Question

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!

Answer 1

A)
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

Answer 2

woah

Answer 3

A and B are right. C is the second derivative

Answer 4

Does that mean for 7 whole years, people got #3 incorrect...?

Answer 5

correct way to do part c

c) To find the rate of change of the slope, you take the second derivative, so take the derivative of (y-2x) / (2y-x) with respect to x:

[(2y-x)(dy/dx - 2)-(y-2x)(2dy/dx - 1)] / (2y-x)^2

To find dy/dx, simply plug in (0,3) into (y-2x) / (2y-x) to get 1/2, so the answer is:

[(6)(1/2- 2)] / (6)^2 = -1/4.