# A2 Math help please

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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 ?

I know i need to find dx/dv, but how ?

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#2

(Original post by

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 ?

**ADotCross**)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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#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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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?

**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 that be dx/dv ?

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#5

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Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2

Would that be dx/dv ?

**ADotCross**)Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2

Would that be dx/dv ?

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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#6

(Original post by

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.

**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.

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#7

**ADotCross**)

Would I find the derivative of Hemispherical which is then (4/6)(pi)r^2

Would that be dx/dv ?

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#8

(Original post by

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.

**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.

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**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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#10

(Original post by

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

**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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#11

(Original post by

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 ?

**ADotCross**)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 ?

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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**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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(Original post by

Oh, I read your reply thinking it was a reply to me since I just refreshed the page and had a notification. I thought you questioned where I had got the expression from.

**Protoxylic**)Oh, I read your reply thinking it was a reply to me since I just refreshed the page and had a notification. I thought you questioned where I had got the expression from.

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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#14

(Original post by

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) ?

**ADotCross**)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) ?

3 is the rate of change of volume, not height

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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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#18

(Original post by

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.

**ADotCross**)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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(Original post by

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?

**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?

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**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?

**ADotCross**)

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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