## Thomas' Calculus 13th Edition

$${\left. {\frac{{dr}}{{dt}}} \right|_{\left( {1,0} \right)}} = - \frac{1}{6}$$
\eqalign{ & r = {\left( {{\theta ^2} + 7} \right)^{1/3}}{\text{ and }}{\theta ^2}t + \theta = 1 \cr & {\text{find }}\frac{{dr}}{{dt}}{\text{ using the chain rule}} \cr & \frac{{dr}}{{dt}} = \frac{{dr}}{{d\theta }}\frac{{d\theta }}{{dt}} \cr & \cr & {\text{Then}}{\text{, find }}\frac{{dr}}{{d\theta }}{\text{ and }}\frac{{d\theta }}{{dt}} \cr & \frac{{dr}}{{d\theta }} = \frac{d}{{dt}}\left[ {{{\left( {{\theta ^2} + 7} \right)}^{1/3}}} \right] \cr & \frac{{dr}}{{d\theta }} = \frac{1}{3}{\left( {{\theta ^2} + 7} \right)^{ - 2/3}}\frac{d}{{dt}}\left[ {{\theta ^2} + 7} \right] \cr & \frac{{dr}}{{d\theta }} = \frac{{2\theta }}{3}{\left( {{\theta ^2} + 7} \right)^{ - 2/3}} \cr & \cr & \frac{d}{{dt}}\left( {{\theta ^2}t} \right) + \frac{d}{{dt}}\left( \theta \right) = \frac{d}{{dt}}\left( 1 \right) \cr & {\theta ^2}\frac{d}{{dt}}\left[ t \right] + t\frac{d}{{dt}}\left[ {{\theta ^2}} \right] + \frac{d}{{dt}}\left( \theta \right) = \frac{d}{{dt}}\left( 1 \right) \cr & {\theta ^2} + 2t\theta \frac{{d\theta }}{{dt}} + \frac{{d\theta }}{{dt}} = 0 \cr & \frac{{d\theta }}{{dt}} = - \frac{{{\theta ^2}}}{{1 + 2t}} \cr & \cr & {\text{then substituting }}\frac{{dr}}{{d\theta }}{\text{ and }}\frac{{d\theta }}{{dt}}{\text{ into }}\frac{{dr}}{{dt}} = \frac{{dr}}{{d\theta }}\frac{{d\theta }}{{dt}} \cr & \frac{{dr}}{{dt}} = \left( {\frac{{2\theta }}{3}{{\left( {{\theta ^2} + 7} \right)}^{ - 2/3}}} \right)\left( { - \frac{{{\theta ^2}}}{{1 + 2t}}} \right) \cr & \cr & {\text{Evaluate at }}\theta = 1{\text{ and }}t = 0 \cr & {\left. {\frac{{dr}}{{dt}}} \right|_{\left( {1,0} \right)}} = \left( {\frac{{2\left( 1 \right)}}{3}{{\left( {{1^2} + 7} \right)}^{ - 2/3}}} \right)\left( { - \frac{{{1^2}}}{{1 + 2\left( 0 \right)}}} \right) \cr & {\left. {\frac{{dr}}{{dt}}} \right|_{\left( {1,0} \right)}} = \left( {\frac{2}{3}\left( {\frac{1}{4}} \right)} \right)\left( { - 1} \right) \cr & {\left. {\frac{{dr}}{{dt}}} \right|_{\left( {1,0} \right)}} = - \frac{1}{6} \cr}