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mlddmlnog

  • 3 years ago

Calculus. Help. :'(

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  1. mlddmlnog
    • 3 years ago
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    for what non-negative valuof b is the line given by \[y=-\frac{ 1 }{ 3 }x+b\] perpendicular to the curve y=x^3?

  2. UnkleRhaukus
    • 3 years ago
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    take the derivative (with respect to x) of both equations , what do you get?

  3. mlddmlnog
    • 3 years ago
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    i got.. y'=-1/3+b and y'=3x^2 ..

  4. mlddmlnog
    • 3 years ago
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    waht do you do next?

  5. UnkleRhaukus
    • 3 years ago
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    try the derivative of the first equation again ( you made a mistake)

  6. mlddmlnog
    • 3 years ago
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    hm... i can't seem to find what my mistake was.. :/ is it the +b part??? because I do not know what to do with that...

  7. UnkleRhaukus
    • 3 years ago
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    b is really bx^0

  8. mlddmlnog
    • 3 years ago
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    so then, it would just be y'=-1/3???

  9. UnkleRhaukus
    • 3 years ago
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    thats better

  10. UnkleRhaukus
    • 3 years ago
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    so you have found the first derivatives of the those lines, the first derivative is the slope of the line

  11. UnkleRhaukus
    • 3 years ago
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    the product of the slopes of perpendicular lines is -1

  12. mlddmlnog
    • 3 years ago
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    yes.. uhm.. give me a few min to think please :)

  13. UnkleRhaukus
    • 3 years ago
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    sure,

  14. mlddmlnog
    • 3 years ago
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    ah, i'm struggling.. please help

  15. UnkleRhaukus
    • 3 years ago
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    ok , so \[m_1\times m_2=-1\] \[m_1=-\frac13\qquad\qquad m_2=3x^2\] solve for x

  16. mlddmlnog
    • 3 years ago
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    x=+1 or -1 .

  17. UnkleRhaukus
    • 3 years ago
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    good

  18. UnkleRhaukus
    • 3 years ago
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    now set \[y_1=y_2\] take the first case (x=-1) and find b

  19. UnkleRhaukus
    • 3 years ago
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    does that make sense?

  20. mlddmlnog
    • 3 years ago
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    yes it does :) sorry. open study was being so slow .but i get it now. thank you! :)

  21. UnkleRhaukus
    • 3 years ago
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    what did you you get for an answer/

  22. mlddmlnog
    • 3 years ago
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    no what do i do..?

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