Ap calc.. Dying!! Please help!
posted by Aparna on .
Given the curve x^2-xy+y^2=9
A) write a general expression for the slope of the curve.
B) find the coordinates of the points on the curve where the tangents are vertical
C) at the point (0,3) find the rate of change in the slope of the curve with respect to x
I don't even know where to start!! Please help!
2x - x dy/dx - y + 2y dy/dx = 0
dy/dx(2y - x) = y - 2x
dy/dx = (y-2x)/(2y-x)
B) for a vertical tangent dy/dx is undefined
For that to happen , the denominator has to be zero
2y - x = 0
or x = 2y
sub back into original:
(2y)^2 - (2y)(y) + y^2 = 9
4y^2 - 2y^2 + y^2 = 9
3y^2 = 9
y^2 = 3
y = ± √3
if y =+√3 , then x = 2√3 ---> point (2√3, √3)
if y = -√3 , then x = -2√3 ---> point (-2√3, -√3)
C) at (0,3)
dy/dx = (3 - 0)/(6-0) = 1/2