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    (Original post by imasillynarb)
    It is trial and error, you just have to keep sticking numbers in until it equals 0
    Great!!!
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    (Original post by Silly Sally)
    Great!!!
    Didnt you do this in P1? The factors arent usually bigger/smaller than 3/-3
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    (Original post by imasillynarb)
    Didnt you do this in P1? The factors arent usually bigger/smaller than 3/-3
    Well yes - but sometimes they are fractions!!! That greatly increases all the possible numbers!!! :eek:
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    (Original post by Silly Sally)
    Ok Bezza - i am probably being really silly here and someone has proably answered this question - but how do you know that (t-2) is a factor?
    I'm just guessing here, but I imagine that Bezza just looked at the original eqn,

    2t^3 - 3t^2 - 4 = 0

    and spotted the solution, or tried out simple values of t to see if they worked, e.g. t=0, t=±1, t=±2, ...

    putting t=2 gives,

    16 - 12 - 4 = 0,

    so t=2 satisfies the eqn!
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    (Original post by Silly Sally)
    Well yes - but sometimes they are fractions!!! That greatly increases all the possible numbers!!! :eek:
    it appears that trial and error is the ONLY way of doing this question.
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    (Original post by Fermat)
    I'm just guessing here, but I imagine that Bezza just looked at the original eqn,

    2t^3 - 3t^2 - 4 = 0

    and spotted the solution, or tried out simple values of t to see if they worked, e.g. t=0, t=±1, t=±2, ...

    putting t=2 gives,

    16 - 12 - 4 = 0,

    so t=2 satisfies the eqn!
    Again sounds trial and error - but i don't mind - as long as i have my trusty calulator to help me. I just wanted to make sure that there wasn't a more "reliable" way to find the value of t

    Thanks fermat and imasillynarb

    Edit: and now mathematican!!!
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    (Original post by Silly Sally)
    Again sounds trial and error - but i don't mind - as long as i have my trusty calulator to help me. I just wanted to make sure that there wasn't a more "reliable" way to find the value of t

    Thanks fermat and imasillynarb

    Edit: and now mathematican!!!
    Well, I doubt theyre going to make it any horrible fractions, if you put a number in and its close(say t = 1) then try t = 0.5 etc..
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    (Original post by imasillynarb)
    Well, I doubt theyre going to make it any horrible fractions, if you put a number in and its close(say t = 1) then try t = 0.5 etc..
    Good point! Thanks!!! I am having a silly sally day, so i am thinking less than I usually do!!!
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    (Original post by Silly Sally)
    Good point! Thanks!!! I am having a silly sally day, so i am thinking less than I usually do!!!

    you know if it was question involving trig would it also be trial and error but instead of numbers u would be putting say Pi/2 or Pi/4

    I.e question like this

    Curve C has parametric equation
    x=4cos2t y=3sint -2<t<2

    A is the point (2, 3/2) and lies on C
    (a) Find the value of t at A (i can do this part)
    (b) Find dy/dx in terms of t (can also do this part)

    (c) show that an equation of the normal to C at A is 6y-16x+23=0 (also can do this)

    BUT...
    The normal at A cuts C again at point B
    (d) Find the y-coordinate of the point B


    double checked everyihtng is right
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    (Original post by lexazver203)
    you know if it was question involving trig would it also be trial and error but instead of numbers u would be putting say Pi/2 or Pi/4

    I.e question like this

    Curve C has parametric equation
    x=4cos2t y=3sint -2<t<2

    A is the point (2, 3/2) and lies on C
    (a) Find the value of t at A (i can do this part)
    (b) Find dy/dx in terms of t (can also do this part)

    (c) show that an equation of the normal to C at A is 6y-16x+23=0 (also can do this)

    BUT...
    The normal at A cuts C again at point B
    (d) Find the y-coordinate of the point B


    double checked everyihtng is right
    How is that trial and error? I think i am missing the point here - sorry
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    (Original post by Silly Sally)
    How is that trial and error? I think i am missing the point here - sorry
    Thats my point - would it be trial and error in this instance too or is there some method.
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    (Original post by lexazver203)
    BUT...
    The normal at A cuts C again at point B
    (d) Find the y-coordinate of the point B
    you don't have to use trial and error for this part. This is one way you could do it:
    First change the original parametric equations into cartesian form.
    x = 4cos2t
    x = 4 ( 1 - 2sin²t)
    x = 4 - 8(sint)²

    you know that y=3sint,
    so sint = y/3

    sub in 'x' to give,
    x = 4 - 8 (y/3)²

    you also know that the equation of the normal is: (6y + 23) / 16
    equating this to the cartesian equation above, and simplifying, you should get this quadratic in 'y':

    128y² + 54y - 369 = 0

    Using the quadratic formula, you'll get the two answers for y:

    y = (-54 + 438) / 256
    or y = (-54 - 438) / 256

    y = 3/2 (which is the solution at A)
    or y = -123 / 64

    so the answer is y = -123 / 64
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    (Original post by Silly Sally)
    Ok Bezza - i am probably being really silly here and someone has proably answered this question - but how do you know that (t-2) is a factor?
    As other people have said it's just trial and error with the factor theorem. You can look at the units term for some guidance on because the numbers you guess will normally be an integer factor of the units term. Sorry, I haven't explained this well! For example, in this question the expression is 2t^3 - t^2 - 4 so the numbers you guess need to be factors of 4 so try ±1, ±2 and putting t=2 makes the expression equal zero so (t - 2) is a factor, and the quadratic factor must have a +2 as the unit term so they multiply to give the right expression. Just trial and error really - I seem to have over-complicated this a bit :rolleyes:
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    (Original post by mockel)
    you don't have to use trial and error for this part. This is one way you could do it:
    First change the original parametric equations into cartesian form.
    x = 4cos2t
    x = 4 ( 1 - 2sin²t)
    x = 4 - 8(sint)²

    you know that y=3sint,
    so sint = y/3

    sub in 'x' to give,
    x = 4 - 8 (y/3)²

    you also know that the equation of the normal is: (6y + 23) / 16
    equating this to the cartesian equation above, and simplifying, you should get this quadratic in 'y':

    128y² + 54y - 369 = 0

    Using the quadratic formula, you'll get the two answers for y:

    y = (-54 + 438) / 256
    or y = (-54 - 438) / 256

    y = 3/2 (which is the solution at A)
    or y = -123 / 64

    so the answer is y = -123 / 64
    thanx for that that is the right answer
 
 
 
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