Question about two-variable function

[Zula110100100]

If I have X(a,t) = .5at² and a and t were both functions of F
a(F) = F-g
t(F) = gT/F

and I wanted to know how much X changes with respect to F, how would I do that?

 

[grzz]

Are g and T constant?

 

[Zula110100100]

yes, they are

 

[grzz]

First step substitute (F - g) for a and gT/F for t to get x in terms of F and the constants.

 

[Zula110100100]

Oh, so is this right?

x(F)= g²T²/2F - g³T²/2F²
x(F) = g²T²/2(F^-1-gF^-2)
x'(F) = g²T²/2(-F^-2+2gF^-3)

 

[grzz]

x(F)= g²T²/2F - g³T²/2F²
QUOTE]

It is better if one leaves the two terms separate as above.

 

[Zula110100100]

Okay, so
x(F) = g²T²/2F - g³T²/2F²
x'(F) = -g²T²/2F² + g³T²/F³

That is correct?

 

[grzz]

I would have a line showing

\frac{g^{2}T^{2}F^{-1}}{2} - \frac{g^{3}T^{2}F^{-2}}{2}

 

[grzz]

Your differentiation is correct.

 

[Zula110100100]

Right, that is before taking the derivative?

I am not sure if I need to take the derivative now, so I guess let me lay it on ya

gT = Ft
gX = Fx

a = F-g
t = gT/F

x = 1/2at²
x = g²T²F^-1/2 - g³T²F^-2/2

If I am trying to find F for a given g, X, and T

I could just put x = gX/F in for x and get

gF^-1X = g²T²F^-1/2 - g³T²F^-2/2
divide by gF^-1
gives me
X=gT²/2 - g²T²F^-1/2
multiply by F and rearrange a bit to get
gT²F/2 - FX = g²T²/2

pull out F for F((gT²-2X)/2) = g²T²/2

divide for

F = g²T²/(gT²-2X)

Any of that look right?

 

[grzz]

I do not know what you are doing now.

Your final line x'(F) = -g2T2/2F2 + g3T2/F3

was ok. That is you were asked to find dX/dF and you found it because your differentiation was ok.

 

[Zula110100100]

Right, the last post was not using the derivative, I might not need to.

If the non-differentiated version is correct, and i know the equation
gX = Fx
can I not solve for x = gX/F
substitute that back into the equation I had gotten for x(F), and then I have F in terms of only g, X, and T, getting rid of x and t?

The resultant equation should be correct to give me F for a given g, X and T, right?

 

[grzz]

From where did you get gX = Fx?

 

[Zula110100100]

Right, so what I am looking at here is g = gravity, X is the total displacement, F is a force that is greater than and opposite to gravity, and x is a distance less than X, at s=0 the velocity is 0 and s=X the velocity is 0, so the potential energy given by gX must equal the energy put into the system Fx

Or to keep this as he math sub-forum, it is a given from post#10

 

[grzz]

Energy put in system will be given by Fx ONLY if F is constant and if F is in the same direction of the displacement x.

 

[Zula110100100]

Which they both indeed are

 

[Zula110100100]

A constant force F whose magnitude is greater than that of gravity, is working against gravity to give a displacement of x, it gains (F-g)x kinetic energy, and gx potential energy is put into the system. At that point F drops to 0, so the system's energy remains constant, it's kinetic energy is put into potential energy by gravity until KE reaches 0, and (F-g)x+gx = gX which reduces to Fx = gX

 

[grzz]

It is getting rather late over here...Sorry.