After a sewage spill, the level of pollution in Sootville is estimated by $f(t)=\frac{450 t^{2}}{\sqrt{t^{2}+19}}$, where $t$ is the time in daysosince the spill occurred. How fast is the level changing after 6 days. Round to the nearest whole number.

Step 1 ：The problem is asking for the rate of change of the pollution level after 6 days. This is a calculus problem, specifically involving derivatives. The derivative of a function gives the rate of change of the function at any given point. So, we need to find the derivative of the function \(f(t)\) and then evaluate it at \(t=6\).

Step 2 ：Given the function \(f(t) = \frac{450 t^{2}}{\sqrt{t^{2}+19}}\), we need to find its derivative \(f'(t)\).

Step 3 ：The derivative of the function \(f(t)\) is \(f'(t) = -\frac{450 t^{3}}{(t^{2} + 19)^{3/2}} + \frac{900 t}{\sqrt{t^{2} + 19}}\).

Step 4 ：Substitute \(t = 6\) into \(f'(t)\) to find the rate of change of the pollution level after 6 days.

Step 5 ：The rate of change of the pollution level after 6 days is 490 units per day.

Step 6 ：Final Answer: The level of pollution in Sootville is changing at a rate of \(\boxed{490}\) units per day after 6 days.