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.