In non-uniform circular motion, the centripetal acceleration and tangential acceleration are perpendicular to each other, but does that mean that they don't affect each other because centripetal acceleration depends on velocity (=$v^2/r$) and tangential acceleration's direction (i.e., whether it will act along the direction of motion or opposite to the direction of motion) also depends upon the change in velocity. So are they dependent on each other or not?
[Physics] In non-uniform circular motion are centripetal acceleration and tangential acceleration independent of each other
kinematicsnewtonian-mechanicsrotational-dynamics
Related Solutions
The initial premise of your question is not, in general, correct.
Consider a 1000 kg car driving at a constant speed of 20 m/s around a flat circular racetrack with a radius of 500 meters. Note the constant speed: the car will take the same time to drive around the track, 157.1 seconds, hour after hour. The only change in the car's velocity is in the direction of the velocity, not its size. The only acceleration is the centripetal acceleration, and the only horizontal force needed is the centripetal force $$a_{cp}=\frac {v^2}{r}=0.8 \frac{m}{sec^2}$$ $$F_{cp}=m \times a_{cp}=800 \ newtons$$
The force exerted by the car's engine serves only to balance the various drag forces.
This force is supplied by the friction between the tires and the pavement; pour out some oil on the track to see what happens when the required centripetal force is not present!
The only acceleration is directed exactly towards the center of the circular track; there is no tangential acceleration. The car, at some moment, is travelling North at 20 m/s, and one half-circuit later, it is travelling South at the same speed. Clearly, it has accelerated.
Assume now that the driver presses on the brake pedal in a manner that she knows will bring the car to a stop in 40 seconds. There is now a tangential acceleration, at $-0.5 \frac{m}{sec^2}$, in addition to the centripetal acceleration above, and a tangential force of 500 newtons directed towards the back of the car. The total force needed from the tires is now the resultant of these two forces: 943.4 newtons directed around 32 degrees aft of inward. Touching the brakes could throw you into a skid! Of course, as the braking changes the car's speed, the centripetal force will decrease...
The proper derivation of the centripetal acceleration—without assuming any kinematic variables are constant—requires a solid understanding of both the stationary Cartesian unit vectors $\hat{i}$ and $\hat{j}$ as well as the rotating polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$. The Cartesian unit vectors $\hat{i}$ and $\hat{j}$ are stationary and always aligned with the X and Y axes respectively, while the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ rotate with an angular velocity of $\omega=\|\dot{\theta}\|$ and point in the directions of increasing radius and angle (respectively). The included graphic below shows the two basis vector pairs overlaid on top of one another.
The position vector of the object is obviously defined as:
$\vec{p}(t)=x\hat{i}+y\hat{j}=rcos(\theta)\hat{i}+rsin(\theta)\hat{j}$,
with
$\|\vec{p}(t)\|=\sqrt{(rcos{\theta})^2+(rsin{\theta})^2}=\sqrt{r^2(sin^2(\theta)+cos^2(\theta))}=r\sqrt{(1)}=r$
Less obviously, it can be shown that the polar unit vectors $\hat{e}_r$ and $\hat{e}_\theta$ can be expressed solely in terms of the Cartesian unit vectors $\hat{i}$ and $\hat{j}$ and the angular position $\theta$ as,
$\boxed{\hat{e}_r=cos(\theta)\hat{i}+sin(\theta)\hat{j}}$ and $\boxed{\hat{e}_\theta=-sin(\theta)\hat{i}+cos(\theta)\hat{j}}$.
These two equations are extremely important, as they will be the key to expressing the Cartesian acceleration in polar coordinates, of which one of the terms will be our desired $v^2/r=\omega^2r$ centripetal acceleration. Moving forward, the vector acceleration of the object in Cartesian coordinates is simply
$\vec{a}(t)=\frac{d^2}{dt^2}\left[\vec{p}(t)\right]=\ddot{x}\hat{i}+\ddot{y}\hat{j}$.
Starting with $x=rcos(\theta)$ and $y=rsin(\theta)$ and differentiating once, we have
$\boxed{\dot{x}=\dot{r}cos(\theta)-r\dot{\theta}sin(\theta)}$ and $\boxed{\dot{y}=\dot{r}sin(\theta)+r\dot{\theta}cos(\theta)}$.
Differentiating again, we will have
$\ddot{x}=\ddot{r}cos(\theta)-\dot{r}\dot{\theta}sin(\theta)-\dot{r}\dot{\theta}sin(\theta)-r\frac{d}{dt}\left[\dot{\theta}sin(\theta)\right]$
$=\ddot{r}cos(\theta)-2\dot{r}\dot{\theta}sin(\theta)-r\left[\ddot{\theta}sin(\theta)+{\dot{\theta}}^2cos(\theta)\right]$, such that
$\boxed{\ddot{x}=(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))}$.
Similarly, the y acceleration $\ddot{y}$ becomes
$\ddot{y}=\ddot{r}sin(\theta)+\dot{r}\dot{\theta}cos(\theta)+\dot{r}\dot{\theta}cos(\theta)+r\frac{d}{dt}\left[\dot{\theta}cos(\theta)\right]$
$=\ddot{r}sin(\theta)+2\dot{r}\dot{\theta}cos(\theta)+r\left[\ddot{\theta}cos(\theta)-{\dot{\theta}}^2sin(\theta)\right]$, such that
$\boxed{\ddot{y}=(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})cos(\theta)}$.
