta)^2\).

Simplify the expression for z

Simplify the above expression to get\(z = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\).Combine like terms:\(z = r^2(\cos^2 \theta + 2\sin^2 \theta)\).

Calculate the partial derivative with respect to r

Take the partial derivative of \(z\) with respect to \(r\), holding \(\theta\) constant.\(\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 (\cos^2 \theta + 2 \sin^2 \theta)] = 2r (\cos^2 \theta + 2 \sin^2 \theta)\).

Identify the new function

Given: \(z = r^2 \tan^2 \theta\). Now, find the partial derivative of this new function with respect to \(r\).

Calculate the partial derivative for the new function

Take the partial derivative of \(z\) with respect to \(r\), holding \(\theta\) constant.\(\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} [r^2 \tan^2 \theta] = 2r \tan^2 \theta\).

Key concepts

Multivariable Calculus Overview

Multivariable calculus deals with functions of more than one variable. In this case, the function z depends on both x and y. When we switch to polar coordinates, z then depends on r and θ. This shift makes certain calculations simpler and provides more insight.

In multivariable calculus, partial derivatives are essential. These derivatives help us understand how the function changes as one variable changes while others stay constant.

Polar Coordinates and Their Importance

Polar coordinates are a way of expressing points in a plane using a radius and an angle. Instead of using x and y coordinates, we use r (the radius) and θ (the angle).

This conversion can simplify problems with circular or rotational symmetry. The given function z depends on x and y. Switching to polar coordinates, we express x and y in terms of r and θ as follows:

• x = r cos θ
• y = r sin θ

With these expressions, we can rewrite the function z in polar coordinates to facilitate easier differentiation.

Partial Differentiation Explained

Partial differentiation allows us to see how a function changes with respect to one variable while keeping others constant. For the given problem, we need to find \(\frac{\partial z}{\partial r}\) while θ is constant.

Let's break it down:
First, rewrite z in terms of r and θ:

• z = (r cos θ)^2 + 2(r sin θ)^2
• z = r^2 (cos^2 θ + 2 sin^2 θ)

Now, take the derivative with respect to r while keeping θ constant: \[ \frac{\partial z}{\partial r} = \frac{\partial}{\partial r} \left[ r^2 (cos^2 θ + 2 sin^2 θ) \right] = 2r (cos^2 θ + 2 sin^2 θ) \] This expression shows how z changes as r changes.

Mathematical Methods: Repeated Calculation

Another part of the exercise asks you to find the partial derivative for a different z function. Given: \(z = r^2 \tan^2 \theta\)

Repeat the steps:

• Calculate the derivative with respect to r:
• \(\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} \left[ r^2 \tan^2 \theta \right] = 2r \tan^2 \theta \)

This derivative shows how z changes with r for the new function. Understanding these steps helps you handle more complex problems with multiple variables.