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# P5 intrinsic co-ords question watch

1. Hi - I thought I was ok on intrinsic co-ordinates...until I met this question! (I'll use p for psi)

s = 12sin^2 p where s is measured from the origin
Show that the cartesian equation of the curve is (x-8)^2/3 + y^2/3 = 4

If someone could point me in the right direction or do it I'd be grateful. Thanks
2. (Original post by Bezza)
Hi - I thought I was ok on intrinsic co-ordinates...until I met this question! (I'll use p for psi)

s = 12sin^2 p where s is measured from the origin
Show that the cartesian equation of the curve is (x-8)^2/3 + y^2/3 = 4

If someone could point me in the right direction or do it I'd be grateful. Thanks
find ds/dp and then using dy/ds=sinp and dx/ds=cosp,u can find the cartesian eqn.
3. (Original post by Bezza)
Hi - I thought I was ok on intrinsic co-ordinates...until I met this question! (I'll use p for psi)

s = 12sin^2 p where s is measured from the origin
Show that the cartesian equation of the curve is (x-8)^2/3 + y^2/3 = 4

If someone could point me in the right direction or do it I'd be grateful. Thanks
I think you'd use sin(p) = dy/ds and then a combination of sin^2 = 1 - cos^2 and cos(p) = dx/ds to find y and x in terms of s.

Then, presumably, combine them as if you were moving from parametric to cartesian and hope you get the right answer.

-Proteus
4. (Original post by IntegralAnomaly)
find ds/dp and then using dy/ds=sinp and dx/ds=cosp,u can find the cartesian eqn.
That was what I was trying but I couldn't get anywhere. ds/dp = 24sinp cosp, but this didn't seem to help. I even tried using the cartesian form of ds/dp for radius of curvature but it got quite nasty.
5. (Original post by IntegralAnomaly)
find ds/dp and then using dy/ds=sinp and dx/ds=cosp,u can find the cartesian eqn.
You have to muck about a bit with the derivatives ie you want dx/dp so you can get x = f(p) through integration. However you only know ds/dp and dx/ds, so use dx/dp = dx/ds . ds/dp. Do the same for y. Looking at the nature of the answer youll probably end up with trig parametric equations which you can eliminate with cos^2A + sin^2A = 1, although this is just a guess
6. here's the first bit:
Attached Images

7. To elaborate the steps are as follows...

- Find ds/dp in terms of p
- Use dx/dp = dx/ds . ds/dp = cosp ds/dp
- Integrate so x = INT cosp ds/dp dp
- Use initial conditions to find constant
- Do the same for y
- Now you have parametric equations x = X(p), y = Y(p)
- Eliminate p to get a cartesian equation involving only x and y
8. Thanks for your help everyone - phils post was the hint I needed. I'd got dy/dp and dx/dp but for some reason didn't think about integrating them. Thanks IA too! I'll try and remember to rep you tomorrow.

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