Chapter 12 - Functions of Several Veriables - 12.5 The Chain Rule - 12.5 Exercises - Page 913: 16

$$\frac{{dV}}{{dt}} = 0$$

$$\eqalign{ & {\text{we have }}V = \frac{{x - y}}{{y + z}},{\text{ with }}x = t{\text{, }}y = 2t{\text{ and }}z = 3t \cr & {\text{find }}\frac{{dV}}{{dt}}{\text{ using the theorem 12}}{\text{.7 }}\left( {{\text{see page 908}}} \right) \cr & \frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial V}}{{\partial z}}\frac{{dz}}{{dt}} \cr & {\text{find the partial derivatives }}\frac{{\partial V}}{{\partial x}}{\text{,}}\frac{{\partial V}}{{\partial y}}{\text{ and }}\frac{{\partial V}}{{\partial z}} \cr & \frac{{\partial V}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{{x - y}}{{y + z}}} \right] \cr & \frac{{\partial V}}{{\partial x}} = \frac{1}{{y + z}}\frac{\partial }{{\partial x}}\left[ {x - y} \right] \cr & \frac{{\partial V}}{{\partial x}} = \frac{1}{{y + z}}, \cr & \frac{{\partial V}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{x - y}}{{y + z}}} \right] \cr & {\text{use quotient rule}} \cr & \frac{{\partial V}}{{\partial y}} = \frac{{\left( {y + z} \right)\left( { - 1} \right) - \left( {x - y} \right)\left( 1 \right)}}{{{{\left( {y + z} \right)}^2}}} \cr & \frac{{\partial V}}{{\partial y}} = \frac{{ - y - z - x + y}}{{{{\left( {y + z} \right)}^2}}} \cr & \frac{{\partial V}}{{\partial y}} = \frac{{ - z - x}}{{{{\left( {y + z} \right)}^2}}} \cr & and \cr & \frac{{\partial V}}{{\partial z}} = \frac{\partial }{{\partial z}}\left[ {\frac{{x - y}}{{y + z}}} \right] = \left( {x - y} \right)\frac{\partial }{{\partial z}}\left[ {\frac{{x - y}}{{y + z}}} \right] \cr & \frac{{\partial V}}{{\partial y}} = \left( {x - y} \right)\left( { - \frac{1}{{{{\left( {y + z} \right)}^2}}}} \right) \cr & \frac{{\partial V}}{{\partial y}} = - \frac{{x - y}}{{{{\left( {y + z} \right)}^2}}} \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[ t \right] = 1 \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {2t} \right] = 2 \cr & \frac{{dz}}{{dt}} = \frac{d}{{dt}}\left[ {3t} \right] = 3 \cr & \cr & {\text{substitute the results of the derivatives in the equation of the theorem 12}}{\text{.7}} \cr & \frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial V}}{{\partial z}}\frac{{dz}}{{dt}} \cr & \frac{{dV}}{{dt}} = \left( {\frac{1}{{y + z}}} \right)\left( 1 \right) + \left( {\frac{{ - z - x}}{{{{\left( {y + z} \right)}^2}}}} \right)\left( 2 \right) + \left( { - \frac{{x - y}}{{{{\left( {y + z} \right)}^2}}}} \right)\left( 3 \right) \cr & \frac{{dV}}{{dt}} = \frac{1}{{y + z}} + \frac{{ - 2z - 2x}}{{{{\left( {y + z} \right)}^2}}} - \frac{{3x - 3y}}{{{{\left( {y + z} \right)}^2}}} \cr & \frac{{dV}}{{dt}} = \frac{{y + z - 2z - 2x - 3x + 3y}}{{{{\left( {y + z} \right)}^2}}} \cr & \frac{{dV}}{{dt}} = \frac{{4y - z - 5x}}{{{{\left( {y + z} \right)}^2}}} \cr & {\text{replace }}x = t{\text{, }}y = 2t{\text{ and }}z = 3t \cr & \frac{{dV}}{{dt}} = \frac{{4\left( {2t} \right) - 3t - 5\left( t \right)}}{{{{\left( {2t + 3t} \right)}^2}}} \cr & \frac{{dV}}{{dt}} = \frac{{8t - 8t}}{{{{\left( {2t + 3t} \right)}^2}}} \cr & \frac{{dV}}{{dt}} = 0 \cr} $$