I am pretty sure this is a really basic question, but after the summer, not using integrals / derivatives at all, I just can not remember how to do math anymore.
I have a function for acceleration that I need to integrate with respect to time $t$ in order to obtain speed.
My acceleration function is (for angle $\phi$)
$$\ddot \phi = -2 \frac{\dot r}{r} \dot \phi – \dot \phi \dot \theta \frac{cos(\theta)}{sin(\theta)}$$
where $$\ddot \phi = \frac{d^2}{dt^2}\phi $$ and $$\dot \theta = \frac{d}{dt}\theta $$ and $$\dot \phi = \frac{d}{dt}\phi$$ and so on.
so I would need to obtain the $\dot \phi$ by integrating the acceleration once.
$$\dot \phi = \int \ddot \phi dt= \int \Big(-2 \frac{\dot r}{r} \dot \phi – \dot \phi \dot \theta \frac{cos(\theta)}{sin(\theta)} \Big) dt$$
I just do not remember at all how to start to process this problem. I know it's almost a bit shameful but indeed could use a helping hand here.
(I have 2 similar equations to be solved for $\ddot \theta$ and $\ddot r$ but I am pretty sure I get these done when I once remember how to calculate things)
Thank you so much if you could help me out.
Those 2 other equations are following : (if someone is interested)
$$\ddot{r} = r \dot \theta ^2 + r\dot\phi^2 sin^2(\theta) -\frac{GM}{r^2}$$
$$\ddot \theta = \dot \phi^2 sin(\theta)cos(\theta) – 2 \frac{\dot r}{r} \theta$$
Best Answer
You can divide the equation by $\dot\phi$ to obtain $$\frac{\ddot\phi}{\dot\phi}=-2\frac{\dot r}{r}-\dot\theta\frac{\cos\theta}{\sin\theta}.$$ Integrating over time yields $$\ln \dot\phi=-2\ln r-\ln(\sin\theta)+C_1,$$ hence $$\dot\phi=\frac{C_0}{r^2\sin\theta}.$$