why does the intensity of light does not vary with time in youngs double slit experiment [
we can see that resultant electric field at a point p is given 
         $E=E0\sin(kx-\omega t+\epsilon)$[equation highlighed with yellow color]
as the resultant field varies with time, why not intensity varies with time
i know that intensity is defined as average over long time,
lets consider waves of very low frequency like 10hz interfere to produce interference pattern. then can we see varying intensity pattern with respect to time?
 A: In a sense it does vary, but the timescale of the variation is so fast that it requires extraordinary measures to develop a measurement system that can respond to these variations, and your eye most certainly can't see the variation.
Consider that the frequency of light is related to the wavelength by 
$$\nu=\frac{c}{\lambda}.$$
Then for green light ($\lambda\approx550\ {\rm nm}$), this gives a frequency of about $5.45\times10^{14}\ {\rm Hz}$, or an angular frequency ($\omega$) of $3.43\times10^{15}\ s^{-1}$.
For comparison, your eye doesn't respond to variations in optical intensity much above 20 Hz. And a very high speed photodiode might respond up to about $5\times 10^{10}\ {\rm Hz}$. 
Therefore we generally measure optical intensity as a time average over many periods of the electromagnetic wave, rather than instantaneously.

lets consider waves of very low frequency like 10hz interfere to produce interference pattern. then can we see varying intensity pattern with respect to time?

Yes, if you did the double slit experiment with a 10 Hz wave, you might just detect the intensity vary (since intensity will be varying at 20 Hz, double the field frequency) by eye. But of course you would need an instrument of some kind to measure the intensity in this experiment, and the apparatus itself would have to be immense to put the "screen" in the far field from the slits (which would have to be spaced by a significant fraction of the wavelength, which would be about $30\times10^6\ {\rm m}$).
