Question

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

Short answer

\(\left(\frac{\partial z}{\partial r}\right)_{\theta} = 2r(\cos^2 \theta + 2 \sin^2 \theta)\)

Step-by-step solution

Substitute x and y in terms of r and θ

Given that \(z = x^2 + 2y^2\), we need to express \(x\) and \(y\) in terms of \(r\) and \(\theta\). Given \(x = r \cos \theta\) and \(y = r \sin \theta\), substitute these into the equation for \(z\): \(z = (r \cos \theta)^2 + 2(r \sin \theta)^2\).

Simplify the expression for z

Simplify the equation: \(z = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\). Combine like terms: \(z = r^2(\cos^2 \theta + 2 \sin^2 \theta)\).

Factor out the common term

Notice that \(r^2\) is a common factor: \(z = r^2 (\cos^2 \theta + 2 \sin^2 \theta)\).

Differentiate z with respect to r

To find \(\left(\frac{\partial z}{\partial r}\right)_{\theta}\), we need to partially differentiate \(z\) with respect to \(r\) while keeping \(\theta\) constant. The derivative is: \(\frac{\partial z}{\partial r} = \frac{\partial}{\partial r} \big[ r^2 (\cos^2 \theta + 2 \sin^2 \theta) \big] = 2r(\cos^2 \theta + 2 \sin^2 \theta)\).

Key concepts

Partial Differentiation

Partial differentiation is a method used in calculus to find the rate at which a function changes with respect to one variable while keeping other variables constant. In simpler terms, it helps us understand how a function behaves as we tweak one parameter at a time.

For instance, if you have a function of two variables like our example where \( z = x^2 + 2y^2 \), partial differentiation will allow us to see how \( z \) changes if we change \( x \) and keep \( y \) constant or vice versa.

In this exercise:

• We converted from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\).
• We then took the partial derivative of \( z \) with respect to \( r \) while keeping \( \theta \) constant.

This helps in studying how the function behaves in different directions, an essential skill in multivariable calculus.

Polar Coordinates

Polar coordinates provide a different way to describe points on a plane using a distance from a reference point and an angle from a reference direction.

Rather than using \((x, y)\) coordinates, polar coordinates use:\[ x = r \cos \theta \] \[ y = r \sin \theta \]

These transformations are useful in various fields, including physics, engineering, and computer graphics, particularly when dealing with circular or rotational symmetry.

In our example:

• We expressed \( x \) and \( y \) in terms of polar coordinates \( r \) and \( \theta \).
• Substituted these into the function \( z \) to write it entirely in terms of polar coordinates.

This converted an otherwise complicated Cartesian function into a simpler form in polar coordinates.

Multivariable Calculus

Multivariable calculus extends calculus to functions of more than one variable. It involves techniques like partial differentiation, multiple integrals, and topics such as vector calculus.

It can be used to:

• Find rates of change in systems with multiple variables.
• Calculate areas, volumes, and other related quantities.
• Solve real-world problems involving functions with several inputs.

In this exercise, converting to polar coordinates and performing partial differentiation are fundamental steps you often use in multivariable calculus to simplify and solve problems.

Trigonometric Transformations

Trigonometric transformations convert between Cartesian and polar coordinates. These are particularly useful for simplifying integrals and derivatives in multivariable calculus.

In our example, we used the identities:

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

This allowed us to take a function initially expressed in terms of \( x \) and \( y \), and rewrite it in terms of \( r \) and \( \theta \).

When we differentiate \( z \) with respect to \( r \), we apply the chain rule, which further demonstrates the power of these transformations in simplifying complex mathematical problems.

These concepts are foundational in calculus and are widely used in numerous practical applications.