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How do you find the cartesian equation of a curve? (C4) Watch

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    Example question:
    A curve is define by the parametric equations x=(t^2/2) + 1 and y=(4/t) -1
    When t= 2, gradient is dy/dx = -1/2
    Find a cartesian equation of the curve.

    We havent learnt how to do it in school. I got this question from a past paper, and was just wondering how you'd do it.
    Thanks!
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    (Original post by Phoebus Apollo)
    Example question:
    A curve is define by the parametric equations x=(t^2/2) + 1 and y=(4/t) -1
    When t= 2, gradient is dy/dx = -1/2
    Find a cartesian equation of the curve.

    We havent learnt how to do it in school. I got this question from a past paper, and was just wondering how you'd do it.
    Thanks!
    Solve the easier equation for t and substitute into the other.
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    Or remove t from the simultaneous equations by other means eg. dividing.
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    (Original post by Phoebus Apollo)
    Example question:
    A curve is define by the parametric equations x=(t^2/2) + 1 and y=(4/t) -1
    When t= 2, gradient is dy/dx = -1/2
    Find a cartesian equation of the curve.

    We havent learnt how to do it in school. I got this question from a past paper, and was just wondering how you'd do it.
    Thanks!
    It's kind of like solving simultaneous equations except trickier. Basically you want to see what you can do with x and y to eliminate t, that'll get you x in terms of y or y in terms of x as required.

    Here, you can kind of brute force it by doing y + 1 = \frac{4}{t} \Rightarrow t = \frac{4}{y+1}

    So x = \frac{\frac{16}{(y+1)^2}}{2} + 1 .

    But generally, you'll need to think about eliminating t and when you have trig parametric equations, you'll need to think about trigonometrical identities, etc...
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    For this type. Find t in terms of x and then substitute this expression of t into the equation for y. This eliminates t from the y equation leaving an equation in terms of x and y only - the Cartesian equation.
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    (Original post by Zacken)
    ...

    Here, you can kind of brute force it by doing y + 1 = \frac{4}{t} \Rightarrow t = \frac{4}{y+1}

    So x = \frac{\frac{16}{(y+1)^2}}{2} + 1 .

    But generally, you'll need to think about eliminating t and when you have trig parametric equations, you'll need to think about trigonometrical identities, etc...
    and you would complete with
    x = \frac{\frac{16}{(y+1)^2}}{2} + 1 = \frac{32}{(y+1)^2}+1  .<<<< The 32 should be 8 but leaving it for the sake of history

    Because you would never get a mark at C4 with x = \frac{\frac{16}{(y+1)^2}}{2} + 1 .

    Zacken knows this but you might not
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    (Original post by ODES_PDES)
    Solve the easier equation for t and substitute into the other.
    (Original post by morgan8002)
    Or remove t from the simultaneous equations by other means eg. dividing.
    (Original post by Zacken)
    It's kind of like solving simultaneous equations except trickier. Basically you want to see what you can do with x and y to eliminate t, that'll get you x in terms of y or y in terms of x as required.

    Here, you can kind of brute force it by doing y + 1 = \frac{4}{t} \Rightarrow t = \frac{4}{y+1}

    So x = \frac{\frac{16}{(y+1)^2}}{2} + 1 .

    But generally, you'll need to think about eliminating t and when you have trig parametric equations, you'll need to think about trigonometrical identities, etc...
    (Original post by B_9710)
    For this type. Find t in terms of x and then substitute this expression of t into the equation for y. This eliminates t from the y equation leaving an equation in terms of x and y only - the Cartesian equation.
    (Original post by nerak99)
    and you would complete with
    x = \frac{\frac{16}{(y+1)^2}}{2} + 1 = \frac{32}{(y+1)^2}+1  .

    Because you would never get a mark at C4 with x = \frac{\frac{16}{(y+1)^2}}{2} + 1 .

    Zacken knows this but you might not
    Thanks everyone
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    (Original post by nerak99)
    and you would complete with
    x = \frac{\frac{16}{(y+1)^2}}{2} + 1 = \frac{32}{(y+1)^2}+1  .
    Surely it'd be x = \frac{8}{(y+1)^2} + 1?
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    (Original post by Zacken)
    Surely it'd be x = \frac{8}{(y+1)^2} + 1?
    Well I am nervous of arguing with you Zacken and you are correct

    a) You kind of make my point and

    b) You are right because I read it as a/(b/c) instead of (a/b)/c
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    (Original post by Zacken)
    Surely it'd be x = \frac{8}{(y+1)^2} + 1?
    Brackets would help for clarity.
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    (Original post by Xenon17)
    Brackets would help for clarity.
    How would it?
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    (Original post by Zacken)
    How would it?
    . Could be written  \frac{\left(\frac{16}{(y+1)^2} \right)}{2} +1 (which is unambiguous)

    Although it still is a massive carbuncle.
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    (Original post by nerak99)
    . Could be written  \frac{\left(\frac{16}{(y+1)^2} \right)}{2} +1
    That's ruins the aesthetic and doesn't really help clear up any ambiguity since there is a visible and marked difference between

    \displaystyle x = \frac{\frac{16}{(y+1)^2}}{2} and \displaystyle x = \frac{16}{\frac{(y+1)^2}{2}}
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    (Original post by Zacken)
    That's ruins the aesthetic and doesn't really help clear up any ambiguity since there is a visible and marked difference between

    \displaystyle x = \frac{\frac{16}{(y+1)^2}}{2} and \displaystyle x = \frac{16}{\frac{(y+1)^2}{2}}
    Well There is no aesthetic in the double fraction form anyway but at least sticking brackets around the toip fraction makes it clear. If this were written by hand it would look like it could be a/b/c or a/b/c (ironic face) Hence the (a/b)/c we have here should be resolved to a/(bc) (smiley face, gritted teeth face, I am going for my tea now face)
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    (Original post by nerak99)
    Well There is no aesthetic in the double fraction form anyway but at least sticking brackets around the toip fraction makes it clear. If this were written by hand it would look like it could be a/b/c or a/b/c (ironic face) Hence the (a/b)/c we have here should be resolved to a/(bc)
    Ah, fair enough. I'll just simplify it down straight away next time, thanks.
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    Perhaps it would have helped if I hadn't £%@%$&ed it up in the first place.
 
 
 
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