91Ó°ÊÓ

We are given the function \(z = \sqrt{r^{2} + s^{2}}\). We need to find the partial derivatives of \(z\) with respect to \(r\) and \(s\). For \(\partial z / \partial r\): $$\frac{\partial z}{\partial r} = \frac{1}{2\sqrt{r^2+s^2}} \cdot (2r) = \frac{r}{\sqrt{r^2+s^2}}$$ For \(\partial z / \partial s\): $$\frac{\partial z}{\partial s} = \frac{1}{2\sqrt{r^2+s^2}} \cdot (2s) = \frac{s}{\sqrt{r^2+s^2}}$$

We are given the functions \(r = \cos 2t\) and \(s = \sin 2t\). We need to find their derivatives with respect to \(t\). For \(d r / d t\): $$\frac{d r}{d t} = \frac{d (\cos 2t)}{d t} = -2\sin 2t$$ For \(d s / d t\): $$\frac{d s}{d t} = \frac{d (\sin 2t)}{d t} = 2\cos 2t$$

Now, we will apply the Chain Rule formula: $$\frac{d z}{d t}=\frac{\partial z}{\partial r} \frac{d r}{d t}+\frac{\partial z}{\partial s} \frac{d s}{d t}$$ Plugging in the values we found in Steps 1 and 2: $$\frac{d z}{d t} = \left(\frac{r}{\sqrt{r^2+s^2}}\right) (-2\sin 2t) + \left(\frac{s}{\sqrt{r^2+s^2}}\right) (2\cos 2t)$$ Now, substitute \(r = \cos 2t\) and \(s = \sin 2t\): $$\frac{d z}{d t} = \left(\frac{\cos 2t}{\sqrt{(\cos 2t)^2+(\sin 2t)^2}}\right) (-2\sin 2t) + \left(\frac{\sin 2t}{\sqrt{(\cos 2t)^2+(\sin 2t)^2}}\right) (2\cos 2t)$$ We know that \(\sin^2(2t) + \cos^2(2t) = 1\), so the square root of the denominator is equal to 1. Therefore: $$\frac{d z}{d t} = -2\cos 2t\sin 2t + 2\sin 2t\cos 2t$$ Finally, simplify the derivative: $$\frac{d z}{d t} = 0$$ So, the derivative of \(z\) with respect to \(t\) is 0.