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Math Help - Normal plane and tangent line

  1. #1
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    Normal plane and tangent line

    Given f(t)=(\cos t,\sin t,t) then find the tangent line and normal plane if f\left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{\pi }{4} \right).

    Let x(t)=\cos t,y(t)=\sin t,z(t)=t so normal plane is given by

    \left\langle \left( x'\left( {{t}_{0}} \right),y'\left( {{t}_{0}} \right),z'\left( {{t}_{0}} \right) \right),\left( x-x\left( {{t}_{0}} \right),y-y\left( {{t}_{0}} \right),z-z\left( {{t}_{0}} \right) \right) \right\rangle =0. Is that correct?

    How to find the tangent line?
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  2. #2
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    Quote Originally Posted by Connected View Post
    Given f(t)=(\cos t,\sin t,t) then find the tangent line and normal plane if f\left( \dfrac{\pi }{4} \right)=\left( \dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},\dfrac{\pi }{4} \right).
    You have the point.
    Now find the vector \nabla f\left( {\frac{\pi }{4}} \right).
    That vector is the normal of the plane and the direction vector of the line.
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  3. #3
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    Haven't seen gradient yet on my class, is there another way to do it?

    By the way, is my procedure correct?
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  4. #4
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    Quote Originally Posted by Connected View Post
    Haven't seen gradient yet on my class, is there another way to do it?
    Because this in a truly international forum, I have no idea where you are located. Nonetheless, you are being cheated if the above quote is in fact true. If I were you then I would complain to the educational authority. There is absolutely no reason to ask questions that are not based in standard practices.
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