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    #1
    Hello!!!

    Show that $n(t)=-g'(t) i+ f'(t) j$ and $-n(t)=g'(t) i- f'(t) j$ are both normal to the curve $r(t)=f(t) i+g(t) j$ at the point $(f(t), g(t))$.

    $n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$

    But is this equal to $0$ ? If so why?

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    #2
    Quote Originally Posted by evinda View Post
    Hello!!!

    Show that $n(t)=-g'(t) i+ f'(t) j$ and $-n(t)=g'(t) i- f'(t) j$ are both normal to the curve $r(t)=f(t) i+g(t) j$ at the point $(f(t), g(t))$.

    $n(t) \cdot r(t)=-g'(t) f(t)+ f'(t) g(t)$

    But is this equal to $0$ ? If so why?
    Hey evinda!!

    That will not generally be 0.
    But then, shouldn't n(t) be normal to the direction or velocity of the curve, which is r'(t)?

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    #3 Thread Author
    Quote Originally Posted by I like Serena View Post
    Hey evinda!!

    That will not generally be 0.
    But then, shouldn't n(t) be normal to the direction or velocity of the curve, which is r'(t)?
    It holds that $r'(t)=f'(t) i+ g'(t) j$.

    And $n(t) \cdot r'(t)=-g'(t) f'(t)+f'(t) g'(t)=0$.

    So $n$ is normal to $r'$. But doesn't this imply that $n$ is parallel to $r$ ?

    Or am I wrong?

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    #4
    Quote Originally Posted by evinda View Post
    It holds that $r'(t)=f'(t) i+ g'(t) j$.

    And $n(t) \cdot r'(t)=-g'(t) f'(t)+f'(t) g'(t)=0$.

    So $n$ is normal to $r'$. But doesn't this imply that $n$ is parallel to $r$ ?

    Or am I wrong?
    No.

    $r$ is merely a vector from the origin to a point on the curve, and $r$ is generally not parallel or perpendicular to the curve.
    See:


    The fact that $n$ is normal to $r'$ means that $n$ is normal to the curve given by $r(t)$.

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    Quote Originally Posted by I like Serena View Post
    No.

    $r$ is merely a vector from the origin to a point on the curve, and $r$ is generally not parallel or perpendicular to the curve.
    See:


    The fact that $n$ is normal to $r'$ means that $n$ is normal to the curve given by $r(t)$.
    Could you explain it further to me? I haven't really understood it...

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    #6
    Quote Originally Posted by evinda View Post
    Could you explain it further to me? I haven't really understood it...
    What is it that you do not understand?

    If we draw a vector $n$ in the drawing that is normal to the curve at point $r(t)$, that vector $n$ will be neither parallel nor perpendicular to the vector $r(t)$.

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    #7 Thread Author
    Quote Originally Posted by I like Serena View Post
    What is it that you do not understand?

    If we draw a vector $n$ in the drawing that is normal to the curve at point $r(t)$, that vector $n$ will be neither parallel nor perpendicular to the vector $r(t)$.
    Ah I see now what you mean... Thank you!!!

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