The issue here is that you are assuming that over a small time interval you are considering, you are assuming that the acceleration on the ball is constant. However, the acceleration changes no matter how small of a 'time interval' you consider. $v(t_0+\delta t)=v(t_0)+\frac{dv}{dt}(t_0)\delta t$ is essentially what you are doing: expanding to first order in the derivative of the velocity to figure out what the velocity is at $t_0+\delta t$, which is fine, but there will be an error if you take finite (non-zero) interval of time. This is an identical issue of following a tangent line at, say point (1,0) to the unit circle centre origin. You can approximate a point on the unit circle that is just above this point by following the tangent line, but you will be off by a certain amount. The position vector to that point in the tangent line will have a length slightly greater than one. That error is because by using the tangent line you ignore that the true path, the circle, is continuously curving.

The issue here is that you are assuming that over a small time interval you are considering, the acceleration on the ball is constant. However, the acceleration changes no matter how small of a 'time interval' you consider. What you are essentially doing is this: $$v(t_0+\delta t)=v(t_0)+\frac{dv}{dt}(t_0)\delta t$$ You're expanding to first order in the derivative of the velocity to figure out what the velocity is at $t_0+\delta t$, which is fine, but there will be an error if you take finite (non-zero) interval of time. This is an identical issue of following a tangent line at, say point (1,0) to the unit circle centre origin. You can approximate a point on the unit circle that is just above this point by following the tangent line, but you will be off by a certain amount. The position vector to that point in the tangent line will have a length slightly greater than one. That error is because by using the tangent line you ignore that the true path, the circle, is continuously curving.