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Math Help - tangent plane to the torus

  1. #1
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    tangent plane to the torus

    Hi everyone, I got a question here and I think I have solved it, but I just feel the my answer is so messy, can anyone tell me if my answer is right please?

    Find the tangent plane to the torus
    x=(2+\cos\phi)\cos\theta
    y=(2+\cos\phi)\sin\theta
    z=\sin\phi
    where  0 \gep \theta \leq 2\pi at the point (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2})

    My answer is (8 \sqrt{3}+6)x-(24+6 \sqrt{3})y-(16+4 \sqrt{3})z=-21 \sqrt{3}+32
    Am I right? It looks so weird.
    Last edited by tsang; October 7th 2010 at 03:31 PM.
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    Quote Originally Posted by tsang View Post
    Hi everyone, I got a question here and I think I have solved it, but I just feel the my answer is so messy, can anyone tell me if my answer is right please?

    Find the tangent plane to the torus
    x=(2+\cos\phi)\cos\theta
    y=(2+\cos\phi)\sin\theta
    z=\sin\phi
    where 0 \lep \theta \geq 2\pi at the point (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2})

    My answer is (8 \sqrt{3}+6)x-(24+6 \sqrt{3})y-(16+4 \sqrt{3})z=-21 \sqrt{3}+32
    Am I right? It looks so weird.
    Something is wrong, because the given point does not lie on that plane. If you change the 32 on the right side of the plane's equation to –32, then the point does lie on the plane. In that case, each of the coefficients in the equation is a multiple of 4+\sqrt3, and you can divide through by that number to get the answer in the form 2\sqrt3x - 6y - 4z = -5-4\sqrt3. That is less weird-looking, but I haven't checked whether it is correct.
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    Quote Originally Posted by Opalg View Post
    Something is wrong, because the given point does not lie on that plane. If you change the 32 on the right side of the plane's equation to –32, then the point does lie on the plane. In that case, each of the coefficients in the equation is a multiple of 4+\sqrt3, and you can divide through by that number to get the answer in the form 2\sqrt3x - 6y - 4z = -5-4\sqrt3. That is less weird-looking, but I haven't checked whether it is correct.
    Thank you very much!
    I just wonder how did you check the answer? Because I was thinking maybe I can subsititute the points (I mean my answer) back to the question and see if it fits the original question, but I didn't know how to do it.
    So how did you check it? Where can you subsititute back and you found it should be -32 instead of 32? Thanks a lot.
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    Quote Originally Posted by Opalg View Post
    Something is wrong, because the given point does not lie on that plane. If you change the 32 on the right side of the plane's equation to –32, then the point does lie on the plane. In that case, each of the coefficients in the equation is a multiple of 4+\sqrt3, and you can divide through by that number to get the answer in the form 2\sqrt3x - 6y - 4z = -5-4\sqrt3. That is less weird-looking, but I haven't checked whether it is correct.
    Hi,I think I have found that as you said,the right side should be -32,I forgot to change the sign when I took off bracket.
    However,after I change it become negative 32,the first two terms in left side is multiple of 3+\sqrt3 instead of 4+\sqrt3,so I still cannot simplyfy.Could you please help me?Thanks a lot.
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    Quote Originally Posted by tsang View Post
    Hi, I think I have found that as you said, the right side should be -32, I forgot to change the sign when I took off bracket.
    However, after I change it become negative 32,the first two terms in left side is multiple of 3+\sqrt3 instead of 4+\sqrt3, so I still cannot simplify. Could you please help me? Thanks a lot.
    I'm not sure how to help, because I don't know what method you used to get the tangent plane equation. What I would do would be to start with the point (x,y,z) = \bigl((2+\cos\phi)\cos\theta, (2+\cos\phi)\sin\theta, \sin\phi\bigr) on the torus, noticing that the given point (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}) corresponds to taking \phi = \pi/6 and \theta = \pi/3. Then I took the partial \frac{\partial}{\partial\phi} and \frac{\partial}{\partial\theta} partial derivatives of (x,y,z), getting \bigl(-\sin\phi\cos\theta, -\sin\phi\sin\theta, \cos\phi\bigr) and \bigl(-(2+\cos\phi)\sin\theta, (2+\cos\phi)\cos\theta, 0\bigr) as the tangent vectors in the \phi and \theta directions respectively. Next, I formed the cross product of the two tangent vectors to get a normal vector to the tangent plane. After cancelling a factor of (2+\cos\phi), I found that the normal vector was \bigl(\cos\phi\cos\theta, \cos\phi\sin\theta, \sin\phi\bigr). Plugging in the values \phi = \pi/6 and \theta = \pi/3 gives the normal vector (\sqrt3,3,2). So the equation of the tangent plane at that point is \sqrt3x + 3y +2z = const. Finally, substituting in the values (x,y,z) = (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}), I got the value 4+4\sqrt3 for the constant. So my equation for the tangent plane is \boxed{\sqrt3x + 3y +2z = 4(1+\sqrt3)}. But don't take that on trust, because I could well have made mistakes in the calculations.
