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Math Help - Finding equation of a tangent line using parametric equations in 3D

  1. #1
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    Finding equation of a tangent line using parametric equations in 3D

    Hello everyone!
    I have been having some trouble with this problem and was hoping that you guys could help me out.

    Find parametric equations for the tangent line at the point
    on the curve

    I dont really know where to start, other than finding the derivative, which is x=-sin(t), y=cos(t), and z = 1. Thanks for any help.
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by palmaas View Post
    Hello everyone!
    I have been having some trouble with this problem and was hoping that you guys could help me out.

    Find parametric equations for the tangent line at the point
    on the curve

    I dont really know where to start, other than finding the derivative, which is x=-sin(t), y=cos(t), and z = 1. Thanks for any help.
    hope this helps.

    You are almost finish

    Next step would be to plug in your t value of \frac{2\pi}{6} into your derivatives

    x = -\sin{\frac{2\pi}{6}}= -\frac{\sqrt{3}}{2}

    y = \cos{\frac{2\pi}{6}} = \frac{1}{2}

    z = 1

    Now what you also need are your original values for x,y,z

    x = \cos{\frac{2\pi}{6}} = \frac{1}{2}

    y = \sin{\frac{2\pi}{6}} = \frac{\sqrt{3}}{2}

    z = \frac{2\pi}{6}

    now we put all this together

    x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t
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  3. #3
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    Thanks a lot! I have another one like it that is giving me trouble as well.
    Find the parametric equations for the tangent line to the curve
    x = t^{2} - 1, y = t^{3} + 1, z = t^{1}
    at the point (8, 28, 3). Use the variable t for your parameter.

    Any help would be appreciated.
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by palmaas View Post
    Thanks a lot! I have another one like it that is giving me trouble as well.
    Find the parametric equations for the tangent line to the curve
    x = t^{2} - 1, y = t^{3} + 1, z = t^{1}
    at the point (8, 28, 3). Use the variable t for your parameter.

    Any help would be appreciated.
    Try the approach I worked above
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  5. #5
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    I tried and it come out incorrect. I end up with x = 16t+63, y = 1587t+12166, and z =t+3. That doesnt seem right. I must be doing something wrong.
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  6. #6
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    Wow, I figured it out. I was making a really stupid mistake. Thanks a lot for your help!
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  7. #7
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    This is great, but what do you do when 2π/3 is not given?

    Quote Originally Posted by 11rdc11 View Post
    hope this helps.

    You are almost finish

    Next step would be to plug in your t value of \frac{2\pi}{6} into your derivatives

    x = -\sin{\frac{2\pi}{6}}= -\frac{\sqrt{3}}{2}

    y = \cos{\frac{2\pi}{6}} = \frac{1}{2}

    z = 1

    Now what you also need are your original values for x,y,z

    x = \cos{\frac{2\pi}{6}} = \frac{1}{2}

    y = \sin{\frac{2\pi}{6}} = \frac{\sqrt{3}}{2}

    z = \frac{2\pi}{6}

    now we put all this together

    x = \frac{1}{2} -\frac{\sqrt{3}}{2}t, y = \frac{\sqrt{3}}{2} +\frac{1}{2}t, z = \frac{2\pi}{6} +t
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  8. #8
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    Whoops, 2π/6
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