Question involving curves/tangent lines.

[WUMmyst]

I am so lost :/
We are learning a new section, and this was on the homework, but it does not look anything like what we did in class today. Any pointers?

 

[WUMmyst]

Okay I have looked at this one again and I was able to get A, but I am stuck on B/C

 

[BigGlenntheHeavy]

$\displaystyle Here \ is \ its \ graph, \ the \ red \ curve \ is \ xy^{2}-x^{3}y \ = \ 6.$

$\displaystyle Try \ and \ take \ it \ from \ here.$

[attachment=0:1brs53gs]def.jpg[/attachment:1brs53gs]

 

[BigGlenntheHeavy]

$\displaystyle WUMmyst, \ did \ you \ get \ b \ and \ c \ from \ the \ graph?$

 

[galactus]

The tangent line is vertical when the slope becomes unbounded.

That is, when the denominator in dy/dx equals 0.

From the equation, $\displaystyle y=\frac{\sqrt{x^{5}+24}+x^{\frac{5}{2}}}{2\sqrt{x}}$

Sub into $\displaystyle 2xy-x^{3}$ and we get: $\displaystyle \sqrt{x^{6}+24x}$

set to 0 and solve for x gives us $\displaystyle x=0, \;\ x= -3^{\frac{1}{5}}\cdot 2^{\frac{3}{5}}\approx -1.888...$

Which can be seen from the graph.

 

[WUMmyst]

Thank you all a ton. I was able to get B but I was struggling with C. But after the latter post I was able to understand. You guys are brilliant! I was able to do the next problem and got all parts correct. Thanks!