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    Quote Originally Posted by Opalg View Post
    I'm not sure how to help, because I don't know what method you used to get the tangent plane equation. What I would do would be to start with the point (x,y,z) = \bigl((2+\cos\phi)\cos\theta, (2+\cos\phi)\sin\theta, \sin\phi\bigr) on the torus, noticing that the given point (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}) corresponds to taking \phi = \pi/6 and \theta = \pi/3. Then I took the partial \frac{\partial}{\partial\phi} and \frac{\partial}{\partial\theta} partial derivatives of (x,y,z), getting \bigl(-\sin\phi\cos\theta, -\sin\phi\sin\theta, \cos\phi\bigr) and \bigl(-(2+\cos\phi)\sin\theta, (2+\cos\phi)\cos\theta, 0\bigr) as the tangent vectors in the \phi and \theta directions respectively. Next, I formed the cross product of the two tangent vectors to get a normal vector to the tangent plane. After cancelling a factor of (2+\cos\phi), I found that the normal vector was \bigl(\cos\phi\cos\theta, \cos\phi\sin\theta, \sin\phi\bigr). Plugging in the values \phi = \pi/6 and \theta = \pi/3 gives the normal vector (\sqrt3,3,2). So the equation of the tangent plane at that point is \sqrt3x + 3y +2z = const. Finally, substituting in the values (x,y,z) = (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}), I got the value 4+4\sqrt3 for the constant. So my equation for the tangent plane is \boxed{\sqrt3x + 3y +2z = 4(1+\sqrt3)}. But don't take that on trust, because I could well have made mistakes in the calculations.
    Dear friend,thank you so much for your details! Thank you for your time, it must took you a while to type all of these.
    I used exactly same methods as you.I calculated two tangent vectors with respect to \phi, and \theta, and I also got exactly same answer when I subsititute to find they are \frac{\pi}{6},and \frac{\pi}{3}
    Except when I calculated normal vector,you end up with something quite simple and nice,but I still had three nasty and complicated numbers, than I used the formula (x-x_0,y-y_0,z-z_0) \cdot n\mid_(\phi,\theta)=0, but did not end up same thing as you.
    I will double check my work and calculation,I may made mistakes since the numbers are weird. Thanks again for your time.
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    Quote Originally Posted by Opalg View Post
    I'm not sure how to help, because I don't know what method you used to get the tangent plane equation. What I would do would be to start with the point (x,y,z) = \bigl((2+\cos\phi)\cos\theta, (2+\cos\phi)\sin\theta, \sin\phi\bigr) on the torus, noticing that the given point (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}) corresponds to taking \phi = \pi/6 and \theta = \pi/3. Then I took the partial \frac{\partial}{\partial\phi} and \frac{\partial}{\partial\theta} partial derivatives of (x,y,z), getting \bigl(-\sin\phi\cos\theta, -\sin\phi\sin\theta, \cos\phi\bigr) and \bigl(-(2+\cos\phi)\sin\theta, (2+\cos\phi)\cos\theta, 0\bigr) as the tangent vectors in the \phi and \theta directions respectively. Next, I formed the cross product of the two tangent vectors to get a normal vector to the tangent plane. After cancelling a factor of (2+\cos\phi), I found that the normal vector was \bigl(\cos\phi\cos\theta, \cos\phi\sin\theta, \sin\phi\bigr). Plugging in the values \phi = \pi/6 and \theta = \pi/3 gives the normal vector (\sqrt3,3,2). So the equation of the tangent plane at that point is \sqrt3x + 3y +2z = const. Finally, substituting in the values (x,y,z) = (1+\frac{\sqrt{3}}{4},\sqrt{3}+\frac{3}{4},\frac{1  }{2}), I got the value 4+4\sqrt3 for the constant. So my equation for the tangent plane is \boxed{\sqrt3x + 3y +2z = 4(1+\sqrt3)}. But don't take that on trust, because I could well have made mistakes in the calculations.

    Sorry that I need a bit more help from you.
    May I ask you that how come the factor (2+cos\phi) can be cancelled? Why can we divide it from the normal vector? I thought we must keep it.
    Thanks a lot.
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  8. #8
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    Quote Originally Posted by tsang View Post
    May I ask you that how come the factor (2+cos\phi) can be cancelled? Why can we divide it from the normal vector? I thought we must keep it.
    The answer is that you only need to know the direction of the normal vector; its length is unimportant. In fact, it would have been even better to have cancelled the factor (2+\cos\phi) at an earlier stage. One of the vectors in the tangent plane (obtained by partial differentiation) is \bigl(-(2+\cos\phi)\sin\theta, (2+\cos\phi)\cos\theta, 0\bigr). You can cancel the factor from that, getting the vector (-\sin\theta, \cos\theta, 0). The cross product calculation then becomes easier, and still gives you the direction of the normal vector.
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