Question

Find \(\partial w / \partial r\) and \(\partial w / \partial \theta\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(r\) and \(\boldsymbol{\theta}\) before differentiating. \(w=\sqrt{25-5 x^{2}-5 y^{2}}, x=r \cos \theta, \quad y=r \sin \theta\)

Short answer

Using the Chain Rule, the derivative of w with respect to r is \(\partial w / \partial r = -5r / \sqrt{25-5r^2}\), and with respect to \(\theta\) is \(\partial w / \partial \theta = 0 \). Upon first converting \(w\) into a function of \(r\) and \(\theta\) before differentiating, the answers remain the same.

Step-by-step solution

Transform w

Change \(w\) into polar coordinates using the given variables. So, it becomes \(w = \sqrt{25 - 5r^{2}}\).

Find \(\partial w / \partial r\) using Chain Rule

The derivative of \(w\) with respect to \(r\) using the chain rule is: \(\partial w / \partial r = -5r / \sqrt{25-5r^2}\).

Find \(\partial w / \partial \theta\) using Chain Rule

As there is no theta in \(w\), the derivative of \(w\) with respect to \(\theta\) is zero. Therefore, \(\partial w / \partial \theta = 0 \).

Differentiation by Converting w First

This part asks to first convert w into a function of \(r\) and \(\theta\), which was done in Step 1, and then find derivatives. Using the solution from Step 1, the derivatives will still remain the same. Therefore, the effort from Step 2 and Step 3 answers this part.