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Implicit Differentiation

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, y′=0  ⟹  y−2x=0y' = 0 \implies y - 2x = 0, 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?
Reply 1
Avatar for joostan
joostan
13
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, y′=0  ⟹  y−2x=0y' = 0 \implies y - 2x = 0, 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 dydx\frac{dy}{dx} that actually lie on the curve.
At all points in the plane with y=2xy=2x, 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 x=18x=\frac{1}{8} lies on the curve, and at this point the derivative is also zero, we have a stationary point there.
(edited 7 years ago)
Original post by joostan
The point of the latter part of the solution the mark scheme gave was to find the zeroes of dydx\frac{dy}{dx} that actually lie on the curve.
Since x=18x=\frac{1}{8} 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 :colondollar:

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?
(edited 7 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by 16Characters....
I'll admit to being quite confused here :colondollar:

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:

y′=0⇒y=2x, 12x2−x≠0\displaystyle y'=0 \Rightarrow y=2x, \ 12x^2-x \neq 0

⇒...⇒x=18\displaystyle \Rightarrow ... \Rightarrow x=\frac{1}{8}


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

y′=0⇒y=2x, 12x2−x≠0\displaystyle y'=0 \Rightarrow y=2x, \ 12x^2-x \neq 0

⇒...⇒x=18\displaystyle \Rightarrow ... \Rightarrow x=\frac{1}{8}


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.
Reply 5
Avatar for joostan
joostan
13
Original post by 16Characters....
I'll admit to being quite confused here :colondollar:

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 hadn't read your post properly, which is why I addressed the wrong issue.
I agree, eliminating the case where both the numerator and the denominator have simple zeroes is a necessary step.

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