At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

need your assistance here guys...

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- Jul 30th 2012, 11:36 PMrcsthe point tangent to line
At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

need your assistance here guys... - Jul 31st 2012, 01:55 AMProve ItRe: the point tangent to line
- Jul 31st 2012, 03:49 AMrcsRe: the point tangent to line
Sir that term i coudnt answer ... im sorry i couldnt get the hint :(

- Jul 31st 2012, 04:35 AMPlatoRe: the point tangent to line
- Jul 31st 2012, 06:04 AMHallsofIvyRe: the point tangent to line
If you do not know what "gradient' (American English: "slope") of a line is, you should not be attempting a problem like this. Finding the derivative is far more advanced that the 'slope' or 'gradient' of a line.

- Jul 31st 2012, 10:08 AMrichard1234Re: the point tangent to line
- Jul 31st 2012, 04:36 PMrcsRe: the point tangent to line
i guess basic for u but im trying to know all these things sir :(

- Jul 31st 2012, 05:03 PMPlatoRe: the point tangent to line
- Jul 31st 2012, 05:29 PMrichard1234Re: the point tangent to line
- Jul 31st 2012, 05:33 PMrichard1234Re: the point tangent to line
The solution would involve differentiating both sides with respect to x:

$\displaystyle y + x \frac{dy}{dx} = 2(1-x-y)(-1 - \frac{dy}{dx})$ (using chain rule)

Then set dy/dx = 0. - Jul 31st 2012, 05:59 PMrcsRe: the point tangent to line
thank you Plato. You have a very positive statement :). Keep it up!