## MarcLeclair 2 years ago A water tank is in the shape of a cone with vertical axis and vertex downward. The tank has a radius of 3m and is high 5m. At first the tank is full of water, but at time t=0 (in seconds), a small hole at the vertex is opened and the water begins to drain. When the height of the water in the tank has dropped to , the water is flowing out at a rate of 2m (cube)/s. At what rate, in meters per seconds, is the water level dropping then? I got this: dv/dt = 2m(cube)/s and we're looking for dh/dt. We also know V=Pi(r^5)(h) /3 and r=3h/5. However I don't know what to do afterwards

1. MarcLeclair

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you can use $$\frac{dh}{dt}=\frac{dh}{dr} \times \frac{dr}{dV}\times \frac{dV}{dt}$$ or just simply sub r=...h into V

do you need further help?

4. MarcLeclair

Sorry I hadn't seen the response, but after substituting r, i would find its derivative? And I don't get the first equation youve done :/

after sub-bing r, you'll get an eq where V is a function of h. the first eq is the chain rule lol for lazy people like me

6. MarcLeclair

how would that give you a function, that's where I always get confuse. I mean its just a volume equation :/ at a certain point in time

yup, that's the function, actually, that eq.

at that particular time where the time change is dt.

9. MarcLeclair

Ahhhhh its like as if the volume decrease, heigh decreases, so we could map a graph and therefore find its derivative.... ahaha thanks :)