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    If I am given curve
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    (Original post by nomad609)
    I...
    So, you have z=h(y), and z=g(x).

    Since you have a diagram plotting x against y, do you have a function for that, or can you easily derive one?
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    (Original post by ghostwalker)
    So, you have z=h(y), and z=g(x).

    Since you have a diagram plotting x against y, do you have a function for that, or can you easily derive one?
    Thank you for your reply, I can try and work it out.
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    (Original post by nomad609)
    Can you check if this is correct please?
    Looks good.

    Name:  Untitled.jpg
Views: 41
Size:  46.5 KB
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    (Original post by ghostwalker)
    Looks good.

    Name:  Untitled.jpg
Views: 41
Size:  46.5 KB
    Thank you sir, for your verification.

    Am I able to ask you what do I do now please?
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    (Original post by nomad609)
    Thank you sir, for your verification.

    Am I able to ask you what do I do now please?
    Well you now have x and y as functions of a parameter t, and it just remains to bring z into the picture.

    You say you have z as a function of x or y - which to be honest, I find rather odd given the x-y plot. I would have thought z depended on both x and y. Is that not the case?
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    (Original post by ghostwalker)
    Well you now have x and y as functions of a parameter t, and it just remains to bring z into the picture.

    You say you have z as a function of x or y - which to be honest, I find rather odd given the x-y plot. I would have thought z depended on both x and y. Is that not the case?
    No it is both x and y. I just articulate poorly.

    So if z is a function of x and y do I try to solve by equating both z terms together and then solving?
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    (Original post by nomad609)
    No it is both x and y. I just articulate poorly.

    So if z is a function of x and y do I try to solve by equating both z terms together and then solving?
    I think I'd like to see the two diagrams and functions you've been given.

    I'd expect z=h(x,y), rather than z=f(x) and z=g(y)
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    (Original post by ghostwalker)
    I think I'd like to see the two diagrams and functions you've been given.

    I'd expect z=h(x,y), rather than z=f(x) and z=g(y)
    certainly sir, I will upload. Thank you for your help.
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    (Original post by nomad609)
    ...
    I see now. The equations you've been given relate to the cross-section of the elliptical cone on which the curve lies and not to the curve iteself.
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    (Original post by nomad609)
    do you need me to take the image again
    No problem, I downloaded and rotated it.

    If we consider the x-z diagram, and the points where the curve touches the line on the right, the x value equals the t value (cos(t) = 1) and we have z=3t/pi

    If we consider the y-z diagram and the points where the curve touches the line on the right the y values is 2t and we have z=3t/pi, again.

    I'd hazard a guess that z=3t/pi is our value for z in terms of the parameter t, and hence the parameterisation of the line becomes (t cos(t), 2t sin(t), 3t/pi)

    Since these diagrams seem to indicate that the envelope for z=constant is an ellipse, we could just check that our parameterisation conforms to that.

    Edit: Last post for today.
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    (Original post by ghostwalker)
    No problem, I downloaded and rotated it.

    If we consider the x-z diagram, and the points where the curve touches the line on the right, the x value equals the t value (cos(t) = 1) and we have z=3t/pi

    If we consider the y-z diagram and the points where the curve touches the line on the right the y values is 2t and we have z=3t/pi, again.

    I'd hazard a guess that z=3t/pi is our value for z in terms of the parameter t, and hence the parameterisation of the line becomes (t cos(t), 2t sin(t), 3t/pi)

    Since these diagrams seem to indicate that the envelope for z=constant is an ellipse, we could just check that our parameterisation conforms to that.

    Edit: Last post for today.
    Thank you so much for this - kind of looks like an Archimdean Spiral.

    I really appreciate your help.

    -nomad609
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    (Original post by ghostwalker)
    No problem, I downloaded and rotated it.

    If we consider the x-z diagram, and the points where the curve touches the line on the right, the x value equals the t value (cos(t) = 1) and we have z=3t/pi

    If we consider the y-z diagram and the points where the curve touches the line on the right the y values is 2t and we have z=3t/pi, again.

    I'd hazard a guess that z=3t/pi is our value for z in terms of the parameter t, and hence the parameterisation of the line becomes (t cos(t), 2t sin(t), 3t/pi)

    Since these diagrams seem to indicate that the envelope for z=constant is an ellipse, we could just check that our parameterisation conforms to that.

    Edit: Last post for today.
    hi ghostwalker, my friend said to use the helix, with its parametric equations. Do you know if I can do it this way too? The way you did it seems cleaner and makes more sense imo.
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    (Original post by nomad609)
    hi ghostwalker, my friend said to use the helix, with its parametric equations. Do you know if I can do it this way too? The way you did it seems cleaner and makes more sense imo.
    Well your curve isn't a helix, though you could use that as a starting point, and you'd end up with the same parameterisation, I suspect. Depends what they did.
 
 
 
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