Math Help - time, velocity, position of a particle

1. time, velocity, position of a particle

at t time a particle is moving along the x axis at position x. the relationship between x and t is given by tx=x^2 +8. at x = 2 the velocity of the particle is:

2. Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

Then:

$t=x^{2}+8$

then

$x=\pm \sqrt[ ]{ t-8}$

then

$velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}$

Then if x = 2, velocity = 1/4

Seem reasonable?

3. Originally Posted by Mentia
Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

Then:

$t=x^{2}+8$

then

$x=\pm \sqrt[ ]{ t-8}$

then

$velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}$

Then if x = 2, velocity = 1/4

Seem reasonable?
I think simpler: from $t= x^2+ 8$, $1= 2x dx/dt$. At x= 2, $1= 4 dx/dt$ so $v= dx/dt= 1/4$.