Rate of increase of a circle area wrt to radius I am trying to solve this through differential equation but the results seems different from calculating with real numbers. the problem is simple 
The radius of a circle is growing at the rate of $d$ units/sec, its initial radius is $R$, find the rate of increase of its area.  The question is transcribed into the simple differential equation:
$Area= \pi.R^2$
$dA/dt = 2.\pi.R.dR/dt$
With an example say the initial radius is 4 units and the rate of increase is 0.5 units/sec:
$dA/dt = 2.\pi.4.(0.5) = 4.\pi$
However when I substitute real values
$Area (4) = \pi.4^2 = 16.\pi$
$Area (4.5) = \pi.(4.5)^2 = (20.25).\pi$
$Diff = (20.25).\pi - 16.\pi = (4.25).\pi$
The difference between these two is $(4.25).\pi$ and the $4.\pi$ as I obtained earlier.  Why this difference?  Shouldn't it be exactly $4.\pi$ ?  What am I missing?
Thanks
vijay
 A: @Vijay 's comment is the key to understanding your problem.
The derivative is the rate of change. If you look back to the definition of the derivative you'll see that it says (more or less) that the value of the difference  is approximately the product of the derivative (the rate of change) and the time (in your example, $1$ second).
$$
 (20.25)\pi−16\pi  = 4.25\pi \approx 4\pi \times 1
$$
That approximation is better for smaller time intervals. You can test that numerically.
A: Remember....that calculus is useful only for small changes....ie the rate of change of a quantity is quite small as compared to the quantity itself...and u can check this out urself by plugging in larger and larger values of dR/dt and u would notice that the answer u get from calculus would start becoming highly inaccurate as compared to what u actually calculate by simply plugging in final and initial values.
So remember that calculus can be used in such situations only when rate of change of a quantity is considerably small as compared to the quantity itself. Although there is no specific limit for its accuracy but a % change of roughly 3-4 % would give u a nearly accurate answer....however the smaller the rate of change....the more accurate the results obtained using calculus are.
