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Exercise Question from Maths book

  1. Dec 3, 2015 #1
    • Member warned about posting with no effort
    1. The problem statement, all variables and given/known data
    "Find all the points on the curve with equation 6y = 2x^3 + 3x^2 + 6x + 5 where the tangent is parallel to the line with equation 3x − y + 2 = 0

    2. Relevant equations
    No idea

    3. The attempt at a solution
    There are no exercises in the book even similar to this example which is fairly frustrating. Its under the differentiation chapter so it obviously involves differentiation. I can do that myself no bother but what is the method to find all points on the curve when a tangent is parallel to the line? Is there some sort of rule?

    NOTE: This is not a graded assignment or even homework for that matter, it is revision for an exam I have in 5 days.
     
  2. jcsd
  3. Dec 3, 2015 #2

    micromass

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    Can you find the tangent line to your curve in an arbitrary point ##(a,b)##?
     
  4. Dec 3, 2015 #3
    Errrrrr.... I might be able to?
    I need to find all points on the curve where the tangent is parallel to the line. I've tried finding a video on YouTube that works with a similar problem but I wasn't able to find anything for it.
     
  5. Dec 3, 2015 #4

    micromass

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    How do you find the tangent line to the curve? Do you see the relation with derivatives?
     
  6. Dec 3, 2015 #5
    You differentiate the 6y - 2x^3 +3x^2 +6x +5
    but I am not given the points on the curve I am asked to find them :/
     
  7. Dec 3, 2015 #6

    SteamKing

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    Well, can you at least find the slope of the line 3x - y + 2 = 0?
     
  8. Dec 3, 2015 #7
    y'(x) = 3 is the slope of the line 3x -y +2 = 0 right? :)
     
  9. Dec 3, 2015 #8

    SteamKing

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    Yes, this is correct.

    Now, can you turn 6y = 2x^3 + 3x^2 + 6x + 5 into y = something ?
     
  10. Dec 3, 2015 #9
    Well if 6y = 2x^3 + 3x^2 +6x +5 then y = (2x^3 + 3x^2 +6x +5)/6 no?
     
  11. Dec 3, 2015 #10

    SteamKing

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    That's correct.

    Now do you understand that the derivative dy/dx for a function can be used to calculate the slope of a tangent to the function when given a particular value of x?
     
  12. Dec 3, 2015 #11
    Yea I understand that I just don't get how to find the points on the curve where the tangent is parallel to the line.
     
  13. Dec 3, 2015 #12

    SteamKing

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    Well, have you calculated dy/dx for y = (2x^3 + 3x^2 +6x +5)/6 yet?

    BTW: I appreciate the likes, but it's not necessary to like every reply.
     
  14. Dec 3, 2015 #13

    micromass

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    How can you see whether two lines are parellel?
     
  15. Dec 3, 2015 #14
    If their slopes are identical? i.e. the exact same value yes?
     
  16. Dec 3, 2015 #15

    micromass

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    Yes. So you need to find the points whose tangent line has slope is 3.
     
  17. Dec 3, 2015 #16
    dy/dx is x^2 + x + 1.
     
  18. Dec 3, 2015 #17

    SteamKing

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    Now that you can find the slope along the function, for what values of x would dy/dx = 3?
     
  19. Dec 3, 2015 #18
    dy/dx would = 3 for....
    x = 1
    (1)^2 + 1 + 1 = 3. What does that prove in terms of the question? :)
     
  20. Dec 3, 2015 #19

    SteamKing

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    Re-read your Post #14 above.

    Also, is x = 1 the only value which satisfies the equation dy/dx = 3 when dy/dx = x2 + x + 1 ?
     
  21. Dec 3, 2015 #20
    x = 1 and x = -2 (because (-2)^2 + (-2) + 1 => 4 -2 + 1 = 3)
    so you have x = 1, x = -2.

    Then you just do when x = 1, y = whatever you solve.
    x = -2, y = whatever you solve
    ?
     
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