2 x dx + 2 y dy = 2 (2 x^2+4 y^2 -x)(4 x dx + 8 y dy -dx)
at (0, 1/4)
0 +(1/2)dy = 2(0+4(1/16)-0)(8(1/4)dy -dx)
(1/2) dy = 2 (1/4)(2 dy-dx) = dy - (1/2) dx
dy = 2 dy - dx
dy = dx
dy/dx = slope = 1
so y = x + b
put in your point to find b
Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2+y^2=(2x^2+4y^2−x)^2
(0, 0.25)
(cardioid)
1 answer
1 answer