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First timer here,

So I have a problem with an oil slick in the form of a circle. I know the rate at which a ruptured pipe pumps oil(gallons per minute), the thickness of the oil on top and the number of days for which spilled oil covers the area. I need volume of oil on top of the water(in galls) and spilled area (in km) after each day.

Example of the known data

                          -> pump rate: 30 gallons/min, 
                          -> thickness: 0.5cm
                          -> number of days: 55

dV/dt = 30 g/min ??, h = 0.5cm

Is this right:

I take the cylinder Volume formula as V = B*h where

  B = V / h

and B = 30 * 3785.412(cubic centimeter) * 60(min/h) * 24(hours/day) * 55(days) etc. From here I take that B = r^2 * PI -> r = sqrt(B/PI) to get the radius and now I'm stuck.

Am I right so far; it'd be nice if you could help in detail since I'm so far away from my high school calculus days. Thank you.

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The standard formulas for volumes and areas do the job; there is no calculus needed here.

Let $V(t)$be the volume and $r(t)$ be the radius of the cylinder at time $t$. The pump rate $\rho$ and the height $h$ of the cylinder are assumed constant; furthermore we are told that at time $T$ the volume is zero. We then have $$\pi r^2(t) h=V(t)=\rho(T-t)\ ,$$ which already gives $V(t)$ explicitly. Solving for $r(t)$ we then obtain $$r(t)=\sqrt{\rho(T-t)\over \pi h}\ .$$ Let us measure time in days d and lengths in meters m, hence $V(t)$ comes in m${}^3$. Then $$\rho=30\cdot{3.785412\over1000}\cdot60\cdot 24=163.66\quad[{\rm m}^3/{\rm d}]\ ,\qquad h=0.005\quad[{\rm m}]\ . $$ It follows that $$r(t)=102.073\>\sqrt{\mathstrut 55-t}\quad[{\rm m}]\ .$$

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  • $\begingroup$ No need for derivatives then? $\endgroup$ – learner Apr 21 '18 at 21:19

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