Question

what is the equation of the tangent to the curve
3x^2 +3y^2 +2xy -2x -2y=0
at the point (0,0)?

Answer 1

use the chain rule to get the derivative:
\[\frac{d}{dx}(3x^2+3y^2+2xy-2x-2y) = 0\]\[6x+6y\frac{dy}{dx}+2y+2x\frac{dy}{dx}-2-2\frac{dy}{dx} = 0\]
Solving for dy/dx gives:
\[\frac{dy}{dx} = -\frac{6x+2y-2}{6y+2x-2}\]

plugging in (0,0) gives a slope of -1. then use point slope form to get the line.

Answer 2

thanks!! you are PRO!!