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FP3 Radius of curvature of curve

Found this in the old P6 papers, adapted to FP3 syllabus. Do explain?

Find the exact value of the radius of curvature of the curve with equation y = arcsin x at the point where x = (sq root 2)/2.

The mark scheme gave: [ 1 + (dy/dx)^2 ] / (d2y/dx2). WHY?
Original post by purplesparks14
Found this in the old P6 papers, adapted to FP3 syllabus. Do explain?

Find the exact value of the radius of curvature of the curve with equation y = arcsin x at the point where x = (sq root 2)/2.

The mark scheme gave: [ 1 + (dy/dx)^2 ] / (d2y/dx2). WHY?


I have no clue!
Reply 2
Original post by GreenLantern1
I have no clue!


Hahaha neither do I! :frown:
Original post by purplesparks14
Found this in the old P6 papers, adapted to FP3 syllabus. Do explain?

Find the exact value of the radius of curvature of the curve with equation y = arcsin x at the point where x = (sq root 2)/2.

The mark scheme gave: [ 1 + (dy/dx)^2 ] / (d2y/dx2). WHY?


I'm pretty certain this isn't on any FP3 syllabus. Which exam board are you on?
Reply 4
Original post by hassi94
I'm pretty certain this isn't on any FP3 syllabus. Which exam board are you on?


Edexcel, sorry! Should have mentioned. Something similar to this is also in FP2 2006 past year. I hope it's an old syllabus thing!
Original post by purplesparks14

The mark scheme gave: [ 1 + (dy/dx)^2 ] / (d2y/dx2). WHY?


Typo.
Original post by purplesparks14
Edexcel, sorry! Should have mentioned. Something similar to this is also in FP2 2006 past year. I hope it's an old syllabus thing!


Well I'm pretty sure the formula should actually have a ^3/2 in there so it's not correct, furthermore I don't think it's on your syllabus. According to the specification there is absolutely no mention of radius of curvature in Edexcel FP2 or 3.

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