# Question about two-variable function

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?

## Answers and Replies

Are g and T constant?

yes, they are

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

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)

x(F)= g2T2/2F - g3T2/2F2
QUOTE]

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

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

That is correct?

I would have a line showing

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

Your differentiation is correct.

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?

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.

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?

From where did you get gX = Fx?

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

Last edited:
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.

Which they both indeed are

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

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