Question

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

Short answer

\( \frac{\partial z}{\partial x} = 2x \)

Step-by-step solution

- Express z in terms of x and y

Given the function, we start by writing it as: \( z = x^2 + 2y^2 \)

- Differentiate with respect to x

We compute the partial derivative of \(z\) with respect to \(x\), treating \(y\) as a constant: \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (x^2 + 2y^2) \]

- Apply the power rule

By applying the power rule, we get: \[ \frac{\partial z}{\partial x} = 2x + 0 = 2x \]

Key concepts

multivariable calculus

Multivariable calculus is the extension of calculus to functions of multiple variables. This means dealing with functions that depend on more than one variable, like our example where the function is given as \( z = x^2 + 2y^2 \). In such settings, we often need to find the rate of change of a function concerning one of its variables while keeping the others constant.
In our example, this is exactly what we did to find \( \frac{\partial z}{\partial x} \). Understanding how these changes happen helps in fields like physics, engineering, and economics.

power rule

The power rule is a basic rule in calculus that makes finding derivatives easier. If you have a function of the form \( f(x) = x^n \), the power rule states that its derivative is given by \( f'(x) = nx^{n-1} \).
In our problem, we applied the power rule to the term \( x^2 \) in the function \( z = x^2 + 2y^2 \). By using the power rule, we get:
\[ \frac{\partial z}{\partial x} = 2x \]
The term \( 2y^2 \) is treated as a constant, so its derivative with respect to \( x \) is zero. Thus, the final partial derivative is \( 2x \).

chain rule

The chain rule is a vital tool for finding the derivatives of composite functions. It helps when a variable is a function of another variable. For example, if you have a function \( z \) in terms of \( x \) and \( y \), and \( x \) and \( y \) are themselves functions of other variables like \( r \) and \( \theta \), you can use the chain rule to find \( \frac{\partial z}{\partial r} \) or \( \frac{\partial z}{\partial \theta} \).
In our exercise, we didn't directly use the chain rule but keeping this rule in mind is helpful for more complex problems that involve multiple layers of functions.

mathematical methods

Different mathematical methods are used to solve problems in multivariable calculus. These methods might include applying basic differentiation rules, using higher-order derivatives, or leveraging the chain rule for function compositions. In more complex scenarios involving multivariable calculus, one might also use tools like gradient vectors, Lagrange multipliers, and Jacobian matrices.
In our solution, we applied a straightforward approach which involved:

• Identifying the function and variables involved
• Applying the power rule
• Evaluating the partial derivative by differentiating with respect to \( x \) while keeping \( y \) constant.

By mastering these basic methods, you can tackle more complex problems with confidence.