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# FP2 Watch

1. Anyone do FP2?
Attachment 206918
What I attempted so far:
Differentated with respect to t to get dy/dt=-2x^-3(dx/dt)
Then I differentiated again to get d2y/dt2=6x^-4(dx/dt) -2x^-3(d2x/dt2)
Then rearranged the subjects so I could sub back into the original equation~got stuck along the way though and I'm not anywhere close to the proof :O
We got given the CD to help with revision over Easter but I can't seem to work out how to open solution bank on my laptop :/
2. I would love to help but there is a problem with the link.
3. (Original post by Gray Wolf)
I would love to help but there is a problem with the link.
Oh did it not attach
4. (Original post by the.cookie.monster)
Oh did it not attach
You've done the tricky bit right. Now it comes down to this: how solid is your algebra?
5. (Original post by Indeterminate)
You've done the tricky bit right. Now it comes down to this: how solid is your algebra?
Okay, well when I rearranged I got:
dx/dt=-0.5x^3(dy/dt)
d2x/dt2=-0.5x^3(d2y/dt2)+3x^-1(dx/dt)
Subbing in I get:
-x^4(d2y/dt2)+6(dx/dt)-6(-0.5x^3(dy/dt))^2=x^2-3x^4
I don't see how I would be able to get rid of (dy/dt)^2
6. (Original post by the.cookie.monster)
Okay, well when I rearranged I got:
dx/dt=-0.5x^3(dy/dt)
d2x/dt2=-0.5x^3(d2y/dt2)+3x^-1(dx/dt)
Subbing in I get:
-x^4(d2y/dt2)+6(dx/dt)-6(-0.5x^3(dy/dt))^2=x^2-3x^4
I don't see how I would be able to get rid of (dy/dt)^2

7. (Original post by Indeterminate)
I've realised what I did wrong now-I differentiated incorrectly d2y/dt2
I was supposed to get 6x^-4(dx/dt)^2 -2x^-3(d2x/dt2)
but thanks anyway
8. (Original post by the.cookie.monster)
I've realised what I did wrong now-I differentiated incorrectly d2y/dt2
I was supposed to get 6x^-4(dx/dt)^2 -2x^-3(d2x/dt2)
but thanks anyway
Was about to say that as I was doing the Q from scratch. You were too quick

By the way, my last post refers to the fact that

d2x/dt2=-0.5x^3(d2y/dt2)+3x^-1(dx/dt)

was an incorrect deduction from the wrong differentiation

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