## University Calculus: Early Transcendentals (3rd Edition)

$\frac{dr}{d\theta}=\frac{-1}{6}$
Given $r=(\theta^2+7)^{(\frac{1}{3})}$ and $\theta^2 t+\theta=1$ On differentiating both sides: $\frac{dr}{d\theta}=\frac{1}{3}(({\theta^2+7})^{\frac{-2}{3}}\frac{d({\theta^2+7})}{d\theta}$ $\frac{dr}{d\theta}=\frac{(2\theta)}{3}(({\theta^2+7})^{\frac{-2}{3}}\frac{d\theta}{dt}$ and $2\theta{t}\frac{d\theta}{dt}+{\theta}^2+\frac{d\theta}{dt}=0$ $(2\theta{t}+1)\frac{d\theta}{dt}=-{\theta}^2$ $\frac{d\theta}{dt}=\frac{-{\theta}^2}{(2\theta{t}+1)}$ at $t=0$, the value of $\theta$ and its differentiation is: $\theta^2 (0)+\theta=1$ $\theta=1$ $d(\theta(0))=\frac{-{\theta}^2}{(2\theta{(0)}+1)}={-{\theta}^2}$ on putting the above value in $\frac{dr}{d\theta}$ $\frac{dr}{d\theta}=\frac{(2)}{3}(({1+7})^{\frac{-2}{3}}(-1)$ $\frac{dr}{d\theta}=\frac{-1}{6}$