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# Implicit Differentiation watch

1. I've just been doing a C3 paper (OCR MEI C3 June 2010 Q5).

Am I losing the plot or is the mark scheme for Q5 missing something here?

It has proven that if (x,y) is a turning point then x = 0 or 1/8. Not that x = 1/8 is actually a turning point.

The first implication, , does not go both ways. The RHS of the implication can still hold even if the LHS does not, if the denominator of the derivative is also 0 (which occurs in the x = 0 case). Surely I must therefore add in an explanation as to why in the x = 1/8 case, y' = 0?
2. (Original post by 16Characters....)
I've just been doing a C3 paper (OCR MEI C3 June 2010 Q5).

Am I losing the plot or is the mark scheme for Q5 missing something here?

It has proven that if (x,y) is a turning point then x = 0 or 1/8. Not that x = 1/8 is actually a turning point.

The first implication, , does not go both ways. The RHS of the implication can still hold even if the LHS does not, if the denominator of the derivative is also 0 (which occurs in the x = 0 case). Surely I must therefore add in an explanation as to why in the x = 1/8 case, y' = 0?
The point of the latter part of the solution the mark scheme gave was to find the zeroes of that actually lie on the curve.
At all points in the plane with , we have that the derivative is zero, the issue is that not all of these solutions are valid, as they do not lie on the curve.
Since lies on the curve, and at this point the derivative is also zero, we have a stationary point there.
3. (Original post by joostan)
The point of the latter part of the solution the mark scheme gave was to find the zeroes of that actually lie on the curve.
Since lies on the curve, and at this point the derivative is also zero, we have a stationary point there.
I'll admit to being quite confused here

I understand what they are doing, solving y' = 0 and y^3 = xy - x^2 simultaneously to find the points which lie on the curve with y' = 0 (i.e. the turning points).

My problem is that I do not understand why their proof actually does this. y - 2x = 0 is not the same as y' = 0 because of the danger of the denominator equalling 0 and complicating matters. So surely we need to justify that the denominator is non-zero at x = 1/8?
4. (Original post by 16Characters....)
I'll admit to being quite confused here

I understand what they are doing, solving y' = 0 and y^3 = xy - x^2 simultaneously to find the points which lie on the curve with y' = 0 (i.e. the turning points).

My problem is that I do not understand why their proof actually does this. y - 2x = 0 is not the same as y' = 0 because of the danger of the denominator equalling 0 and complicating matters. So surely we need to justify that the denominator is non-zero at x = 1/8?
I agree with you. A better answer would be:

Sometimes mark schemes don't give perfect solutions but just show you what you need to get the marks.
5. (Original post by notnek)
I agree with you. A better answer would be:

Sometimes mark schemes don't give perfect solutions but just show you what you need to get the marks.
Thanks. And thank you joostan also.
6. (Original post by 16Characters....)
I'll admit to being quite confused here

I understand what they are doing, solving y' = 0 and y^3 = xy - x^2 simultaneously to find the points which lie on the curve with y' = 0 (i.e. the turning points).

My problem is that I do not understand why their proof actually does this. y - 2x = 0 is not the same as y' = 0 because of the danger of the denominator equalling 0 and complicating matters. So surely we need to justify that the denominator is non-zero at x = 1/8?
I agree, eliminating the case where both the numerator and the denominator have simple zeroes is a necessary step.

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