Related Rates

[dydxx]

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This is the question ==> http://puu.sh/lziha/8fd449bff1.png

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What I have attempted;

we know dv/dt = 3000

we are trying to find dSA/dt

so dh/dt = dh/dv * dv/dt

dSA/dt = dSA/dh * dh/dt

Volume looks like 1/3*h^(2)*h

V = h^(3)/3

dv/dh = h^(2)

I see a similar triangle there as well so h/20 = 0.5h/b (I think)
and I'm stuck trying to find the surface area of the water...

 

[Deleted member 4993]

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This is the question ==> http://puu.sh/lziha/8fd449bff1.png

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What I have attempted;

we know dv/dt = 3000

we are trying to find dSA/dt

so dh/dt = dh/dv * dv/dt

dSA/dt = dSA/dh * dh/dt

Volume looks like 1/3*h^(2)*h

V = h^(3)/3

dv/dh = h^(2)

I see a similar triangle there as well so h/20 = 0.5h/b (I think)
and I'm stuck trying to find the surface area of the water...

You need to copy the image and paste it here. Most of us are unwilling to go to an unknown site in search of a problem.

 

[Steven G]

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This is the question ==> http://puu.sh/lziha/8fd449bff1.png

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What I have attempted;

we know dv/dt = 3000

we are trying to find dSA/dt

so dh/dt = dh/dv * dv/dt

dSA/dt = dSA/dh * dh/dt

Volume looks like 1/3*h^(2)*h

V = h^(3)/3

dv/dh = h^(2)

I see a similar triangle there as well so h/20 = 0.5h/b (I think)
and I'm stuck trying to find the surface area of the water...

You can use any variables you like but you need to define it. What is a b????

 

[stapel]

This is the question ==> http://puu.sh/lziha/8fd449bff1.png

Here's the text:

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e. A water container is constructed in the shape of a square-based pyramid. The height h of the pyramid is the same as the length of each side of its base.

A vertical height of 20 cm is then cut off the top of the pyramid, and a new flat top added. The pyramid is then inverted, and water is poured in at a rate 3,000 cm³ per minute.

Find the rate at which the surface area of the water is increasing when the depth of the water is 15 cm.