Asked by SHEK
Find an equation of the line tangent to the curve defined by x3+2xy+y3=4 at the point (1,1).
Answers
Answered by
oobleck
The slope of the tangent line at any point is y'
3x^2 + 2y + 2xy' + 3y^2 y' = 0
y' = -(3x^2+2y)/(2x+3y^2)
At (1,1) y' = -1
So the line is
y-1 = -1(x-1)
See the graphs at
www.wolframalpha.com/input/?i=plot+x%5E3%2B2xy%2By%5E3%3D4%2C+y+%3D+2-x
3x^2 + 2y + 2xy' + 3y^2 y' = 0
y' = -(3x^2+2y)/(2x+3y^2)
At (1,1) y' = -1
So the line is
y-1 = -1(x-1)
See the graphs at
www.wolframalpha.com/input/?i=plot+x%5E3%2B2xy%2By%5E3%3D4%2C+y+%3D+2-x
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