Question

find $\frac{dr}{d\theta}$ for $\theta^{-8/9}-r^{-8/9}=6$. $\frac{dr}{d\theta}=(\frac{\theta}{r})^{\frac{17}{9}}$

Explanation:

Step1: Differentiate both sides with respect to $\theta$

Differentiating $\theta^{-\frac{8}{9}}-r^{-\frac{8}{9}} = 6$ term - by - term. The derivative of $\theta^{-\frac{8}{9}}$ with respect to $\theta$ using the power rule $\frac{d}{d\theta}(x^n)=nx^{n - 1}$ is $-\frac{8}{9}\theta^{-\frac{8}{9}-1}=-\frac{8}{9}\theta^{-\frac{17}{9}}$. For the second term $-r^{-\frac{8}{9}}$, we use the chain - rule. Let $u = r$, then $\frac{d}{d\theta}(-r^{-\frac{8}{9}})=-(-\frac{8}{9})r^{-\frac{8}{9}-1}\frac{dr}{d\theta}=\frac{8}{9}r^{-\frac{17}{9}}\frac{dr}{d\theta}$. The derivative of the constant 6 with respect to $\theta$ is 0. So we have $-\frac{8}{9}\theta^{-\frac{17}{9}}+\frac{8}{9}r^{-\frac{17}{9}}\frac{dr}{d\theta}=0$.

Step2: Solve for $\frac{dr}{d\theta}$

First, move $-\frac{8}{9}\theta^{-\frac{17}{9}}$ to the other side of the equation: $\frac{8}{9}r^{-\frac{17}{9}}\frac{dr}{d\theta}=\frac{8}{9}\theta^{-\frac{17}{9}}$. Then multiply both sides by $\frac{9}{8}$ to get $r^{-\frac{17}{9}}\frac{dr}{d\theta}=\theta^{-\frac{17}{9}}$. Multiply both sides by $r^{\frac{17}{9}}$ to isolate $\frac{dr}{d\theta}$. We obtain $\frac{dr}{d\theta}=(\frac{\theta}{r})^{\frac{17}{9}}$.

Answer:

$\frac{dr}{d\theta}=(\frac{\theta}{r})^{\frac{17}{9}}$