A 5-gram drop of thick red paint is added to a large can of white paint.A red disk forms and spreads outward, growing lighter at the edges.Since the amount of red paint stays constant through time, the density of the red paint in the disk must vary with time.Suppose that its density p in gm/cm² is of the form $p=k(t) f(r)$ for some functions k(t)of time and f(r)of the distance to the center of the disk.For fixed r, which of the following integrals gives the average density of red paint at a distance r from the center of the disk from 0 to T seconds? A) $\frac{k(t)}{T} \int_{0}^{T} f(r) d r$ B) $\frac{f(r)}{T} \int_{0}^{T} k(t) d t$ C) $\frac{k(t)}{5 T} \int_{0}^{T} f(r) d r$ D) $\frac{f(r)}{5 T} \int_{0}^{T} k(t) d t$

Question

Quizplus · Accepted Answer

The average density of red paint at a distance r from the center over time is obtained by averaging the time-dependent part of the density function, $k(t)$, over the interval from 0 to T seconds. Since $p=k(t)f(r)$, and we are looking for the average density at a fixed r, we integrate $k(t)$ over time and divide by T, while $f(r)$ remains constant for the fixed distance r. This is correctly represented by $\frac{f(r)}{T} \int_{0}^{T} k(t) d t$.

A 5-Gram Drop of Thick Red Paint Is Added to a Large