Answer
$$\frac{{dz}}{{dt}} = 0$$
Work Step by Step
$$\eqalign{
& {\text{we have }}z = \sqrt {{r^2} + {s^2}} ,{\text{ with }}r = \cos 2t{\text{ and }}s = \sin 2t \cr
& {\text{find }}\frac{{dz}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial r}}\frac{{dr}}{{dt}} + \frac{{\partial z}}{{\partial s}}\frac{{ds}}{{dt}} \cr
& {\text{find the partial derivatives }}\frac{{\partial z}}{{\partial r}}{\text{ and }}\frac{{\partial z}}{{\partial s}} \cr
& \frac{{\partial z}}{{\partial r}} = \frac{\partial }{{\partial r}}\left[ {\sqrt {{r^2} + {s^2}} } \right] = \frac{{2r}}{{2\sqrt {{r^2} + {s^2}} }} \cr
& \frac{{\partial z}}{{\partial r}} = \frac{r}{{\sqrt {{r^2} + {s^2}} }} \cr
& \frac{{\partial z}}{{\partial s}} = \frac{\partial }{{\partial s}}\left[ {\sqrt {{r^2} + {s^2}} } \right] = \frac{s}{{\sqrt {{r^2} + {s^2}} }} \cr
& {\text{find the partial derivatives }}\frac{{dr}}{{dt}}{\text{ and }}\frac{{ds}}{{dt}} \cr
& \frac{{dr}}{{dt}} = \frac{d}{{dt}}\left[ {\cos 2t} \right] = - 2\sin 2t \cr
& \frac{{ds}}{{dt}} = \frac{d}{{dt}}\left[ {\sin 2t} \right] = 2\cos 2t \cr
& \cr
& {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr
& \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial r}}\frac{{dr}}{{dt}} + \frac{{\partial z}}{{\partial s}}\frac{{ds}}{{dt}} \cr
& \frac{{dz}}{{dt}} = \left( {\frac{r}{{\sqrt {{r^2} + {s^2}} }}} \right)\left( { - 2\sin 2t} \right) + \left( {\frac{s}{{\sqrt {{r^2} + {s^2}} }}} \right)\left( {2\cos 2t} \right) \cr
& {\text{replace }}r = \cos 2t{\text{ and }}s = \sin 2t \cr
& \frac{{dz}}{{dt}} = \left( {\frac{{\cos 2t}}{{\sqrt {{{\left( {\cos 2t} \right)}^2} + {{\left( {\sin 2t} \right)}^2}} }}} \right)\left( { - 2\sin 2t} \right) + \left( {\frac{{\sin 2t}}{{\sqrt {{{\left( {\cos 2t} \right)}^2} + {{\left( {\sin 2t} \right)}^2}} }}} \right)\left( {2\cos 2t} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dz}}{{dt}} = \frac{{ - 2\cos 2t\sin 2t}}{{\sqrt {{{\cos }^2}2t + {{\sin }^2}2t} }} + \frac{{2\sin 2t\cos 2t}}{{\sqrt {{{\cos }^2}2t + {{\sin }^2}2t} }} \cr
& \frac{{dz}}{{dt}} = 0 \cr} $$
