Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 14

$$\frac{{dQ}}{{dt}} = - \frac{{\sin t\cos t}}{{\sqrt {1 + {{\cos }^2}t} }}$$

$$\eqalign{ & {\text{we have }}Q = \sqrt {{x^2} + {y^2} + {z^2}} ,{\text{ with }}x = \sin t{\text{, }}y = \cos t{\text{ and }}z = \cos t \cr & {\text{find }}\frac{{dQ}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr & \frac{{dQ}}{{dt}} = \frac{{\partial Q}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial Q}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial Q}}{{\partial z}}\frac{{dz}}{{dt}} \cr & {\text{find the partial derivatives }}\frac{{\partial Q}}{{\partial x}}{\text{,}}\frac{{\partial Q}}{{\partial y}}{\text{ and }}\frac{{\partial Q}}{{\partial z}} \cr & \frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\sqrt {{x^2} + {y^2} + {z^2}} } \right] \cr & use\,\,\left( {\sqrt u } \right)' = \frac{1}{{2\sqrt u }}\left( {u'} \right) \cr & \frac{{\partial Q}}{{\partial x}} = \frac{1}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}\frac{\partial }{{\partial x}}\left[ {2x} \right] \cr & \frac{{\partial Q}}{{\partial x}} = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr & {\text{then similarly}} \cr & \frac{{\partial Q}}{{\partial y}} = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\text{ and }}\frac{{\partial Q}}{{\partial z}} = \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr & \cr & {\text{find the partial derivatives }}\frac{{dx}}{{dt}}{\text{,}}\frac{{dy}}{{dt}}{\text{ and }}\frac{{dz}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sin t} \right] = \cos t \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\cos t} \right] = - \sin t \cr & \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {\cos t} \right] = - \sin t \cr & \cr & {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr & \frac{{dQ}}{{dt}} = \frac{{\partial Q}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial Q}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial Q}}{{\partial z}}\frac{{dz}}{{dt}} \cr & \frac{{dQ}}{{dt}} = \left( {\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)\left( {\cos t} \right) + \left( {\frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)\left( { - \sin t} \right) + \left( {\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right)\left( { - \sin t} \right) \cr & {\text{factoring }}\frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }} \cr & \frac{{dQ}}{{dt}} = \frac{1}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\left( {x\cos t - y\sin t - z\sin t} \right) \cr & {\text{replace }}x = \sin t{\text{, }}y = \cos t{\text{ and }}z = \cos t \cr & \frac{{dQ}}{{dt}} = \frac{1}{{\sqrt {{{\sin }^2}t + {{\cos }^2}t + {{\cos }^2}t} }}\left( {\left( {\sin t} \right)\cos t - \left( {\cos t} \right)\sin t - \left( {\cos t} \right)\sin t} \right) \cr & \frac{{dQ}}{{dt}} = \frac{1}{{\sqrt {1 + {{\cos }^2}t} }}\left( {\sin t\cos t - \sin t\sin t - \sin t\cos t} \right) \cr & \frac{{dQ}}{{dt}} = - \frac{{\sin t\cos t}}{{\sqrt {1 + {{\cos }^2}t} }} \cr} $$