chain rule with partial derivatives

[kelly070280]

Recall Newton's Law of Gravitation, which asserts that the magnitude F of
the force of attraction between objects of masses M and m is
F=((GMm)/(r²)), where r is the distance between them and G is a
universal constant. Let an object of mass M be located at the origin, and
suppose that a second object of changing mass m(t) (say from fuel
consumption) is moving away from the origin so that its position vector is
r=<x,y,z>.

Obtain a formula for ((dF)/(dt)) in terms of G, M, m, r, x, y, z, and the
time derivatives of m, x, y, and z.


If I use the chain rule what will the partial derivatives be? I thought it would be F'(t) * r'(t) but then I can't figure out how to obtain those. Do I take the derivative of F with respect to each term involved? And then do the same with r?

Any help with where to go with this problem is greatly appreciated!

 

[Unco]

Perhaps replace r straight off, Kelly.

\(\displaystyle \, F \, = \, \frac{GMm}{x^2 \, + \, y^2 \, + z^2 \, }\) with x, y, z, m functions of t, agreed?

The chain rule says: \(\displaystyle \frac{dF}{dt} \, = \, \frac{\partial F}{\partial x}\,\frac{dx}{dt} \, + \, \frac{\partial F}{\partial y}\,\frac{dy}{dt} \, + \, \frac{\partial F}{\partial z}\,\frac{dz}{dt} \, + \, \frac{\partial F}{\partial m}\,\frac{dm}{dt}\)

 

[kelly070280]

what happens to dM/dt? I understand that dG/dt is 0 because it's a constant, but I thought M still had to be included?

Thanks!

 

[skeeter]

M is constant also ... dM/dt = 0.

 

[kelly070280]

thank you!