further rates of change (1 Viewer)

AFGHAN22

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a horizontal trough 10 m long, has a cross section in the shape of a right-angled isosceles triange, if water is poured in at the rate of 8m^3/min, at what rate is the water level rising when the depth of the water is 2m.

the answer is 1/5m/min but i got 2/5m/min
 

sasquatch

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yeah i did it and got the same answer as you! so i think your book is wrong..
 

alcalder

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OK, so the volume of this trough:

If h is the height of the water and x is the length of the sides of the right isosceles triangle, the length of the top side of the trough is 2h.

Pythagorus says: x2 + x2 = (2h)2

2x2 = 4h2
x2 = 2h2

So, V = area of cross-section . length of trough

V = x2/2 x 10

V = 5x2 BUT x2 = 2h2

V = 10h2

dV/dh = 20h

When h = 2, dV/dh = 40 OR dh/DV = 1/40

dh/dt = dh/dV x dV/dt

= 1/40 x 8

dh/dt = 1/5 m/min
 

sasquatch

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haha yeah alcalder is right. What i think we all did (well i did this)

i got dv/dx = 20x, and dV/dt = 8 m^3/min

So dx/dt = (dv/dt) / (dV/dx) = 2/5x.

I did most of it in my head so i accidentally left out the x.

AFGHAN22 said:
at what rate is the water level rising when the depth of the water is 2m.
x = 2.

so dx/dt = 1/5

I think by putting that answer 2/5.. it kinda gave us a bias.. hehe
 
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AFGHAN22

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can someone please draw up a diagram of this trough becuase i have difficulty understanding what it looks like....i understand how to do everything except i don't quite understand how x^2 +x^2=(2h)^2...as soon as someone can explain that then i can understand the question...but i think the main problem is how i think the diagram looks like..thanks for all your replies
 

Mountain.Dew

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ur right AFGHAN22. main reason why we got 2/5 instead of 1/5 was because the f**ken diagram was wrong. attached are both the incorrect and the correct diagrams.

hope this helps AFGHAN22

alcalder, thank you for clarifying our mistakes :)
 

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