find $\frac{dr}{d\theta}$ for $\theta^{-8/9}-r^{-8/9}=6$.\n$\frac{dr}{d\theta}=square$
find $\frac{dr}{d\theta}$ for $\theta^{-8/9}-r^{-8/9}=6$.\n$\frac{dr}{d\theta}=square$
Answer
Explanation:
Step1: Differentiate both sides
Differentiate $\theta^{-\frac{8}{9}}-r^{-\frac{8}{9}} = 6$ with respect to $\theta$. The derivative of a constant 6 is 0. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $\frac{d}{d\theta}(\theta^{-\frac{8}{9}})-\frac{d}{d\theta}(r^{-\frac{8}{9}})=0$. $-\frac{8}{9}\theta^{-\frac{8}{9}-1}-\left(-\frac{8}{9}r^{-\frac{8}{9}-1}\frac{dr}{d\theta}\right)=0$.
Step2: Simplify the equation
Simplify the exponents: $-\frac{8}{9}\theta^{-\frac{17}{9}}+\frac{8}{9}r^{-\frac{17}{9}}\frac{dr}{d\theta}=0$.
Step3: Isolate $\frac{dr}{d\theta}$
Add $\frac{8}{9}\theta^{-\frac{17}{9}}$ to both sides: $\frac{8}{9}r^{-\frac{17}{9}}\frac{dr}{d\theta}=\frac{8}{9}\theta^{-\frac{17}{9}}$. Then divide both sides by $\frac{8}{9}r^{-\frac{17}{9}}$: $\frac{dr}{d\theta}=\left(\frac{\theta}{r}\right)^{\frac{17}{9}}$.
Answer:
$\left(\frac{\theta}{r}\right)^{\frac{17}{9}}$