# another intrinsic questionWatch

This discussion is closed.
#1
you have

s = sec^3 psi - a

and how do you prove that

y = a tan^3 psi

and find a similar expression for x in terms of psi???
0
13 years ago
#2
(Original post by lesser weevil)
you have

s = sec^3 psi - a

and how do you prove that

y = a tan^3 psi

and find a similar expression for x in terms of psi???
I'll pressume you mean

For the expression in terms of x use
0
#3
thanks mate, but still rather confusing. I hate intrinsic coordinates.!!!
0
13 years ago
#4
(Original post by lesser weevil)
thanks mate, but still rather confusing. I hate intrinsic coordinates.!!!
What part do you dislike?
0
13 years ago
#5
(Original post by Gaz031)
I'll pressume you mean

For the expression in terms of x use
i don't get the part
dy/ds=sin(psi)

where did you get that from?
0
13 years ago
#6
(Original post by hihogyu)
i don't get the part
dy/ds=sin(psi)

where did you get that from?
dy/ds=sin(psi) and dx/ds=cos(psi) are standard results.

You know that dy/dx=tan(psi), and recall that:
(ds/dx)Â² = 1 + (dy/dx)Â²
(ds/dx)Â² = 1 + tanÂ²(psi) = secÂ²(psi)
=> ds/dx = sec(psi)

dy/dx = (ds/dx)(dy/ds) = sec(psi) (dy/ds) = tan(psi)
=> dy/ds = sin(psi)
0
13 years ago
#7
(Original post by dvs)
dy/ds=sin(psi) and dx/ds=cos(psi) are standard results.

You know that dy/dx=tan(psi), and recall that:
(ds/dx)Â² = 1 + (dy/dx)Â²
(ds/dx)Â² = 1 + tanÂ²(psi) = secÂ²(psi)
=> ds/dx = sec(psi)

dy/dx = (ds/dx)(dy/ds) = sec(psi) (dy/ds) = tan(psi)
=> dy/ds = sin(psi)
so i see..!

cheers
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