Question

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

Short answer

First part: \(4y\), Second part: \(2y\).

Step-by-step solution

Define the given functions

Start with the given equations: \[z = x^2 + 2y^2\]\[x = r\cos\theta\]\[y = r\sin\theta\]

Express z in terms of x and y

Substitute the expressions for x and y into the equation for z:\[z = (r\cos\theta)^2 + 2(r\sin\theta)^2\]\[z = r^2 \cos^2 \theta + 2r^2 \sin^2 \theta\]Combine like terms:\[z = r^2(\cos^2 \theta + 2\sin^2 \theta)\]

Calculate the partial derivative

To find \( \left( \frac{\partial z}{\partial y} \right)_x \), we start with the expression for z in terms of x and y. From the original equation:\[z = x^2 + 2y^2\]Differentiate z with respect to y while treating x as a constant:\[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(x^2 + 2y^2) = 0 + 4y = 4y\]Hence, \( \left( \frac{\partial z}{\partial y} \right)_x = 4y \)

Define the second given function

For the second part of the problem, start with the given equation: \[z = r^2 \tan^2 \theta\]

Express y in terms of r and θ

We know that\[ y = r \sin \theta \].

Rewrite z in terms of y

We know that\[r = \sqrt{x^2 + y^2}\]. Substitute y for r\sin \theta and use the above result to get:\[ \tan \theta =\frac{y}{\sqrt{x^2+ y^2}}\]. Therefore, substitute y into z \[z = (x^2+y^2)\frac{y^2}{x^2+y^2} =y^2 \]

Calculate the partial derivative

Now, differentiate z with respect to y while treating x as a constant:\[ \frac{\partial z}{\partial y} = \frac {\partial }{\partial y}y^2 = 2y\].

Key concepts

multivariable calculus

Multivariable calculus deals with functions of multiple variables. Unlike single-variable calculus, which only involves one independent variable, multivariable calculus considers functions involving two or more variables.
For example, in our exercise, we have functions like \(z = x^2 + 2y^2\) and \(z = r^2 \tan^2 \theta\) with variables \(x\), \(y\), \(r\), and \(\theta\).
This allows us to explore concepts such as partial derivatives and gradients, which are essential in fields like physics and engineering.

When dealing with multiple variables, you often need to use specialized techniques like partial differentiation and the chain rule to find how changes in these variables affect the function.

differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function changes. In multivariable calculus, we often focus on partial differentiation, which deals with finding the derivative of a function with respect to one variable while keeping the other variables constant.

In our exercise, we aim to find the partial derivative \( \left( \frac{\partial z}{\partial y} \right)_x \).
The process involves:

• Starting with the function \( z = x^2 + 2y^2 \)
• Treating \( x \) as constant while differentiating with respect to \( y \)
• Calculating \( \left( \frac{\partial z}{\partial y} \right)_x = 4y \)

Understanding partial derivatives is crucial for functions that depend on more than one variable.

coordinate transformation

Coordinate transformations are used to switch between different coordinate systems without changing the function's properties.
For example, in our exercise, we move from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\).
This is done using the transformations:

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

By substituting these into our original function, we can express \(z\) in terms of polar coordinates. This is useful for simplifying the calculations or when the problem is naturally expressed in different coordinates.
In polar coordinates, for instance, angles and radii can often simplify differentiation and integration tasks.

chain rule

The chain rule is a fundamental tool for differentiation when dealing with composite functions. It allows you to differentiate a function that is defined in terms of another function.
In our exercise, we use the chain rule to differentiate \( z \) with respect to \( y \) when \( z \) is in terms of \( r \) and \( \theta \).
For example, if \( z = r^2 \tan^2 \theta \) and \( y = r \sin \theta \), we use the chain rule to express \( z \) in terms of \( y \).

Here's a simplified step-by-step approach to using the chain rule:

• Identify the outer function and the inner function.
• Differentiate the outer function with respect to the inner function's variable.
• Multiply by the derivative of the inner function with respect to the original variable.

This technique is vital for working through complex differentiation problems in multivariable calculus.