Kepler vector problem

Allie · #1 $Old$ October 6th, 2008, 03:10 PM

What is the force required so that a particle of mass m has the position function r(t)=t^3i+t^2j+t^3k
can anyone walk me through how to do this?

icemanfan · #2 $Old$ October 6th, 2008, 03:27 PM

$r(t) = <t^3, t^2, t^3>$

$r'(t) = <3t^2, 2t, 3t^2>$

$a(t) = r''(t) = <6t, 2, 6t>$

The force required is F = ma:

$F(t) = m(a(t)) = m<6t, 2, 6t> = <6mt, 2m, 6mt>$

The actual magnitude of the force is:

$\sqrt{(6mt)^2 + (2m)^2 + (6mt)^2}$

$= \sqrt{36m^2t^2 + 4m^2 + 36m^2t^2}$

$= \sqrt{72m^2t^2 + 4m^2}$

$= \sqrt{(4m^2)(18t^2 + 1)}$

$= 2m\sqrt{18t^2 + 1}$

Allie · #3 $Old$ October 6th, 2008, 03:41 PM

why is a=r''?

thank you for helping me btw

icemanfan · #4 $Old$ October 6th, 2008, 03:49 PM

r is position.

The derivative of position is velocity, and the derivative of velocity (the second derivative of position) is acceleration.

Allie · #5 $Old$ October 6th, 2008, 03:50 PM

ohh of course thank you so much