Uniform Circular Motion and Centripetal Acceleration In introductory physics books (or at least mine) it limits the equation $a_c=v^2/r$ to the situation where the speed around the circular path is constant. It enforces the idea that the speed is CONSTANT. 
But wouldn't the equation also apply to non-constant speeds? ($a_c$ would just change from being a constant to being a function of the speed)
It would be very counter-intuitive to me if this equation did not apply to variable speeds (because why does this instant in time care about the speed of the next instant in time?)
So my question is, can you also use this equation for variable speeds?
 A: In variable speeds also, the centripetal force does apply, for example, the earth moves around the sun in an ellipse with speeds varying at different points, but we can take it at any particular instant, and use the formula, BUT, there is an additional component of acceleration when it comes to circle of varying speeds, the tangential component, which is absent in the uniform circular motion, which means;

*

*for a planet of mass $m$ around sun of mass $M$ at distance $r$, following a perfect circle we have at any instant $\dfrac{mv^2}{r}=G\dfrac{Mm}{r^2}$.
BUT

*for a planet with varying speed orbit which can only be an ellipse will have at any instant $\left(G\dfrac{m_1m_2}{r^2}\right)^2= \left(\dfrac{mv^2}{r}\right)^2 + \left(\dfrac{dv}{dt}\right)^2$(at that instant).
You can easily see that the second equation reduces to the first when $v$ is constant,
and as for that question about why particle at one instant cares about the other instant, well, it DOESN'T, you can very well right at any instant in any orbit as $\left(G\dfrac{m_1m_2}{r^2}\right) \cos x = \dfrac{mv^2}{r}$ where $x$ is the angle between gravity and the normal, the only thing is there is an additional component (the ($\sin x$) one ) which causes change in speed and hence alters the expression.

A: See this answer on how a particle moving along a path will have acceleration tangent to the path equal to the scalar $a_t = \frac{{\rm d}v(t)}{{\rm d}t}$ and acceleration perpendicular to the path equal to the scalar $a_c = \frac{{v(t)}^2}{r(t)}$.
So you are correct to assume that this applies for all cases when looking at the component of acceleration normal to the path of motion. As a total acceleration state the complete picture is given by
$$ \vec{a}(t) = \dot{v}(t)\, \vec{e} + \frac{{v(t)}^2}{r(t)} \vec{n} $$
This can be used to find the instantaneous radius of curvature of a path as
$$ \boxed{ r(t) = \dfrac{ \left( \dot{x}(t)^2 + \dot{y}(t)^2 \right)^\frac{3}{2} }
{\ddot{x}(t) \dot{y}(t) - \ddot{y}(t) \dot{x}(t) } }$$
A: Let's check the intuition. We have:
$$\vec{v}(t) = v(t)\hat{\theta}$$
where v(t) is the magnitude of the velocity and $\hat{\theta}$ is the unit vector in the tangential direction. This unit vector is obviously a function of the position and this then implicitely depends on the time. Differentiating again gives:
$$\vec{a}(t) = \frac{dv}{dt}\hat{\theta} + v\frac{\partial\hat{\theta}}{\partial\theta} \frac{d\theta}{dt}$$
Here we've used that the particle moves along a circle, so $\hat{\theta}$ can be copnsidered as a function of only $\theta$ and the total derivative w.r.t. time is then obtained by multiplying that with the derivative of $\theta$ w.r.t. time.
Then we have:
$$\frac{\partial\hat{\theta}}{\partial\theta} = -\hat{r}$$
and:
$$\frac{d\theta}{dt}= \frac{v}{r}$$
Therefore the acceleration is:
$$\vec{a}(t) = \frac{dv}{dt}\hat{\theta} - \frac{v^2}{r}\hat{r}$$
So, there is now a tangential component due the the derivative of the magnitude of the acceleration, but the radial component is given by the usual centripetal acceleration formula.
