first find the point where y = 1
x^2 + 2x(1) + 1 = 4
x^2 + 2x - 3 = 0
(x+3)(x-1) = 0
x = -3 or x = 1
but in quad I, the point is (1,1)
use implicit differentiation:
2x + 2x(dy/dx) + 2y + 3y^2 (dy/dx) = 0
dy/dx (2x + 3y^2) = -2x - 2y
dy/dx = (-2x - 2y)/(2x + 3y^2)
= (-2-2)/(2+3) = -4/5
equation of tangent:
y - 1 = (-4/5)(x-1)
5y - 5 = -4x + 4
4x + 5y = 9
check:
http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+2xy+%2B+y%5E3+%3D+4,+4x+%2B+5y+%3D+9
see what happens at (1,1)
I'm not sure how to go about this problem:
Find the equation of the line tangent to the curve x^2 + 2xy + y^3 = 4 at y=1 in the 1st quadrant.
2 answers
Thank you so much :)