A cylindrical water trough has a diameter of 6 ft and a height of 2 feet. It is being filled at the rate of 4 ft^3/min, so how fast is the water level rising when the water is 1 foot deep?

Question

anonymous · Answer

|dw:1391200269105:dw|

jigglypuff314 · Answer

V = pi r^2 h

given info >,<
V' = 4       h = 1       r' = 0      d = 6  so  r = 3
looking for   h' = ?

anyways, first find the Derivative of   V = pi r^2 h

anonymous · Answer

so, 0

jigglypuff314 · Answer

wait... ... ... but dat not make sense >,<
idk DX

anonymous · Answer

I'm confused. :(

jigglypuff314 · Answer

I'm confused too sorry >,<
@ranga can you help please???

jigglypuff314 · Answer

@amoodarya could you help please?

jigglypuff314 · Answer

@agent0smith @annas could you help please?

jigglypuff314 · Answer

@shamil98 could you help please?

jigglypuff314 · Answer

@agent0smith can you help? please?

anonymous · Answer

@CGGURUMANJUNATH

anonymous · Answer

@austinL

agent0smith · Answer

Your problem is your drawing. It's oriented wrongly. That's not a trough. This is. Think of a horse or pig's trough |dw:1391202564700:dw|

anonymous · Answer

alright but how woud i go about solving the problem? the picture on my homework is a cylinder like a swimming pool

agent0smith · Answer

find dh/dt

V = pi r^2 h

you didn't differentiate w.r.t. time which is why you got zero.

agent0smith · Answer

$$\Large \frac{ dV }{ dt } =  \pi r^2 \frac{ dh }{ dt }$$

jigglypuff314 · Answer

why wouldn't it be  V' = 2 pi r r' h' ?

agent0smith · Answer

r isn't changing (how can it, it's a constant radius throughout the vertical cylinder)

jigglypuff314 · Answer

ohhhhhh right k

anonymous · Answer

jiggly, i don't get the one we're workin' on

agent0smith · Answer

$$\Large \frac{ dV }{ dt } =  \pi r^2 \frac{ dh }{ dt }$$solve for dh/dt. You know dv/dt and r.

agent0smith · Answer

Guessing you solved this....?

anonymous · Answer

no

anonymous · Answer

im confused

anonymous · Answer

I still need help