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eroshea

  • 2 years ago

find the equation of the tangent and normal line to the curve y=2x^3 - 3x^2 -2x +5, where x =1

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  1. terenzreignz
    • 2 years ago
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    To get a line, you need a slope, and a point, right? Or, knowing the slope of a line and a point where it passes through, that's sufficient to find the equation of the line, yes?

  2. eroshea
    • 2 years ago
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    yes.. so how will i find the slope?

  3. terenzreignz
    • 2 years ago
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    Magic. Or, you can differentiate this function first... go ahead now :) Remember, the slope of the tangent line at the point would be its derivative, while the slope of the normal line would be the negative reciprocal of the derivative.

  4. terenzreignz
    • 2 years ago
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    Start by differentiating, though.

  5. eroshea
    • 2 years ago
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    dy/dx = 6x^2 - 6x -2

  6. terenzreignz
    • 2 years ago
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    Quick to the point, are we? :) Now, when x = 1, what's dy/dx?

  7. eroshea
    • 2 years ago
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    dy/dx = -2

  8. terenzreignz
    • 2 years ago
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    And that is the slope of your tangent line :) What's the slope of your normal line? (Remember, the normal line is *perpendicular* to the tangent line)

  9. eroshea
    • 2 years ago
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    negative reciprocal of it.. 1/2

  10. terenzreignz
    • 2 years ago
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    And that is the slope of the normal line. So you have the slopes for both lines, you now only need a point. Any idea how to get one?

  11. eroshea
    • 2 years ago
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    hmm.. where they would intersect? but i don't know how

  12. terenzreignz
    • 2 years ago
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    Well, both the tangent and the normal would intersect your curve at the point WHERE x = 1. So at the curve itself, when x = 1, what is y? :P

  13. eroshea
    • 2 years ago
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    y = -2

  14. eroshea
    • 2 years ago
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    then i just need to use slope intercept form , right? with P(1,-2) and a slope of 1/2?

  15. terenzreignz
    • 2 years ago
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    No... that's dy/dx I meant y. You know... your original curve? :P

  16. terenzreignz
    • 2 years ago
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    y=2x^3 - 3x^2 -2x +5

  17. eroshea
    • 2 years ago
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    y = -8.. sorry

  18. eroshea
    • 2 years ago
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    wrong again!

  19. terenzreignz
    • 2 years ago
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    LOL don't worry about that :) So you have a point (1 , -8) and you have a slope (-2 for your tangent, 1/2 for your normal) You can't use a slope intercept form, you need the more general point-slope form. When you have a point \(\large (\color{red}p,\color{green}q)\) and a slope \(\large \color{blue}m\) then the equation of the line is: \[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]

  20. eroshea
    • 2 years ago
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    OWH.. i got -2 again

  21. eroshea
    • 2 years ago
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    its 2 right?

  22. terenzreignz
    • 2 years ago
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    What is? Your slope? No, you were right, it's -2 for the tangent.

  23. eroshea
    • 2 years ago
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    no .. the coordinates of the point (1,2)

  24. terenzreignz
    • 2 years ago
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    Ah well then, it won't be (1,-8) but (1,2) instead :) Work with that then. :D

  25. terenzreignz
    • 2 years ago
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    \[\Large y = \color{blue}m(x-\color{red}p)+\color{green}q\]

  26. eroshea
    • 2 years ago
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    y = -2x + 4 ?

  27. terenzreignz
    • 2 years ago
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    correct :) Now the normal line? the only difference is the slope :)

  28. eroshea
    • 2 years ago
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    2y= x + 3

  29. terenzreignz
    • 2 years ago
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    and you're done :P

  30. eroshea
    • 2 years ago
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    :D thanks kiddo .. but you're smarter than me

  31. terenzreignz
    • 2 years ago
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    Nah. I've just read in advance :P

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