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"A circular oil spill grows at a rate given by the differential equation $dr/dt = k/r$, where $r$ represents the radius of the spill in feet, and time $t$ is measured in hours. If the radius of the spill is $400$ feet $16$ hours after the spill begins, what is the value of $k$? Include units in your answer.:

So spill is spreading at $400/16$ per hour which is $25$ per hour. \begin{align*} \frac{dr}{dt} = 25 &= \frac{k}{r} \\ \frac{k}{r}& = 25\\ r &= \frac{1}{25}k\\ \frac{dr}{dt}& = 0 \end{align*} $\frac{dr}{dt} = 0$ only when $\frac{k}{r} = 0$ so by logic the answer is when $k = 0$

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Solving the differential equation you get $$\frac{1}{2}r^2=kt+C$$ where $C$ is a integration constant. Let´s assume $t=0$ implies $r=0$, then $C=0$, so $r=\sqrt{2kt}$. Then, since $r=400$ when $t=16$, it follows $$k=\frac{400^2}{2(16)}=5000$$

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