# Rates of Change

• Apr 9th 2010, 01:17 AM
Sunyata
Rates 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 AM
ICanFly
Sorry, this "9/32pi m/min" is velocity? Level of velocity?
• Apr 9th 2010, 01:50 AM
Quote:

Originally Posted by Sunyata
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,

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

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

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

As the water flows in, the height and radius of the water vary, but they
still have the same ratio

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

Hence

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

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

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

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

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

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