# Diffrential equation question on tangents.

• Jan 10th 2009, 12:51 AM
AshleyT
Diffrential equation question on tangents.
Ok,
$\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}$
1. Show that curve has no tangent parallel to y axis.
Well, when thats the case, dy/dx = 0.
Giving
$2 - y^2 = 0$
And therefore $y = +-/sqrt{2}$
Well you were meant to do
2xy - 3 = 0 and sub into original equation to show there is no solution.

And my question is Why? Do you always do 1/(dy/dx) and equate that to 0 to find a line parallel to y axis? Why do you do that?
• Jan 10th 2009, 01:23 AM
mr fantastic
Quote:

Originally Posted by AshleyT
Ok,
$\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}$
1. Show that curve has no tangent parallel to y axis.
Well, when thats the case, dy/dx = 0.
Giving
$2 - y^2 = 0$
And therefore $y = +-/sqrt{2}$
Well you were meant to do
2xy - 3 = 0 and sub into original equation to show there is no solution.

And my question is Why? Do you always do 1/(dy/dx) and equate that to 0 to find a line parallel to y axis? Why do you do that?

If the tangent is parallel to the vertical axis it is a vertical line. What's the gradient of a vertical line ....?

You're getting mixed up with parallel to the x-axis, in which case the tangent is a horizontal line with gradient zero ....
• Jan 10th 2009, 02:07 AM
AshleyT
Quote:

Originally Posted by mr fantastic
If the tangent is parallel to the vertical axis it is a vertical line. What's the gradient of a vertical line ....?

You're getting mixed up with parallel to the x-axis, in which case the tangent is a horizontal line with gradient zero ....

Oh...I see...Thank-you!