A2 Math help please

Original post by ADotCross

You basically need to relate dx/dt to dV/dt using the chain rule - this will involve dV/dx.

Reply 2

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?

Reply 3

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 ?

Reply 4

Original post by ADotCross

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.

Reply 5

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.

Reply 6

Original post by ADotCross

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].

Reply 7

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 :smile:

Reply 8

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 ?

Reply 9

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 :smile:

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.

Reply 10

Original post by 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 ?

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.

Reply 11

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 :smile:

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

Reply 12

Original post by 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.

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) ?

Reply 13

TenOfThem

Original post by 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) ?

No

3 is the rate of change of volume, not height

Reply 14

Original post by TenOfThem

No

3 is the rate of change of volume, not height

How do I find dx/dt ?

Reply 15

TenOfThem

Original post by ADotCross

How do I find dx/dt ?

\dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt}

I though that you knew this

Reply 16

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.

Reply 17

Original post by 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.

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?

Reply 18

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.

Reply 19