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Math Help - Normal to a space curve.

  1. #1
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    Normal to a space curve.

    The question is:

    Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
    \mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}
    at the point P(32, -32, 16).
    -----------

    Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via \mathbf{n} \cdot \mathbf{v} = 0 where \mathbf{v} = \mathbf{Q}-\mathbf{P} (the vector from the plane's origin (P) to any point Q).

    Of course, I can find the derivative of r(t), which is:

    \dot{\mathbf{r}}(t) = 4t^{-3/4}\mathbf{i}-3t^{-1/4}\mathbf{j}+\mathbf{k}

    That will give me the tangent vector at any time t.

    I know that t = 16, from the 'k' component, so I can evaluate \dot{\mathbf{r}}(16) = (\frac{1}{2}, -\frac{3}{2}, 1)

    But I need the normal vector (perpendicular to the curve).

    Any tips?
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    The question is:

    Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
    \mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}
    at the point P(32, -32, 16).
    -----------

    Now the simple part of course, once I have the normal (n) to the curve at P, I can construct a plane equation via \mathbf{n} \cdot \mathbf{v} = 0 where \mathbf{v} = \mathbf{Q}-\mathbf{P} (the vector from the plane's origin (P) to any point Q).

    Of course, I can find the derivative of r(t), which is:

    \dot{\mathbf{r}}(t) = 4t^{-3/4}\mathbf{i}-3t^{-1/4}\mathbf{j}+\mathbf{k}

    That will give me the tangent vector at any time t.

    I know that t = 16, from the 'k' component, so I can evaluate \dot{\mathbf{r}}(16) = (\frac{1}{2}, -\frac{3}{2}, 1)

    But I need the normal vector (perpendicular to the curve).

    Any tips?
    A unit normal of a surface S, at point P is:

    Try this:

     \frac{\vec{u}\text{ }'(t)}{|\vec{u}\text{ }'(t)|}

    Where \vec{u(t)} is the unit tangent vector.

    If you have found the tangent vector, then to get the unit tangent vector you must divide the tangent vector by it's magnitude.
    Last edited by Mush; March 17th 2009 at 04:46 AM.
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  3. #3
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    So if I understand correctly, u(t) = r'(t)/|r'(t)| ? (Which is the unit tangent vector at 't').

    Then u'(t) = r''(t)? (The derivative of r'(t)).

    So you want me to find, \frac{\ddot{\mathbf{r}}(t)}{|\ddot{\mathbf{r}}(t)|  }?

    EDIT:

    Yes, I know how to find a unit vector from a vector, but I'm interested in the perpendicular vector to the tangent, not the tangent.
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    So if I understand correctly, u(s) = r'(t)/|r'(t)| ? (Which is the unit tangent vector at 't').

    Then u'(t) = r''(t)? (The derivative of r'(t)).

    So you want me to find, \frac{\ddot{\mathbf{r}}(t)}{|\ddot{\mathbf{r}}(t)|  }?

    EDIT:

    Yes, I know how to find a unit vector from a vector, but I'm interested in the perpendicular vector to the tangent, not the tangent.
    I know what you're trying to find. But in order to find the unit normal vector, you must find the unit tangent vector, as is shown below:

     \vec{n} (t) = \frac{\vec{u}\text{ } ' (t)}{|\vec{u}\text{ } '(t)|}

    Where

     \vec{u} (t) = \frac{\vec{r}\text{ }'(t)}{|\vec{r}\text{ }'(t)|}
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  5. #5
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    Ok, so the unit tangent vector is simply \frac{\sqrt{7}}{\sqrt{2}}(1/2,-3/2,1), right?
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  6. #6
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    Hmmm.... all the expressions I get are really complex. The book has the answer (for the plane equation) that is nice and simple looking (no square roots, or fancy powers, or fractions, or anything like that!):

    x-3y+2z=160

    Could someone explain how they got that?
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  7. #7
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    You do not need a vector that is normal to the curve (such a vector would be of no use in this problem). You only need one that is normal to the desired plane.

    Quote Originally Posted by scorpion007 View Post
    Find the Cartesian form of the equation for the plane that is perpendicular to the space curve
    \mathbf{r}(t)=16t^{\frac{1}{4}}\mathbf{i} - 4t^{\frac{3}{4}}\mathbf{j}+t\mathbf{k}
    at the point P(32, -32, 16).
    \mathbf r' is tangent to the curve, so we find it to be

    \mathbf r'(t)=4t^{-3/4}\mathbf i-3t^{-1/4}\mathbf j+\mathbf k.

    The point P(32,\,-32,\,16) corresponds to t=16 and

    \mathbf r'(16)=\frac12\mathbf i-\frac32\mathbf j+\mathbf k.

    This vector is tangent to the curve, so it must be normal to the plane perpendicular to the curve. Using the point (32,\,-32,\,16) and normal vector 2\mathbf r'(16), the equation of the plane is

    a(x-x_0)+b(y-y_0)+c(z-z_0)=0

    \Rightarrow (x-32)-3(y+32)+2(z-16)=0

    \Rightarrow x-3y+2z-160=0.
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  8. #8
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    Oh my! Thank you so much! It was way easier than I thought. I misunderstood the question initially, as you can see.

    I didn't realise that the derivative of r(t) was in fact going to give me normal vectors for planes at each 't' that are perpendicular to r(t). I see that now, thanks to you.

    Once I understood that, the rest is easy!
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