### Dr. Wilson

5. Find the slope of the line which is tangent to the curve consisting of points in the plane satisfying the following equation

xy2 - y = x + 1.

Find the slope of the line which is tangent to the curve at the following points.

 a) (1, 2) b) (-1/2, -1)

To find dy/dx, differentiate both sides of the equation with respect to x.

1.y2 + x(2y)(dy/dx) - dy/dx = 1

Transpose the terms that do not invove the unknown dy/dx to the other side of the equation

2xy(dy/dx) - dy/dx = 1 - y2

Factor out the unknown dy/dx from the terms on the left

(2xy - 1)dy/dx = 1 - y2

Divide both sides by the coefficient of the unknown.

dy/dx = (1 - y2)/(2xy - 1)

a) At (1, 2)

dy/dx = (1 - (2)2)/(2(1)(2) - 1)

= (1 - 4)/(4 - 1)

= -3/3

= -1

dy/dx = (1 - (1)2)/(2(2)(1) - 1)

= (1 - 1)/(4 - 1)

= 0/3

= 0

In this case, since the equation is a quadratic in y, it is relatively easy to solve for y as a function of x.

Transpose all terms to one side

xy2 - y - x - 1 = 0

Incredibly, the square root comes out even.

Indeed the equation factors

xy2 - y - x - 1 = (xy - x - 1)(y + 1) = 0

When we set the factors equal to 0, we get,

 xy - x - 1 = 0 xy = x + 1 y + 1 = 0 y = -1

If we draw the graph, we see that (1, 2) is on the portion of the graph that is given by

y = (x + 1)/x

and that the slope of the tangent line at that point is -1.

If we differentiate y = (x + 1)/x, we get, using the chain rule

and we see that when x = 1, the slope of the tangent line is -1.

b) Also looking at the graph, we see that the point (-1/2, -1) is on both parts of the graph. The question becomes, for which part of the graph will the formula for dy/dx give us the slope of the tangent line. If you look at our formula for

dy/dx = (1 - y2)/(2xy - 1)

if we substitute in the coordinates of (-1/2, -1) we get

(1 - (-1)2)/(2(-1/2)(-1) - 1)

= (1 - 1)/(1 - 1)

= 0/0

which is what we often get when the answer can be any one of several possibilities.