Question

Find \(\partial w / \partial s\) and \(\partial w / \partial t\) using the appropriate Chain Rule, and evaluate each partial derivative at the given values of \(s\) and \(t\) $$ \begin{array}{l} \text { Function } \\ \hline w=x^{2}-y^{2} \\ x=s \cos t, \quad y=s \sin t \end{array} $$ $$ \frac{\text { Point }}{s=3, \quad t=\frac{\pi}{4}} $$

Short answer

\(\partial w / \partial s = 0\) and \(\partial w / \partial t = -18\).

Step-by-step solution

Differentiate \(w\) with respect to \(x\) and \(y\)

First we need to find the derivatives of \(w\) with respect to \(x\) and \(y\). These turn out to be \(\partial w / \partial x = 2x\) and \(\partial w / \partial y = -2y\).

Differentiate \(x\) and \(y\) with respect to \(s\) and \(t\)

Next we differentiate \(x = s\cos(t)\) and \(y = s\sin(t)\) with respect to \(s\) and \(t\). This gives us \(\partial x / \partial s = \cos(t)\), \(\partial x / \partial t = -s\sin(t)\), \(\partial y / \partial s = \sin(t)\), and \(\partial y / \partial t = s\cos(t)\).

Use the Chain Rule to find \(\partial w / \partial s\) and \(\partial w / \partial t\)

Now we use the chain rule to find \(\partial w / \partial s\) and \(\partial w / \partial t\). By the chain rule, \(\partial w / \partial s = (\partial w / \partial x)(\partial x / \partial s) + (\partial w / \partial y)(\partial y / \partial s) = 2s\cos^2(t) - 2s\sin^2(t) = 2s(\cos^2(t) - \sin^2(t))\). Similarly, \(\partial w / \partial t = (\partial w / \partial x)(\partial x / \partial t) + (\partial w / \partial y)(\partial y / \partial t) = -2s^2\sin(t)\cos(t) - 2s^2\sin(t)\cos(t) = -4s^2\sin(t)\cos(t)\).

Evaluate at given point

Lastly we evaluate these derivatives at \(s=3\) and \(t=\pi/4\). At this point, \(\partial w / \partial s = 2(3)(\cos^2(\pi/4) - \sin^2(\pi/4)) = 0\) and \(\partial w / \partial t = -4(3)^2\sin(\pi/4)\cos(\pi/4) = -18\).