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So a right circular cone is 25m deep with a radius of 9m. It's being filled with water at a rate of $3m^3$ per minute. a) find dh/dt when its height is 10m.

What I did first was use a proportion to find the new radius: $9/25 = r/10$ therefore $r = 3.6$. Next, since I don't have dr/dt, I put $r^2$ in terms of $h$ in the volume equation $V = (π/3) * r^2 * h$. The new equation, in context should be $V = (π/3)*1.296h^2$.

Deriving yields $dv/dt = 1.296(π/3) * 2h *dh/dt$. Plugging $h$ in and solving should give $dh/dt = .111$

This answer seems incorrect. Is it? If so, how?

now b): when the radius of the water surface is 4m, what is $dr/dt$? Can I use a similar process to what I did with the height?

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  • $\begingroup$ Please use parentheses to show that by $V = π/3 * r^2 * h$ you mean $V = (π/3) * r^2 * h$. Many times when people write $a/b * c$ they mean $a/(b * c) $\endgroup$ Commented Nov 13, 2012 at 4:25
  • $\begingroup$ Oops. I'll make the change. Also, another issue, the title says "cylinder" not cone. I don't think I can change the title though. $\endgroup$ Commented Nov 13, 2012 at 4:44

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Note that the units are wrong in $V = (π/3)*1.296h^2$. You have $r=0.36h$, so $V=h^3(0.36)^2\pi/3=0.1296h^3\pi/3$. Now try the derivative. A check is that the area of the upper surface is $\pi(3.6)^2$, so the rate of rise is $3/(\pi(3.6)^2)$

For b, yes, you can do the same thing-write $V$ as a function of $r$, take the derivative, and use the fact that $V'=3 m^3/min$

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