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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    @ganeshie8 @IrishBoy123 any ideas?

  2. zepdrix
    • one year ago
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    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? :)

  3. anonymous
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    i feel i messed up somewhere though so im redoing that right now

  5. anonymous
    • one year ago
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    im getting \[y'= \frac{ 4x(x^2+9) }{ 2(y+x) }\]

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

  7. anonymous
    • one year ago
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    no i just simplified from 4x^3 + 36x

  8. zepdrix
    • one year ago
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    But where is the -2y in the numerator? :o hmm

  9. anonymous
    • one year ago
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    oh your totally right

  10. anonymous
    • one year ago
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    argh your so clever :P

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

  12. zepdrix
    • one year ago
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    you're* that's gonna bug me, i had to lol

  13. anonymous
    • one year ago
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    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

  14. anonymous
    • one year ago
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    anyway so its \[y'= \frac{ 4x^3 + 36x + 2y }{ 2(x+y) }\]

  15. zepdrix
    • one year ago
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    Woops, -2y on top I think ya?

  16. zepdrix
    • one year ago
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    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.

  17. anonymous
    • one year ago
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    :( yes , okay so now what?

  18. zepdrix
    • one year ago
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    This derivative function is undefined when the denominator is zero. (This is also when we're getting vertical tangents.)

  19. zepdrix
    • one year ago
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    So vertical tangent when the denominator is zero, \(\large\rm x+y=0\)

  20. zepdrix
    • one year ago
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    Overheat again? :) LOL

  21. Jhannybean
    • one year ago
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    |dw:1441787497363:dw| where its undefined? :P

  22. anonymous
    • one year ago
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    yep :( i need a new computer

  23. zepdrix
    • one year ago
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    Hmm ya that's a weird answer :o I do something wrong?

  24. anonymous
    • one year ago
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    anyway one of the answer choices is x = -y so i think that's the answer right?

  25. zepdrix
    • one year ago
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    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

  26. Jhannybean
    • one year ago
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    I think that's about right, it's either \(\sf y=-x\) or \(\sf x=-y\).

  27. anonymous
    • one year ago
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    thank you once more :D

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