Now, we must plug these scalar derivatives into our formulation for the vector acceleration. In Cartesian coordinates, this is
$\vec{a}(t)=\ddot{x}\hat{i}+\ddot{y}\hat{j}=\{(\ddot{r}-r\dot{\theta}^2)cos(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(-sin(\theta))\}\hat{i}+\{(\ddot{r}-r\dot{\theta}^2)sin(\theta)+(r\ddot{\theta}+2\dot{r}\dot{\theta})(cos(\theta))\}\hat{j}$
which can be rearranged into the following form:
$\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\{cos(\theta)\hat{i}+sin(\theta)\hat{j}\}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\{-sin(\theta)\hat{i}+cos(\theta)\hat{j}\}$
But as we have already seen, this is simply equal to
$\boxed{\boxed{\vec{a}(t)=(\ddot{r}-r\dot{\theta}^2)\hat{e}_r+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{e}_\theta}}$
As we can now appreciate from carrying out the full derivation, there are actually two components each to both the radial and tangential accelerations. The $\ddot{r}$ term is straightforwardly equal to the second derivative of the position vector magnitude. The second term, $r\dot{\theta}^2$, is our long sought-after centripetal acceleration $r\dot{\theta}^2=\omega^2r=v^2/r$, and (as expected) it points in the negative radial direction. The tangential terms are perhaps a bit less intuitive. The $r\ddot{\theta}$ term is the acceleration that occurs whenever the radius and angular acceleration $\ddot{\theta}$ are both non-zero (imagine the tangential acceleration of a turbine blade of a jet engine as the engine spools up). The final term $2\dot{r}\dot{\theta}$ is what's commonly known as the Coriolis acceleration, and it occurs whenever the radius and angle change simultaneously. It arises because, for a given angular velocity, the arc length travelled every second increases with radius (tangential velocity increases with radius). Thus, an object with a given angular velocity will have different tangential velocities at different local radii of rotation. If the radius changes with time ($\dot{r}\not=0$) and the angular velocity $\dot{\theta}$ is not equal to zero, then the tangential velocity will change with time, which is by definition a tangential acceleration.
Best Answer
Let's look at the general acceleration vector in polar coordinates$^*$:
$$\mathbf a=\left(\ddot r-r\dot\theta^2\right)\hat r+\left(r\ddot\theta+2\dot r\dot\theta\right)\hat\theta$$
If we want our object to remain on the same circle, which I'm assuming this is what you are interested in, we must have $\dot r=0$ and $\ddot r=0$. This means our acceleration must have the form: $$\mathbf a=-r\dot\theta^2\hat r+r\ddot\theta\hat\theta$$
Since, by Newton's laws, the acceleration vector is proportional to the force via the mass $m$ of our particle, we see that we need a radial force magnitude$^{**}$ of $$F_r=-mr\dot\theta^2$$ and a tangential force magnitude of $$F_\theta=mr\ddot\theta$$
Because we are confined to move around a single circle we can determine the role of each force. The radial force must only be responsible for changing the direction of the velocity, since if it could effect the speed then this means the object would have to change its $r$ coordinate, hence knocking us off the circle. Similarly, the tangential force must only be responsible for changing the speed of the particle as it moves around the circle, since if it could effect the direction of the velocity then it would do so by knocking us off the circle.
You can probably tell by now that if we want to stay on the circle, these forces are required to be "linked", in a sense. Indeed, if you take the time derivative of $F_r$ and use the requirement that $\dot r=0$, you will find that $$\frac{\text dF_r}{\text d t}=-2\dot\theta F_\theta$$ showing that the presence of a tangential force requires a change in the magnitude of the radial force in order for the particle to remain on the circle (or, on the flip side, a change in the magnitude of the radial force must be accompanied by a tangential force).
What if this condition is not met? Well, looking at the derivative of $F_r$ without the condition that $\dot r=0$, we must have $$-m\dot r\dot\theta^2\neq 0$$ This means that $\dot r\neq0$, which means we are no longer engaging in circular motion. Therefore, we need these two force components to be linked in this way to keep the motion circular.
A subtle point remains to be cleared up (as realized in comments to other answers). This does not necessarily mean that these forces are physically linked in general. This answer assumes we have an object undergoing non-uniform circular motion, and then investigates what must be true about the forces acting on the object. However, there could be situations where the radial and tangential force are not physically linked, at which point to achieve non-uniform circular motion you would have to make these forces act in such a way so that non-uniform circular motion is achieved.
$^*$ dots represent a rate of change with respect to time if you are not familiar with calculus. for example, $\dot r$ is the rate of change of the variable $r$ with respect to time. The derivation of this equation can be found here.
$^{**}$ Note that this is what you usually encounter in your introductory physics classes as $F_{r}=mv^2/r$, since for motion along a circle of radius $r$, $\dot\theta=v/r$. The negative sign in this answer is to keep track of the direction of increasing/decreasing $r$, but if you are working a question where you only care about the magnitude of the radial force then this is irrelevant, hence why you usually don't see the negative sign.