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A water tank shaped like a cone pointing downwards is $10$ metres high. $2$ metres above the tip the radius is $1$ metre. Water is pouring from the tank into a cylindrical barrel with vertical axis and a diameter of $8$ metres. Assume that the height of the water in the tank is $4$ metres, and is decreasing at a rate of $0.2$ metres per second. How fast is the height of the water in the barrel changing?

Made a function for the volume of water that drains from the cone: $\frac {16 \pi} 3 - \frac \pi 3 \frac {(4 - 0,2 t)^3} 4$. And the volume of the cylinder is: $\pi r^2 h$.

Where do I go from here?

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Your volume expression is not correct. You have not defined the variable $t$, but it appears to be time (since when?) You need to compute the volume of water as a function of $h$, the height above the tip. Then we are given $h=4, \frac {dh}{dt}=0.2$ You can use $\frac {dV}{dt}=\dfrac {\frac {dh}{dt}}{\frac {dV}{dh}}$ to get $\frac {dV}{dh}$, the rate water is flowing out. Then divide by the area of the top surface of the barrel to get the rate the water is rising.

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  • $\begingroup$ The volume expression is correct for what's left in the cone at all times (t from 0 to 20). And when I take the derivative of my function, I get the volume of water that flows out at any second. 4-0.2t is the height of the water in the cone. $\endgroup$
    – ccipher
    Sep 17 '13 at 19:30
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Answer is $1/320(4-0.2t)^2$

Or am I terribly wrong?

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