Given the curve of C defined by the equation x^2 - 2xy +y^4 =4.
1) Find the derivative dy/dx of the curve C by implicit differentiation at the point P= (1,-1).
2) Find a line through the origin that meets the curve perpendicularly.
x^2 - 2xy + y^4 =4
2x - 2y - 2xy' + 4y^3 y' = 0
y' = (x-y)/(x-2y^3)
Plug in (1,-1)
Now find a line through (0,0) whose slope at (h,k) is -1/y' = (h-2k^3)/(k-h)
But such a line has equation y=mx, so we want a line through (h,mh) with slope (h-2m^3h^3)/(mh-h) = (1-2m^3h^2)/(m-1)
Hmmm. I've been having no joy here. Any thoughts?
To find the derivative dy/dx of the curve C using implicit differentiation, we will differentiate both sides of the equation with respect to x and then solve for dy/dx.
1) Differentiating both sides of the equation x^2 - 2xy + y^4 = 4 with respect to x:
d/dx (x^2) - d/dx (2xy) + d/dx (y^4) = d/dx (4)
Applying the chain rule and product rule, we get:
2x - 2y(dy/dx) + 4y^3(dy/dx) = 0
Now, we can solve for dy/dx by isolating it:
dy/dx(2y + 4y^3) = 2x
dy/dx = 2x / (2y + 4y^3)
To find the derivative at the point P (1,-1), substitute the values of x and y into the derivative equation:
dy/dx = 2(1) / (2(-1) + 4(-1)^3)
= 2 / (-2 - 4)
= -2/6
= -1/3
Therefore, the derivative dy/dx of the curve C at the point P (1,-1) is -1/3.
2) To find a line through the origin that meets the curve C perpendicularly, we need to find the slope of the line perpendicular to the curve at the point P (1,-1). The slope of the perpendicular line is the negative reciprocal of the derivative dy/dx at that point.
The negative reciprocal of -1/3 is 3. Therefore, the slope of the perpendicular line is 3.
Now, we have the slope of the line and a point it passes through (the origin), so we can use the point-slope form of a line to find its equation.
Using the point-slope form with the slope = 3 and the point (0,0):
y - y1 = m(x - x1)
y - 0 = 3(x - 0)
y = 3x
Therefore, the line through the origin that meets the curve C perpendicularly is given by the equation y = 3x.