another intrinsic question Watch

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lesser weevil
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#1
Report Thread starter 13 years ago
#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???
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Gaz031
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#2
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#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 s=asec^3(\psi) - a

\frac{ds}{d\psi}=3asec^3(\psi)ta  n(\psi)\\

ds=3sec^3(\psi)tan(\psi) d\psi\\

\frac{dy}{ds}=sin{\psi}\\

\int 1 dy = \int sin(\psi) ds\\

y=\int 3asin(\psi)sec^3(\psi)tan(\psi) d(\psi)\\

y=\int 3atan^2(\psi)sec^2(\psi) d(\psi)\\

y=atan^3(\psi)+C

For the expression in terms of x use \frac{dx}{ds}=cos(\psi)
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lesser weevil
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#3
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thanks mate, but still rather confusing. I hate intrinsic coordinates.!!!
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Gaz031
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#4
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(Original post by lesser weevil)
thanks mate, but still rather confusing. I hate intrinsic coordinates.!!!
What part do you dislike?
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hihogyu
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#5
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#5
(Original post by Gaz031)
I'll pressume you mean s=asec^3(\psi) - a

\frac{ds}{d\psi}=3asec^3(\psi)ta  n(\psi)\\

ds=3sec^3(\psi)tan(\psi) d\psi\\

\frac{dy}{ds}=sin{\psi}\\

\int 1 dy = \int sin(\psi) ds\\

y=\int 3asin(\psi)sec^3(\psi)tan(\psi) d(\psi)\\

y=\int 3atan^2(\psi)sec^2(\psi) d(\psi)\\

y=atan^3(\psi)+C

For the expression in terms of x use \frac{dx}{ds}=cos(\psi)
i don't get the part
dy/ds=sin(psi)

where did you get that from?
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dvs
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#6
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(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)² = 1 + (dy/dx)²
(ds/dx)² = 1 + tan²(psi) = sec²(psi)
=> ds/dx = sec(psi)

dy/dx = (ds/dx)(dy/ds) = sec(psi) (dy/ds) = tan(psi)
=> dy/ds = sin(psi)
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hihogyu
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#7
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#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)² = 1 + (dy/dx)²
(ds/dx)² = 1 + tan²(psi) = sec²(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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