# Thread: connected rates of change

1. ## connected rates of change

A inverted conical vessel contains water to a height of 12cm initially. A hole is opened at the vertex and water leaks away in such a manner that after t minutes, the rate of decrease of the height, h cm, of the water in the vessel is given by dh/dt = -2t/3 cm/min

a) calculate the time taken for the vessel to empty
b)Given the volume of water in the vessel after t minutes is ( $\frac{\pi}{6}$ h^3) cm^3 , calculate the rate of change of the volume of water in the vessel when t =3

I am already stumped by the first part. As dh/dt is constantly changing due to the vessel not being uniform in base area as the water level decreases i can't just divide the figures. Can anyone point me in the right direction for this question?

2. ## Re: connected rates of change

The ratio of the height to the top radius if the water is constant. You can use this to your advantage. Think "Similar Triangles".

3. ## Re: connected rates of change

Originally Posted by TKHunny
The ratio of the height to the top radius if the water is constant. You can use this to your advantage. Think "Similar Triangles".
I have thought of that. But which is the new radius and height value which i have to take to compare with the initial h=12 and radius = r?

4. ## Re: connected rates of change

h = h(r) -- How can you express that?

Since we have $Volume(h,r) = \frac{1}{3}\pi r^{2}h$, that volume hint should be able to provide the required relationship.

5. ## Re: connected rates of change

what are the information in the question that i need to use for this?
am i supposed to express h as a function of r? if so, i don't really know how to do that with only one equation for volume and with V in the equation :/

6. ## Re: connected rates of change

a) Integrate -2t/3

you get : h = -t^2/3 + c. Now, since t=0, h=12, Therefore, h = -t^2 / 3 + 12

Therefore time taken to empty is to put h = 0, t = 6

b) V= 1/3 {\pi} r^2 h

put V = ( h^3) = 1/3 {\pi} r^2 h
Solve and you get, 1/2 h^2 = r^2

Therefore, V = 1/6 {\pi} h^3

From (a), t = 3, h = 9

dV/dh = 1/2 {\pi} h^2

Use the rate of change: dV/dt = dV/dh x dh/dt

= 1/2 {\pi| (9)^2 x (-2)
=81pi

I am a teacher and joined this forum recently. btw, can someone help me with how to type the codes for symbols?

7. ## Re: connected rates of change

Originally Posted by mithgar
a) Integrate -2t/3

you get : h = -t^2/3 + c. Now, since t=0, h=12, Therefore, h = -t^2 / 3 + 12

Therefore time taken to empty is to put h = 0, t = 6

b) V= 1/3 {\pi} r^2 h

put V = ( h^3) = 1/3 {\pi} r^2 h
Solve and you get, 1/2 h^2 = r^2

Therefore, V = 1/6 {\pi} h^3

From (a), t = 3, h = 9

dV/dh = 1/2 {\pi} h^2

Use the rate of change: dV/dt = dV/dh x dh/dt

= 1/2 {\pi| (9)^2 x (-2)
=81pi

I am a teacher and joined this forum recently. btw, can someone help me with how to type the codes for symbols?

well, found the tutorial on how to use math tag.
you get : h = $\frac{-t^2}{3} + c$

tried but doesnt seem to work

8. ## Re: connected rates of change

You have to use:
[ tex ] ... [ / tex ]
if you want to work with latex.

9. ## Re: connected rates of change

Originally Posted by Siron
You have to use:
[ tex ] ... [ / tex ]
if you want to work with latex.

Thanks!