# integration with respect to time

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)

$$\ddot{r} = r \dot \theta ^2 + r\dot\phi^2 sin^2(\theta) -\frac{GM}{r^2}$$ $$\ddot \theta = \dot \phi^2 sin(\theta)cos(\theta) - 2 \frac{\dot r}{r} \theta$$

• I'm assuming $r$, $\theta$, and $\phi$ functions of $t$, right? And when you integrate acceleration, you get velocity as a result, not necessarily speed because speed is just how fast something's going regardless of its direction. So if I'm correct, you need to find the velocity function. Aug 25, 2022 at 17:54
• Yes you are correct. I have solved other parts of my problem (so far), and got to this problem where I needed some help. I will post the whole problem/task tomorrow and tag you so you can see it. Shortly, I want to solve initial theta and phi velocoties so that the particle stays on the 'surface of sphere' while orbiting center/mass under potential (i.e. Newtonian). This also means that the 'r' (radial) acceleration and velocities are zero as the distance r stays constant (radius of sphere). Aug 25, 2022 at 19:29

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 $$\dot\phi=\frac{C_0}{r^2\sin\theta}.$$
• Just a side note: In the solution above we obviously "lose" the trivial solution $\dot\phi=0$ when taking the logarithm. Also, the constant $C_0=e^{C_1}$ is always positive. Symmetrically, $\dot\phi$ can also be negative; then, the second equation would be modified to $\ln(-\dot\phi)=...$ leading to $\dot\phi=\frac{-C_0}{r^2\sin\theta}$. Putting all this together, we see that $\phi\in C^2((a,b))$ is a solution if and only if there is $C\in\mathbb{R}$ such that $\dot\phi=\frac{C}{r^2\sin\theta}$ in $(a,b)$. Aug 25, 2022 at 10:59