Question

Find the vertices of :
\[x ^{4}+y ^{4}=1\]
vertices is:point where derivative of kurvature is 0

Answer 1

graph it

Answer 2

:). I need analitic solution...

Answer 3

yeah well you can work out the details once you have the solution,

Answer 4

i am more intrested in a proces then in a solution itself

Answer 5

well it might be less intuative

Answer 6

hmm what do you mean by curvature is zero? how do you measure curvature?

Answer 7

f''(x) = 0?

Answer 8

curvature is a mesure of change of the direction of unit tangent to arc length. It's one of geometric invariants of a curve

Answer 9

f''(x) = 0? no

Answer 10

have you taken the derivatives o the formula

Answer 11

if it helps anyone: the curvature function can be obtained by implicit differentiation

Answer 12

thinking of x ^{4}+y ^{4}=1 as F(x,y) = x ^{4}+y ^{4}-1=0
and y(x)

Answer 13

ohh it's calc iii or beyond me :/ ...sorry
@dumbcow @satellite73 maybe they know how to do it.

Answer 14

my aprouch was:
F(x,y(x))
F'(1,y')
F''(0,y'')
so curvature function wouold be: k=y''/[(1+y'^2)^(3/2)]
but not sure if this is ok...

Answer 15

@Zarkon ?

Answer 16

any sugestions @dumbcow ?

Answer 17

do you have to use curvature function (k) ?
wouldn't the vertices be where the slope is 0 at (0,+-1)

Answer 18

btw you are correct with curvature equation
http://en.wikipedia.org/wiki/Curvature#Curvature_of_a_graph

Answer 19

not really. If you look at the graph of this function http://www.wolframalpha.com/input/?i=x+%5E%7B4%7D%2By+%5E%7B4%7D%3D1 the slope is not 0 at the vertices.

Answer 20

ya but i am getting nowhere to find the actual values.....

Answer 21

http://www.wolframalpha.com/input/?i=x+%5E4%2By+%5E4-1%3D0
this is the right one, sry

Answer 22

\[y' = -\frac{x^{3}}{y^{3}}\]
\[y'' = -\frac{3x^{2}y^{4}+3x^{6}}{y^{7}}\]
correct?

Answer 23

1º is correct, checking the 2º...:)

Answer 24

oh it can be simplified if you factor out a 3x^2
and x^4+y^4 = 1
\[y''=-\frac{3x^{2}}{y^{7}}\]

Answer 25

y'' is correct i think , just i am getting oposit sign.... maybe i am wrong

Answer 26

i am getting \[y''=3x ^{2}/y ^{7}\]

Answer 27

when you plugged in y' did you include the neg sign?
it is neg though, look at implicit derivatives on wolfram

Answer 28

yes

Answer 29

but, it's ok continue. I will check my steps later

Answer 30

ok, if we now look at curvature function
after plugging everything in and simplifying i get
\[k = -\frac{3x^{2}y^{2}}{(x^{6}+y^{6})^{3/2}}\]

Answer 31

so the x derivative of this should be = 0

Answer 32

ok that derivative was a little messy, after setting =0
i was able to get to here
\[2(x^{10}-y^{10}) = 7x^{4}y^{4}(y^{2}-x^{2})\]
there should be 4 solutions right?

Answer 33

http://www.wolframalpha.com/input/?i=2%28x^10-y^10%29+%3D+7x^4y^4%28y^2-x^2%29
according to wolfram
solution is y = +-x
\[\pm x = \pm \sqrt[4]{1-x^{4}}\]
\[x = \pm \sqrt[4]{\frac{1}{2}} \]

Answer 34

thx a lot @dumbcow