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#1
If a hemispherical bowl of radius 6cm contains water to depth xcm, the volume of the water is 1/3(pie)x^2*(18-x).water is poured into the bowl at a rate of 3cm^3 per second. find rate at which water level is rising when depth is 2cm.

I know i need to find dx/dv, but how ?
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5 years ago
#2
If a hemispherical bowl of radius 6cm contains water to depth xcm, the volume of the water is 1/3(pie)x^2*(18-x).water is poured into the bowl at a rate of 3cm^3 per second. find rate at which water level is rising when depth is 2cm.

I know i need to find dx/dv, but how ?
You basically need to relate dx/dt to dV/dt using the chain rule - this will involve dV/dx.
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5 years ago
#3
You are given that the volume of water is 1/3(pi)x^2*(18-x) you can find the derivative of this, dV/dx. You need dx/dV, what do you think you should do to find this?
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#4
(Original post by Protoxylic)
You are given that the volume of water is 1/3(pi)x^2*(18-x) you can find the derivative of this, dV/dx. You need dx/dV, what do you think you should do to find this?
Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2
Would that be dx/dv ?
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5 years ago
#5
Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2
Would that be dx/dv ?
Where did you get that expression from?

You're given V in terms of x, so what is dV/dx?

You're told what dV/dt is and you need to find dx/dt using the chain rule.
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5 years ago
#6
(Original post by davros)
Where did you get that expression from?

You're given V in terms of x, so what is dV/dx?

You're told what dV/dt is and you need to find dx/dt using the chain rule.
Read, you are given volume in terms of X (the water level). You can find dV/dx and inverse what you get to find dx/dV.
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5 years ago
#7
Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2
Would that be dx/dv ?
Essentially, find your derivative dV/dx. You can find dx/dV by finding the inverse of dV/dx which is 1/[dV/dx].
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5 years ago
#8
(Original post by Protoxylic)
Read, you are given volume in terms of X (the water level). You can find dV/dx and inverse what you get to find dx/dV.
Did you mean to quote me or the OP in the last post? I'm not sure whether I'd misread something or pointed the OP in the wrong direction
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#9
(Original post by davros)
Where did you get that expression from?

You're given V in terms of x, so what is dV/dx?

You're told what dV/dt is and you need to find dx/dt using the chain rule.

Dv/Dx is the derivative of the volume of water which is = (pi)(12x-x^2) ?

In addition to that I would have to find dv/dx = dv/dt . dx/dv ?
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5 years ago
#10
(Original post by davros)
Did you mean to quote me or the OP in the last post? I'm not sure whether I'd misread something or pointed the OP in the wrong direction
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5 years ago
#11
Dv/Dx is the derivative of the volume of water which is = (pi)(12x-x^2) ?

In addition to that I would have to find dv/dx = dv/dt . dx/dv ?
No, your final line doesn't make sense!

You mean dv/dt = (dv/dx) . (dx/dt) and you know (or can work out) some of the info from what's given in the question.
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#12
(Original post by davros)
Did you mean to quote me or the OP in the last post? I'm not sure whether I'd misread something or pointed the OP in the wrong direction
Was answering the other person's question. I've got two sheets of paper doing what you and the other person's requested and, seeing where to go next with the question with you guys help
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#13
(Original post by Protoxylic)
would it be Dv/Dx . Dx/Dt

so dv/dx = 1/(pi)(12x-x^2)

dx/dt = 3

Is that the answer if i multiply them both together ?

which gives 3/20(pi) ?
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5 years ago
#14
would it be Dv/Dx . Dx/Dt

so dv/dx = 1/(pi)(12x-x^2)

dx/dt = 3

Is that the answer if i multiply them both together ?

which gives 3/20(pi) ?
No

3 is the rate of change of volume, not height
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#15
(Original post by TenOfThem)
No

3 is the rate of change of volume, not height
How do I find dx/dt ?
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5 years ago
#16
How do I find dx/dt ?

I though that you knew this
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#17
I'm going to continue with this later, been stressing over one question for ages and I'm getting frustrated, because I'm not understanding this new topic yet.
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5 years ago
#18
I'm going to continue with this later, been stressing over one question for ages and I'm getting frustrated, because I'm not understanding this new topic yet.
Which bit of this is "new" to you? Have you encountered the chain rule before, or are you having trouble applying it to this particular physical situation?
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#19
(Original post by davros)
Which bit of this is "new" to you? Have you encountered the chain rule before, or are you having trouble applying it to this particular physical situation?
I can do simple chain rule, but applying it seems all confusing.
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#20
(Original post by davros)
Which bit of this is "new" to you? Have you encountered the chain rule before, or are you having trouble applying it to this particular physical situation?
would it be Dv/Dx . Dx/Dt

so dv/dx = 1/(pi)(12x-x^2)

dx/dt = 3

Is that the answer if i multiply them both together ?

which gives 3/20(pi) ?

Well, thank you for your help. I asked a teacher and the answer was 3/20(pi) but I just wrote the function for 3 before (as you can see in the quote) wrong as it should had been dv/dt=3.

I appreciate everyone's input to help me achieve the answer
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