Homework Help: Algebraic Manipulation

1. May 24, 2012

Nano-Passion

1. The problem statement, all variables and given/known data
Simplify so that there is no square root function. Leave it in terms of S

S = xc^2 / (1 - 2Gy/rc^2)^1/2

$$s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}}$$

Attempt

I've tried too many things, it would be too cumbersome to copy all of it down.

2. May 24, 2012

rock.freak667

Try squaring both sides and making the fraction simpler. Not sure which term you want as the subject though.

3. May 24, 2012

Nano-Passion

I've done that, but ultimately we need it to be an explicit function of E, so you will still have to take the square root of the right side and it doesn't seem to be possible to simplify it.

4. May 24, 2012

Staff: Mentor

You've been around here long enough to know the drill. You don't need to copy all of it down, but you do need to show some effort. I cut you some slack this time.

E? There's no E in this problem.

5. May 25, 2012

Nano-Passion

Sorry, mistype. I meant S.

6. May 26, 2012

HallsofIvy

What, exactly, do you mean by "no square root function"? You can, of course, immediately replace the square root by a 1/2 power, as you have done. Why is that not a correct answer? You can, of course, proceed to combine the fraction and simplify but you will still have 1/2 powers whatever you do.

7. May 26, 2012

Nano-Passion

The 1/2 power is the equivalent of the square root so that defeats the purpose.

8. May 26, 2012

Staff: Mentor

Is the goal to get rid of the square root in the denominator?

9. May 26, 2012

Nano-Passion

Yes, but within the condition that equation is left as an explicit function of E.

10. May 26, 2012

SOA Andrew

Multiply the quotient on the right-hand side of the equation by a convenient form of 1; in this case, $$\frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}$$

This technique is called rationalizing the denominator.

Last edited: May 26, 2012
11. May 26, 2012

SammyS

Staff Emeritus
There's that "E" again !

Perhaps ...
Are you supposed to be solving for y ?​

12. May 26, 2012

Nano-Passion

$$s = \frac{xc^2}{(1-\frac{2Gy}{rc^2})^{1/2}} \frac{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}{(1 - \frac{2Gy}{rc^2})^\frac{1}{2}}$$

$$s = xc^2 \frac{\sqrt{1 - \frac{2Gy}{rc^2}}} {1-\frac{2Gy}{rc^2}}$$
$$s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} (1 - \frac{rc^2}{2Gy})$$
Whoops.

No, for S.

Last edited: May 26, 2012
13. May 26, 2012

SOA Andrew

There is no longer a square root in the denominator, so you have succeeded in eliminating the square root in the denominator. However, I am afraid the last step is incorrect because

$$\frac{1}{1 - \frac{2Gy}{rc^2}} \neq 1 - \frac{rc^2}{2Gy}$$

The penultimate step is the last correct step.

14. May 26, 2012

Nano-Passion

True, I've rewrote it below.

$$s = xc^2 \sqrt{1 - \frac{2G}{rc^2}} \frac{1}{1 - \frac{2Gy}{rc^2}}$$