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startanewww

Application of differentiation. Find the equation of the normal to the curve y=(x+2)e^x-3x at the point (0,2).

  • one year ago
  • one year ago

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  1. startanewww
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    dy/dx at (0,2) = 0 i don't know what to do next.

    • one year ago
  2. mark_o.
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    @ startanewww what did you get on your derivative at (0,2)?

    • one year ago
  3. startanewww
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    isn't it 0?

    • one year ago
  4. mark_o.
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    after getting the slope at (0,2) find the normal means perpendicular to the line at (0,2)

    • one year ago
  5. Yahoo!
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    1st find the equation of tangent to the curve...

    • one year ago
  6. mark_o.
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    slope perpendicular to it is m2= - 1/m1

    • one year ago
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    y-y1 = m(x-x1)

    • one year ago
  8. startanewww
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    but the slope I found for dy/dx at (0,2) is zero. What can i do?

    • one year ago
  9. startanewww
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    then, the slope of the normal will be -1/0 = infinity how can i find the equation of normal?

    • one year ago
  10. Yahoo!
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    Diffrentiate y=(x+2)e^x-3x

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  11. Yahoo!
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    u know Product Rule ?

    • one year ago
  12. Yahoo!
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    u get slope = -1

    • one year ago
  13. Yahoo!
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    SO...Slope of Normal = -1 / m1 = -1/-1 = 1

    • one year ago
  14. Yahoo!
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    Nw just write the equation using points (0,2) m = 1 y-y1 = m(x-x1)

    • one year ago
  15. startanewww
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    y'=e^x+(x+2)e^x-3 dy/dx at (0,2) = e^0 + (0+2)e^0 - 3 = 0

    • one year ago
  16. startanewww
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    am i correct?

    • one year ago
  17. Yahoo!
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    y=(x+2)e^x-3x y' = (x+2) *1 + e^x *1 - 3

    • one year ago
  18. startanewww
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    differentiating e^x gives e^x right?

    • one year ago
  19. startanewww
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    am i correct?

    • one year ago
  20. mark_o.
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    lets say y'=0, can you form a line perpendicular to the curve? try it :D

    • one year ago
  21. mark_o.
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    first fint an equation of the line tangent to the curve at (0,2)

    • one year ago
  22. startanewww
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    (y-2)/(x-0)=0 y=2?

    • one year ago
  23. startanewww
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    what about the equation of normal?

    • one year ago
  24. mark_o.
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    yes correct :D then ?

    • one year ago
  25. startanewww
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    what can i do?

    • one year ago
  26. startanewww
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    the slope will be infinity....

    • one year ago
  27. mark_o.
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    what is a line perpendicular to line y=2?

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  28. mark_o.
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    at (0,2) ?

    • one year ago
  29. startanewww
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    will the slope of the normal be infinity ?

    • one year ago
  30. mark_o.
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    |dw:1356675309579:dw|

    • one year ago
  31. startanewww
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    but -1/0 = infinity?

    • one year ago
  32. startanewww
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    x=0 is the line

    • one year ago
  33. startanewww
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    i don't know the answer....

    • one year ago
  34. startanewww
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    i think you are right

    • one year ago
  35. mark_o.
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    ah ok theres no answer in the book? :D

    • one year ago
  36. startanewww
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    my classmates said the answer was x=0

    • one year ago
  37. mark_o.
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    hmm the line perpendicular to y=2 is the y axis or the vertical line

    • one year ago
  38. startanewww
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    ok so the answer should be x=0 only?

    • one year ago
  39. mark_o.
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    hmm it can be x=0

    • one year ago
  40. startanewww
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    x = 0 to infinity?

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  41. mark_o.
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    it could be y= 0 to 2, or 2 to infinity, or y= 0 to negative infinity

    • one year ago
  42. startanewww
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    ?????? how come

    • one year ago
  43. startanewww
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    equation of the normal at (0,2)...? should there be only one answer?

    • one year ago
  44. mark_o.
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    because line tangent to the curve is y=2 or the horizontal line, now the line normal to it is the line perpendicular to y=2,, w/c are ________ ?

    • one year ago
  45. startanewww
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    vertical lines

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  46. startanewww
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    ?

    • one year ago
  47. mark_o.
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    the vertical line y axis

    • one year ago
  48. mark_o.
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    yess correct :D

    • one year ago
  49. startanewww
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    so what is the final answer? x=0 ?

    • one year ago
  50. mark_o.
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    i would say at point (0,2) it is y= from 0 to 2 vertical line

    • one year ago
  51. startanewww
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    ok thanks

    • one year ago
  52. mark_o.
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    ok YW,,, good luck now :D

    • one year ago
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