Messed up somewhere in my integration

1. Mar 6, 2006

mr bob

$$\frac{dv}{dt}= -x^{-3}$$
when t=0, the particle is at rest with x=1
Therefore by integrating i get
$$v = \sqrt(x^-2 - 1)$$

$$\frac{dx}{dt}= \sqrt(\frac{1 - x^2}{x^2})$$

$$dx\frac{x}{(/sqrt(1 - x^2))} = dt$$

$$-\sqrt(1 - x^2) = t + C$$

$$C=-1$$

Therefore:-

$$t = 1 - \sqrt(1 - x^2)$$

However i cant get the answer $$t = \sqrt(15)$$ when $$x = 1/4$$. I think i messed up somewhere in my integration. I would really appreciate any help with this.

Thank you,
Bob

Last edited: Mar 6, 2006
2. Mar 6, 2006

vaishakh

Did you add the constants well after integration? If you show your works of integration it would be easy to check.

3. Mar 6, 2006

mr bob

$$\frac{dv}{dt}= -x^{-3}$$
$$\frac{1}{2}v^2 = \frac{x^-2}{2} + C$$

As at rest when x = 1, v = 0
then C = -1

$$v = \sqrt(x^-2 - 1)$$

Then i continue to integrate as above.

Last edited: Mar 6, 2006
4. Mar 6, 2006

vaishakh

You didn't multiply C by 2 in the third step.

5. Mar 6, 2006

mr bob

But isn't C just a constant. Which wouldnt matter if i multiplied it by 2 or not.

6. Mar 6, 2006

vaishakh

Sorry it was a typo. I preesed enter by mistake.
Check the sign at t = -(1 - x^2)^-2. It is correct till there.

7. Mar 6, 2006

mr bob

Thank you very much. My maths has been pretty bad over the last couple of days (you can tell by the number of posts).

8. Mar 6, 2006

benorin

By thw way, you should use \sqrt{} rather than \sqrt()