Finding 'a' when 'x' is given as a function of 't'.

1. Oct 9, 2013

Noob245

Hello everyone! I have this question given to me by a friend and the question is:
Position of a particle x is given by x^2=t^2 + 1, t is time in second, find acceleration for time t.

So this is what I decide to do:
x^2 = t^2+1
so
x = sqrt(t^2+1)
if you double differentiate the function on RHS
You get:
(t^2+1)^-3/2 or 1/(t^2+1)^3/2

I'd be grateful if you give me the steps for finding the answer (including the steps for finding out the derivative, because I used an online calculator).

2. Oct 9, 2013

Staff: Mentor

Are you saying you don't know how to determine the derivative of the right hand side? If not, show us how you get the first derivative?

3. Oct 9, 2013

Noob245

No, I do not know how. :(

4. Oct 9, 2013

tia89

It's the standard derivation of a function... hint: think how you can express roots as powers and think what the derivative of a power is...
Good work

5. Oct 9, 2013

6. Oct 9, 2013

Noob245

Yes; I tried double differentiating it but my answer didn't match the one given in the calc. the answer I got was

-1/(t^2+1)^-3/2

This mistake is all the more reason I want the steps. I know that if you double differentiate 'x' I get 'a' but for some reason by answer does not even come close to the original answer. It'd be really kind of you guys if you could just tell me if I was correct or wrong, and if wrong where I went wrong and how I should proceed further.

7. Oct 9, 2013

SteamKing

Staff Emeritus
Have you heard of the chain rule?

8. Oct 9, 2013

Staff: Mentor

From this answer, it just looks like all you did was make a mistake in algebra. So us the details so we can see if we can spot it.

chet

9. Oct 9, 2013

Noob245

GUYS GUYS! Forget about the calculus, please tell me if the algorithm I gave for sum is correct or not? If not, tell me the correct algorithm and the solution so I can try it out and match my answer.

10. Oct 9, 2013

CompuChip

What algorithm?

How about you try the first derivative first?
$x = \sqrt{t^2 + 1}$ - what is dx/dt? You are right that the derivative of $\sqrt{u} = u^{\tfrac12}$ is $\tfrac12 u^{-\tfrac12} = \frac{1}{2 \sqrt{u}}$. But as the square root is over t² + 1, not u, you will need the chain rule like SteamKing said.

11. Oct 9, 2013

tia89

You didn't give any detail about what you did... haw are we supposed to know where and why and what you did wrong????