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.
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.