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    Bah I never thought i'd be this hard!

    2 questions which I can't answer with all my life:

    1)The parametric equations of a curve

    x= t + e^-t
    y=1 - e^-t

    Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

    I found dy/dx but I can't make it into a log form >.<

    2) Curve is defined parametrically

    x= 2(1+costt)
    y=4sin^2t

    Find cartesian equation

    At first I change 4sin^2t using the identities and made it into 4-4cos^2t
    hence there are no sins and only cos.

    BUT I CANT DO THE CARTESIAN BIT >.<

    Help :'(
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    (Original post by shan735)
    Bah I never thought i'd be this hard!

    2 questions which I can't answer with all my life:

    1)The parametric equations of a curve

    x= t + e^-t
    y=1 - e^-t

    Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

    I found dy/dx but I can't make it into a log form >.<

    2) Curve is defined parametrically

    x= 2(1+costt)
    y=4sin^2t

    Find cartesian equation

    At first I change 4sin^2t using the identities and made it into 4-4cos^2t
    hence there are no sins and only cos.

    BUT I CANT DO THE CARTESIAN BIT >.<

    Help :'(
    For the first part, what do you get when you set dy/dx=1?

    For the second one, note that when you want to get something in cartesian form, you want to eliminate all variables (namely 't' in this case) leaving only x and y terms. Rearrange your x-t equation for cos(t) and substitute this into your y-t equation.
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    (Original post by shan735)
    Bah I never thought i'd be this hard!

    2 questions which I can't answer with all my life:

    1)The parametric equations of a curve

    x= t + e^-t
    y=1 - e^-t

    Find dy/dx and hence find the value of t if the gradient is 1, giving answer in log form

    I found dy/dx but I can't make it into a log form >.<

    2) Curve is defined parametrically

    x= 2(1+costt)
    y=4sin^2t

    Find cartesian equation

    At first I change 4sin^2t using the identities and made it into 4-4cos^2t
    hence there are no sins and only cos.

    BUT I CANT DO THE CARTESIAN BIT >.<

    Help :'(

    Q1.
    dx/dt= 1-e^t
    dy/dt=e^-t

    so dy/dx = (e^-t)/(1-e^-t)

    dy/dx=1
    so

    (e^-t)=(1-e^-t)

    1=2e^-t
    e^-t=1/2
    takin logs
    -t=ln 1/2
    i.e
    -t= -ln2 (as ln1/2=ln1-ln2)

    so t=ln2

    Q2.
    x=2(1+cost)
    y=4sin^2t

    y=4(1-cos^2t)
    x^2=4(1+2cost+cos^2t)

    so y+x^2=8(1+cost)

    i.e y+x^2=4x

    so....

    y=x(4-x)
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    (Original post by catjaum)
    q1.
    Dx/dt= 1-e^t
    dy/dt=e^-t

    so dy/dx = (e^-t)/(1-e^-t)

    dy/dx=1
    so

    (e^-t)=(1-e^-t)

    1=2e^-t
    e^-t=1/2
    takin logs
    -t=ln 1/2
    i.e
    -t= -ln2 (as ln1/2=ln1-ln2)

    so t=ln2

    q2.
    X=2(1+cost)
    y=4sin^2t

    y=4(1-cos^2t)
    x^2=4(1+2cost+cos^2t)

    so y+x^2=8(1+cost)

    i.e y+x^2=4x

    so....

    Y=x(4-x)


    edit:why neg dis???
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    (Original post by catjaum)
    edit:why neg dis???
    Probably because you didn't read the forum guidelines and posted a full solution straight off.

    See http://www.thestudentroom.co.uk/showthread.php?t=403989 and the thread referenced in it.
 
 
 
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