Hi

How would I do this question:

Water is running into a conical reservoir, 15 m deep and 8 m in diameter at the top at the rate of 1 cubic m/min. What is the depth when the level is rising 9/32pi m/min?

Thanks a lot.

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- Apr 9th 2010, 01:17 AMSunyataRates of Change
Hi

How would I do this question:

Water is running into a conical reservoir, 15 m deep and 8 m in diameter at the top at the rate of 1 cubic m/min. What is the depth when the level is rising 9/32pi m/min?

Thanks a lot. - Apr 9th 2010, 01:45 AMICanFly
Sorry, this "9/32pi m/min" is velocity? Level of velocity?

- Apr 9th 2010, 01:50 AMArchie Meade
Hi Sunyata,

$\displaystyle V_{cone}=\frac{{\pi}r^2h}{3}$

Write r in terms of h.

If you split the cone vertically, you have a right-angled triangle,

height H=15 and width R=4.

Hence $\displaystyle \frac{R}{H}=\frac{4}{15}$

$\displaystyle R=\frac{4H}{15}$

As the water flows in, the height and radius of the water vary, but they

still have the same ratio

$\displaystyle r=\frac{4h}{15}$

Hence

$\displaystyle v_{water}=\frac{\pi}{3}\frac{16h^2}{225}h=\frac{16 {\pi}h^3}{3(225)}$

As the water enters at $\displaystyle 1\ m^3$ per minute

$\displaystyle \frac{dv}{dt}=1=\frac{d}{dt}\frac{16{\pi}h^3}{3(22 5)}=\frac{16{\pi}}{3(225)}\frac{dh^3}{dt}$

$\displaystyle \frac{dh^3}{dt}=\frac{3(225)}{16{\pi}}\ \Rightarrow\ 3h^2\frac{dh}{dt}=\frac{3(225)}{16{\pi}}$

Now substitute your value for $\displaystyle \frac{dh}{dt}=\frac{9}{32{\pi}}$

to find "h squared" and take the square root.