# the point tangent to line

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

• Jul 31st 2012, 01:55 AM
Prove It
Re: the point tangent to line
Quote:

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

What is the gradient of the x axis? Hint: It's horizontal...
• Jul 31st 2012, 03:49 AM
rcs
Re: the point tangent to line
Sir that term i coudnt answer ... im sorry i couldnt get the hint :(
• Jul 31st 2012, 04:35 AM
Plato
Re: the point tangent to line
Quote:

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

$y+xy'=2(1-x-y)(-1-y')$ now find a point $(x,y)$ so that $y'=0$.
• Jul 31st 2012, 06:04 AM
HallsofIvy
Re: 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 AM
richard1234
Re: the point tangent to line
Quote:

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

You've posted a lot of algebra questions on this forum, and this question might be too advanced for you. How much calculus do you know?
• Jul 31st 2012, 04:36 PM
rcs
Re: the point tangent to line
i guess basic for u but im trying to know all these things sir :(
• Jul 31st 2012, 05:03 PM
Plato
Re: the point tangent to line
Quote:

Originally Posted by rcs
i guess basic for u but im trying to know all these things sir :(

What the hell does that mean?
That is not even a well formed English sentence
If you cannot post an proper statement or question then post nothing !
• Jul 31st 2012, 05:29 PM
richard1234
Re: the point tangent to line
Quote:

Originally Posted by rcs
i guess basic for u but im trying to know all these things sir :(

I don't think you're quite ready for derivatives and calculus, especially since you've been posting elementary algebra questions. You need to be proficient in algebra to do well in calculus.
• Jul 31st 2012, 05:33 PM
richard1234
Re: the point tangent to line
The solution would involve differentiating both sides with respect to x:

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

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