# Homework Help: 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()