Find all vertical tangent lines of a curve

[mwilks]

curve : xy^2 - x^3y = 6
derivative : (3x^2y - y^2) / (2xy - x^3)
question : find the x coordinate of each point on the curve where the tangent line is vertical.

i can find derivatives and stuff but i dont know how to answer this question. your a genius if you can figure this one out! thanks a lot in advance

 

[galactus]

curve : $\displaystyle xy^{2} - x^{3}y = 6$
derivative : $\displaystyle y'=\frac{3x^{2}y -y^{2}}{2xy - x^{3}}$
question : find the x coordinate of each point on the curve where the tangent line is vertical.

I can find derivatives and stuff but I dont know how to answer this question. you're a genius if you can figure this one out!. Thanks a lot in advance

If the tangent line is vertical, then the slope heads toward infinity.

If we differentiate $\displaystyle xy^{2}-x^{3}y=6$

we get $\displaystyle 2xyy'+y^{2}-x^{3}y'-3x^{2}y=0$

If we divide by y', then the terms without a y' will tend to 0 as $\displaystyle y'\to {\infty}$.

So, eliminate the terms without a y' and we're left with:

$\displaystyle 2xyy'-x^{3}y'=0$

Divide out the y' and we are left with:

$\displaystyle 2xy-x^{3}=0$

Solve for y:

$\displaystyle y=\frac{x^{2}}{2}$

Now, sub this back into the original function and solve for x.

That will be your points of vertical tangency.

You could also solve the equation for y and graph it. Then, you can see any vertical tangents.

Be careful, though, when doing so. Vertical tangents and vertical asymptotes are two different critters.