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Differentiating $r = f(\theta)$,\ $r = \frac{2}{2 - \cos\theta} = 2(2 - \cos\theta)^{-1}$ $\dot{r} = -2(2 - \cos\theta)^{-2}\sin\theta\dot{\theta} = \frac{-2\dot{\theta}\sin\theta}{(2 - \cos\theta)^2} + 2$ $\ddot{r} = 4(2 - \cos\theta)^{-3}(\sin\theta\dot{\theta})^2 - 2(2 - \cos\theta)^{-2}\cos\theta\dot{\theta}^2 - 2(2 - \cos\theta)^{-2}\sin\theta\ddot{\theta} = \frac{4(\sin\theta\dot{\theta})^2}{(2 - \cos\theta)^3} - \frac{2(\cos\theta\dot{\theta}^2 + \sin\theta\ddot{\theta})}{(2 - \cos\theta)^2} + 2$

          Differentiating $r = f(\theta)$,\
$r = \frac{2}{2 - \cos\theta} = 2(2 - \cos\theta)^{-1}$ 
$\dot{r} = -2(2 - \cos\theta)^{-2}\sin\theta\dot{\theta} = \frac{-2\dot{\theta}\sin\theta}{(2 - \cos\theta)^2} + 2$
$\ddot{r} = 4(2 - \cos\theta)^{-3}(\sin\theta\dot{\theta})^2 - 2(2 - \cos\theta)^{-2}\cos\theta\dot{\theta}^2 - 2(2 - \cos\theta)^{-2}\sin\theta\ddot{\theta} = \frac{4(\sin\theta\dot{\theta})^2}{(2 - \cos\theta)^3} - \frac{2(\cos\theta\dot{\theta}^2 + \sin\theta\ddot{\theta})}{(2 - \cos\theta)^2} + 2$

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Differentiating r = f(θ),r = (2)/(2 - cosθ) = 2(2 - cosθ)^-1
ṙ = -2(2 - cosθ)^-2sinθθ̇ = (-2θ̇sinθ)/((2 - cosθ)^2) + 2
r̈ = 4(2 - cosθ)^-3(sinθθ̇)^2 - 2(2 - cosθ)^-2cosθθ̇^2 - 2(2 - cosθ)^-2sinθθ̈ = (4(sinθθ̇)^2)/((2 - cosθ)^3) - (2(cosθθ̇^2 + sinθθ̈))/((2 - cosθ)^2) + 2

Added by Susana W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Brian Ketelobeter

08/04/2019

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Step 1: Differentiate $r$ with respect to $\theta$ using the power rule and chain rule. Show more…

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Differentiating $r=f(\theta)$, $$r = \frac{2}{2-\cos\theta} = 2(2-\cos\theta)^{-1}$$ $$\dot{r} = -2(2-\cos\theta)^{-2}\sin\theta\dot{\theta} = \frac{-2\dot{\theta}\sin\theta}{(2-\cos\theta)^2} + 2$$ $$\ddot{r} = 4(2-\cos\theta)^{-3}(\sin\theta\dot{\theta})^2 - 2(2-\cos\theta)^{-2}\cos\theta\dot{\theta}^2 - 2(2-\cos\theta)^{-2}\sin\theta\ddot{\theta} = \frac{4(\sin\theta\dot{\theta})^2}{(2-\cos\theta)^3} - \frac{2(\cos\theta\dot{\theta}^2 + \sin\theta\ddot{\theta})}{(2-\cos\theta)^2} + 2$$

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