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anonymous

  • 5 years ago

This is a question on my math homework, but I can't make any sense of it. Here it is: find the equation of the normal to the curve with equation y=(x^3)+1. Is there a typo or am I just missing something?

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  1. anonymous
    • 5 years ago
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    The normal is a line perpendicular to the tangent line

  2. anonymous
    • 5 years ago
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    Could you explain how to do this problem? It has been a while since I have done this kind of problem.

  3. anonymous
    • 5 years ago
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    for what point are you trying to find the normal to?

  4. anonymous
    • 5 years ago
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    Also if the way I have put in the equation is confusing it should be: \[y=x^3+1\]

  5. myininaya
    • 5 years ago
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    thats what i was fixing to ask lol

  6. anonymous
    • 5 years ago
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    (1,2)

  7. anonymous
    • 5 years ago
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    k... now take the derivative of you function.... plug in the values (1,2) this will give you the slope, Let me know what you get

  8. myininaya
    • 5 years ago
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    find f'(x) and evaluate it at x=1 then take the negative reciprocal of that this gives you the slope of the normal line

  9. anonymous
    • 5 years ago
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    for the derivative I got \[y=3x^2\]

  10. anonymous
    • 5 years ago
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    and now I put in 2 for y and 1 for x, such that \[2=3(1)^2\]

  11. anonymous
    • 5 years ago
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    correct?

  12. anonymous
    • 5 years ago
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    no..... sorry just 1 for x, bc y' is the slope

  13. anonymous
    • 5 years ago
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    so then y=3

  14. anonymous
    • 5 years ago
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    and to get x i plug in 3 for y in the derivative?

  15. anonymous
    • 5 years ago
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    yeah thats your slop for the tanget, now you have to find the slope for the normal

  16. anonymous
    • 5 years ago
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    since the normal is perpendicular to the tangent, take the reciprocal of the slope of the tangent to get the slope of the normal

  17. anonymous
    • 5 years ago
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    So if the sope of the tangent is 3, then the slope of the normal is 1/3, correct?

  18. anonymous
    • 5 years ago
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    close... -1/3

  19. anonymous
    • 5 years ago
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    If the reciprocal is the number written as a fraction then flipped why did the sign change?

  20. anonymous
    • 5 years ago
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    sorry it is 1/3, I'm being retarded

  21. anonymous
    • 5 years ago
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    and then thats it? the answer is 1/3?

  22. anonymous
    • 5 years ago
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    now use the point slope formula to get the equation of the normal

  23. anonymous
    • 5 years ago
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    refresh my memory, what is the point slope formula? Or can i just use y=mx+b?

  24. anonymous
    • 5 years ago
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    wait the slope is -1/3 bc perpendicular slope have opposite signs and the point slope formula is \[y-y_1=m(x-x_1)\]

  25. anonymous
    • 5 years ago
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    perpendicular lines are negative reciprocals

  26. anonymous
    • 5 years ago
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    ok. y1 and x1 are from the points given at the start, right?

  27. anonymous
    • 5 years ago
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    yes

  28. anonymous
    • 5 years ago
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    and m=\[-1/3\]

  29. anonymous
    • 5 years ago
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    yes

  30. anonymous
    • 5 years ago
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    so plugging everything in gets\[3-2=-1/3(1-1)\]

  31. anonymous
    • 5 years ago
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    wich solves to 1=0

  32. anonymous
    • 5 years ago
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    no.... you are looking y, the values you have are y1, x1, and m

  33. anonymous
    • 5 years ago
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    y1=2, x1=1 and m=-1/3

  34. anonymous
    • 5 years ago
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    so then it should be \[y-2=-1/3(x-1)\]

  35. anonymous
    • 5 years ago
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    yeah... now put it in terms of y=mx+b

  36. anonymous
    • 5 years ago
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    \[2=-1/3(1)+b\]

  37. anonymous
    • 5 years ago
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    and solve for b?

  38. anonymous
    • 5 years ago
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    no... so you have \[y-2=-\frac{1}{3}(x-1)\] Expand this and you get \[y-2=-\frac{1}{3}x+\frac{1}{3}\] now combine like terms \[y=-\frac{1}{3}x+\frac{1}{3}+2\] \[y=-\frac{1}{3}x+\frac{7}{3}\] which is in the form of y=mx+b

  39. anonymous
    • 5 years ago
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    could you also distribute the negative and move the 1/3 from \[y-2=-1/3(x-1)\] to give 3y-6=-x+1

  40. anonymous
    • 5 years ago
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    then move -x+1 over to give x+3y-7=0

  41. anonymous
    • 5 years ago
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    yes you could

  42. anonymous
    • 5 years ago
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    ok thank you for the help. yo just got a new fan!

  43. anonymous
    • 5 years ago
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    also now i'll use this site for help with math if the teacher is unavailable. Thanks again!

  44. anonymous
    • 5 years ago
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    Cheers , hope I didin't confuse you

  45. anonymous
    • 5 years ago
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    no you did fine. Have to go now bye

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