I'm trying to solve a kinematic problem involving two angles. The second angle ($\theta_2$), can be expressed in the first angle ($theta_1$) by the equation shown in the title. However, after I have determined the equation for the position, I also have to compute the equations for the velocity and acceleration. This requires me to take the time derivative of the following equation, with $L_1$ and $L_2$ being constants and $\theta_1$ the position variable:
$$\frac{\mathrm d}{\mathrm dt}\sin^{-1}((L_1/L_2)\sin(\theta_1))$$ $$\frac{\mathrm d^2}{\mathrm dt^2}(\sin^{-1}((L_1/L_2)\sin(\theta_1)))$$
Does anyone know the solution to this?
It's easiest to do this implicitly. Let $$y=\sin^{-1}\frac{L_1}{L_2}\sin\theta \implies \sin y = \frac{L_1}{L_{2}}\sin\theta$$ You want to find $\frac{dy}{dt}$ and $\frac{d^{2}y}{dt^{2}}$. Differentiating, $$\cos y \frac{dy}{dt}=\frac{L_1}{L_2}\cos\theta \frac{d\theta}{dt}$$ Now, $$\cos y= \sqrt{1-\sin^{2}y}=\sqrt{1-\frac{L_{1}^{2}}{L_{2}^{2}}\sin^{2}\theta}$$ Bringing this together, $$\frac{dy}{dt}=\frac{r\cos\theta}{\sqrt{1-r^{2}\sin^{2}\theta}}\frac{d\theta}{dt}$$ Where $r=L_{1}/L_{2}$.
Differentiating again, you may find it helpful to differentiate first with respect to the variable $u=r\cos\theta$ and apply the chain rule. That is, write $$f(x)=\frac{x}{\sqrt{(1-r^{2})+x^{2}}}$$ Then $$\frac{dy}{dt}=f(r\cos\theta)\frac{d\theta}{dt}$$ $$\frac{d^{2}y}{dt^{2}}=-r\sin\theta \frac{d\theta}{dt} f'(r\cos\theta)\frac{d\theta}{dt}+f(r\cos\theta)\frac{d^{2}\theta}{dt^{2}}$$ Now observing that $$f'(x)=\frac{(1-r^{2})}{((1-r^{2})+x^{2})^{3/2}}$$ So $$f'(r\cos\theta)=\frac{(1-r^{2})}{(1-r^{2}\sin^{2}\theta)^{3/2}}$$ And hence, $$\frac{d^{2}y}{dt^{2}}=-r\sin\theta\frac{(1-r^{2})}{(1-r^{2}\sin^{2}\theta)^{3/2}}\left(\frac{d\theta}{dt}\right)^{2}+\frac{r\cos\theta}{\sqrt{1-r^{2}\sin^{2}\theta}}\frac{d^{2}\theta}{dt^{2}}\\=\frac{r}{(1-r^{2}\sin^{2}\theta)^{3/2}}\left(-(1-r^{2})\sin\theta \left(\frac{d\theta}{dt}\right)^{2}+r\cos\theta(1-r^{2}\sin^{2}\theta)\frac{d^{2}\theta}{dt^{2}}\right)$$