Question

show y''=-y^-3 on the circle x^2+y^2=1

Answer 1

\[x^2+y^2=1\]

\[y''=-y^{-3}\]

Answer 2

need to prove

Answer 3

have you tried to differentiate x^2+y^2=1?

Answer 4

yes

Answer 5

well you first need to find y'
then find y''

Answer 6

let me know if you come back

Answer 7

and need any help

Answer 8

2x+2yy'=0

2x=-2yy'

x=-yy'

Answer 9

so y'=x/-y

Answer 10

so find y'' by differentiating both sides

Answer 11

use quotient rule for the (-x/y) part

Answer 12

i did

Answer 13

but didnt get the answer

Answer 14

what did you get

Answer 15

can you show me your steps?

Answer 16

anyways if you found the correct y''
then you will replace y' with -x/y
and then get rid of the compound fraction
then use that x^2+y^2=1

Answer 17

then you will have your -1/y^3=y''`

Answer 18

did you want to show me what you got for y''

Answer 19

because as I was saying you have to manipulate what you have for y'' to get the -1/y^3

Answer 20

im still not getting it

Answer 21

@SithsAndGiggles

Answer 22

@freckles has everything you need. Differentiating the circle equation, you have
\[\frac{d}{dx}[x^2+y^2=1]~~\implies~~2x+2yy'=0~~\implies~~y'=-\frac{x}{y}\]
Differentiating again, you have (via the quotient rule)
\[y''=-\frac{y-xy'}{y^2}\]
Substitute \(y'=-\dfrac{x}{y}\), and you have
\[y''=-\frac{y+x\dfrac{x}{y}}{y^2}~~\iff~~y''=-\frac{y+\dfrac{x^2}{y}}{y^2}\]
Multiply the RHS by \(\dfrac{y}{y}\), and you have
\[y''=-\frac{y^2+x^2}{y^3}\]
You know that \(x^2+y^2=1\), so
\[y''=-\frac{1}{y^3}=-y^{-3}\]
Notice that I've done nothing different from above.