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    x = 4t + cos2t
    y= sin3t
    show that dy/dx = 1/√2 when t=π/12

    my dx/dt = 4 - 2sin2t
    dt/dx = 1/ (4 - 2sin2t)

    dy/dt = 3cos3t

    so dy/dx= (3cos3t)/(4-2sin2t)

    even substituting in π/12 gets me nowhere. the only thing i can possibly take out from the dy/dx is cos/sin = 1/tan and then im lost.
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    (Original post by ihatePE)
    x = 4t + cos2t
    y= sin3t
    show that dy/dx = 1/√2 when t=π/12

    my dx/dt = 4 - 2sin2t
    dt/dx = 1/ (4 - 2sin2t)

    dy/dt = 3cos3t

    so dy/dx= (3cos3t)/(4-2sin2t)

    even substituting in π/12 gets me nowhere. the only thing i can possibly take out from the dy/dx is cos/sin = 1/tan and then im lost.
    just substitute in π/12 for t & make sure your calculator is in radians....

    remember sin π/6 = 1/2

    cos π/4 = 1/√2
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    (Original post by ihatePE)
    ...

    even substituting in π/12 gets me nowhere
    Why...?

    You do know that \frac{\sqrt{2}}{2}=\frac{1}{ \sqrt{2}}, right? Or did you not enter stuff into your calc correctly?
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    (Original post by the bear)
    just substitute in π/12 for t & make sure your calculator is in radians....

    remember sin π/6 = 1/2

    cos π/4 = 1/√2
    i completely forgot about that, c2 is so far away
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    (Original post by RDKGames)
    Why...?

    You do know that \frac{\sqrt{2}}{2}=\frac{1}{ \sqrt{2}}, right? Or did you not enter stuff into your calc correctly?
    i dont get it, how does \frac{\sqrt{2}}{2}=\frac{1}{ \sqrt{2}}
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    (Original post by ihatePE)
    i dont get it, how does \frac{\sqrt{2}}{2}=\frac{1}{ \sqrt{2}}
    From C1 you (should) know that \sqrt{2}=2^{1/2}. And 2=2^1. Divide one by the other and the indices subtract leaving you with 2^{1/2-1}=2^{-1/2} which is 1/root2
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    (Original post by ihatePE)
    i dont get it, how does \frac{\sqrt{2}}{2}=\frac{1}{ \sqrt{2}}
    (Original post by RDKGames)
    From C1 you (should) know that \sqrt{2}=2^{1/2}. And 2=2^1. Divide one by the other and the indices subtract leaving you with 2^{1/2-1}=2^{-1/2} which is 1/root2
    in other words \dfrac{1}{\sqrt 2} \times \dfrac{\sqrt 2}{\sqrt 2} and i'm sure you can do the rest
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    (Original post by will'o'wisp2)
    in other words \dfrac{1}{\sqrt 2} \times \dfrac{\sqrt 2}{\sqrt 2} and i'm sure you can do the rest
    that is if i wanted to rationalise the surd. here it asked me to reverse that which i never thought about before.
 
 
 
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