a rectangular water tank has a length 20 ft, width 10 ft , and depth 15 ft . if the tank is full , how much work does it take to pump all the water out (using calculus)

- anonymous

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- anonymous

so this is another combination of mathematical calculus and the physical energy problem.

- anonymous

this one is much easier than the previous one, don't you think so? because there's no dripping water here~! :)

- anonymous

but im still confused

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## More answers

- anonymous

this is what i got 62.4(density of water)(200,the volume) integral 15-h limit 15 to zero

- anonymous

the measurement is ft , not meter.

- anonymous

i know that why u use 62.4 as the density

- anonymous

youd better change to ISU, more convenient to calculate, as 1 g/cm^3 the density of the water/

- anonymous

im sorry i had to go to bed, its midnight here

- anonymous

oh okay thank you for ur help

- anonymous

anwar di u think u can help me

- anonymous

Is the water leaving the tank from the top or the bottom? I'll assume it's leaving from the bottom.
W = F * d; draw the tank with the dimensions. Find the density of an infinitely thin "slab" of water. Its volume is going to be 20*10*dy = 200dy. The force is going to be the volume times the density, so 200dy * 62.4 = 12480dy. The distance that this slab is going to have to travel to leave the tank is y, so the work done in moving that slab is 12480ydy. Sum all the work done by all the slabs with...
\[\int\limits_{0}^{15}12480ydy\]
I got 1404000 foot-pounds of work.

- anonymous

thank u for coming to the rescue , but the answer is 220,500 ft lb

- anonymous

Integrate 12480ydy from 0 to 15 is what I meant to say...

- anonymous

is the distance 15-y

- anonymous

The distance would be 15 - y if the water was leaving from the top... I don't think that wouldn't get the amount of work to be 220500, though...

- anonymous

it just says how much work would it take to pump the water, it doesn"t say top and bottom

- anonymous

Well, for water to be pumped out of a tank it needs to be pumped out from somewhere. The water needs to move to a part of the tank to be pumped out.

- anonymous

here what I have, I am not sure about it though.. haven't done physics in a while:
we have the following formula:
\[W=\int\limits_{v_i}^{v_f}pdv\]
where p is pressure of water (can be found known depth and temperature of water),
vi and vf are the initial and final volume respectively (both known)

- anonymous

knowing depth..*

- anonymous

so it doesn't matter if we use y or 15-y to represent the distance

- anonymous

well we have the initial volume=20*10*15,
final volume=0..
we will get a negative result for the work.

- anonymous

if we use 15-y?

- anonymous

hey im a little confused. can someone please explain to me how the distance become 15-y

- anonymous

I have no idea why my answer is incorrect. If you draw the height of the tank and pick an arbitary point along the height, the distance between the top of the tank and that point is 15 - y.

- anonymous

and when is it just y

- anonymous

Distance betwen the arbitary point and the bottom of the tank.

- anonymous

but in the triangle is the top is just y, and the bottom is 15-y

- anonymous

Triangle?

- anonymous

this is another question , when u slice a triangle and want to use similar triangles the top portion is y and the bottom half is 15-y ,( this just an example)

- anonymous

Yes, I suppose you could think of it that way if you positioned the tank such that the x-axis was the top of the tank and 15 feet below that is the bottom of the tank.

- anonymous

k ty

- anonymous

hey i have a question can i ask

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