# Word prob.

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• Oct 12th 2007, 01:41 PM
kwivo
Word prob.
A particle in a magnetic field travels in a straight line. The force (in Newtons) acting upon it is given by the function
F(s) = square root of (s) s^2, where s = f(t), the position of the particle in meters, changes with time t. At what rate is the force changing with respect to time at a time T for which the particle is at position s = 1 and has velocity 4 m/s.

Please explain this prob and how to do it!

• Oct 12th 2007, 01:54 PM
ticbol
Quote:

Originally Posted by kwivo
A particle in a magnetic field travels in a straight line. The force (in Newtons) acting upon it is given by the function
F(s) = pss2, where s = f(t), the position of the particle in meters, changes with time t. At what rate is the force changing with respect to time at a time T for which the particle is at position s = 1 and has velocity 4 m/s.

Please explain this prob and how to do it!

Is that
F(s) = p*s -s^2 ?

s = f(t), okay, but what is p?
• Oct 12th 2007, 01:56 PM
kwivo
Sorry, I fixed it.
• Oct 12th 2007, 03:24 PM
topsquark
Quote:

Originally Posted by kwivo
A particle in a magnetic field travels in a straight line. The force (in Newtons) acting upon it is given by the function
F(s) = square root of (s) s^2, where s = f(t), the position of the particle in meters, changes with time t. At what rate is the force changing with respect to time at a time T for which the particle is at position s = 1 and has velocity 4 m/s.

Please explain this prob and how to do it!

$\displaystyle F(s) = \sqrt{s} - s^2$
where s is a function of t.

What is
$\displaystyle \frac{dF}{dt}$?

Looks like a chain rule problem to me.

For example, the first term:
$\displaystyle \frac{d}{dt}\sqrt{s} = \frac{1}{2\sqrt{s}} \cdot \frac{ds}{dt}$

You do the rest.

-Dan