- anonymous

more implicit differentials
for what values of x does the curve y^2 -x^4 + 2xy -18x^2 = 10 have vertical tangent lines?

- chestercat

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- anonymous

@ganeshie8 @IrishBoy123 any ideas?

- zepdrix

Horizontal tangent lines when: \(\large\rm y'=0\)
Vertical tangent lines when: \(\large\rm y'=\frac{stuff}{0}\)
Have you tried finding your y' yet? :)

- anonymous

yes

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## More answers

- anonymous

i feel i messed up somewhere though so im redoing that right now

- anonymous

im getting
\[y'= \frac{ 4x(x^2+9) }{ 2(y+x) }\]

- zepdrix

Hmm I think there is another term on top.
Did you forget to product rule again? :)\[\large\rm y^2 -x^4 + \color{orangered}{2xy} -18x^2 = 10\]

- anonymous

no i just simplified from 4x^3 + 36x

- zepdrix

But where is the -2y in the numerator? :o hmm

- anonymous

oh your totally right

- anonymous

argh your so clever :P

- zepdrix

\[\large\rm y^2 -x^4 + 2xy -18x^2 = 10\]Differentiating gives,\[\large\rm 2yy'-4x^3+2y+2xy'-36x=0\]That's your first step ya? :D

- zepdrix

you're*
that's gonna bug me, i had to lol

- anonymous

yeah i even crossed it out and all i guess it just slipped my mind when i was rewriting it on the other side of the equal sign

- anonymous

anyway so its \[y'= \frac{ 4x^3 + 36x + 2y }{ 2(x+y) }\]

- zepdrix

Woops, -2y on top I think ya?

- zepdrix

Anyway, let's just get rid of all the 2's I guess,\[\large\rm y'=\frac{2x^3+18x-y}{x+y}\]That's the only simplification that really cleans it up nicely.

- anonymous

:( yes , okay so now what?

- zepdrix

This derivative function is undefined when the denominator is zero.
(This is also when we're getting vertical tangents.)

- zepdrix

So vertical tangent when the denominator is zero,
\(\large\rm x+y=0\)

- zepdrix

Overheat again? :) LOL

- Jhannybean

|dw:1441787497363:dw| where its undefined? :P

- anonymous

yep :( i need a new computer

- zepdrix

Hmm ya that's a weird answer :o I do something wrong?

- anonymous

anyway one of the answer choices is x = -y so i think that's the answer right?

- zepdrix

yay team \c:/
it just doesn't make a whole lot of sense with the graph of the function :D
I guess I just need to think about it a sec lol

- Jhannybean

I think that's about right, it's either \(\sf y=-x\) or \(\sf x=-y\).

- anonymous

thank you once more :D

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