# Water is flowing at a rate of 5 m^3/s into the conical tank

• Feb 6th 2009, 12:55 PM
Jasmina8
Water is flowing at a rate of 5 m^3/s into the conical tank
Water is flowing at a rate of 5 m^3/s into the conical tank. (The conical tank looks like this: the whole height of the conical tank is 120 m, the height of the water is h, the width of the top of the conical tank is 60 m)

a. find the volume V of the water as a function of the water level h.

b. find h as a function of the time t during which water has been flowing into the tank.

Help?(Speechless)
• Feb 6th 2009, 01:01 PM
Jhevon
Quote:

Originally Posted by Jasmina8
Water is flowing at a rate of 5 m^3/s into the conical tank. (The conical tank looks like this: the whole height of the conical tank is 120 m, the height of the water is h, the width of the top of the conical tank is 60 m)

a. find the volume V of the water as a function of the water level h.

see post #3 here and see if you can do this part.

Quote:

b. find h as a function of the time t during which water has been flowing into the tank.
this is for calculus, right? find dh/dt so you can see how fast the height is changing per unit time, you should be able to come up with a formula from there
• Feb 6th 2009, 01:14 PM
Jasmina8
this is for pre calcc....i really dont get how to do that problem that u told me to see...and i've been trying to do my problem for like 2 hrs. and i dont get it...like i'm having a really hard time starting it off...
• Feb 6th 2009, 01:19 PM
galactus
A handy trick to remember with related rates when you have an area involved is to know that "The rate of change of volume is equal to the cross sectional area at that instant times the rate of change of the height."

$\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}$

You know that dV/dt=5.

A(t)=the area of the circular surface of the water at height h.

By similar triangles, $\displaystyle \frac{1}{4}=\frac{r}{h}\Rightarrow r=\frac{h}{4}$

The area of the surface at height h, A(t), is then $\displaystyle \frac{{\pi}h^{2}}{16}$

So, $\displaystyle \frac{dh}{dt}=\frac{80}{{\pi}h^{2}}$
• Feb 6th 2009, 01:31 PM
Jasmina8
.....
but what do i plug into the D's...idk where did u get d from?
• Feb 6th 2009, 01:35 PM
galactus
You are given dV/dt=5 and you need dh/dt(which we just showed you how to find). What else are you talking about?.

Please, take it easy on the IM type writing.
• Feb 6th 2009, 01:39 PM
Jasmina8
.... :)
ohhh...sorry i think that i was over analyzing the whole problem and confusing myself more than i should be.... thank you so much for your help though...

and sorry about the IM typing.
• Feb 6th 2009, 01:45 PM
galactus
Good, I am glad you have it. No need to apologize for the IM. It's just that there is a place for that. Personally, I hate to see it creeping into the mainstream way of writing.