Finding where two curves share tangent lines

slamminsammya
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Homework Statement


Find all points for which the curves x^2+y^2+z^2=3 and x^3+y^3+z^3=3 share the same tangent line.


Homework Equations


Sharing the same tangent line amounts to having the same derivative. The constraint then is that 3x^2+3y^2+3z^2=2x+2y+2z. The points must obviously also lie on the original curves.


The Attempt at a Solution


Combining the constraint on the derivatives (3(x^2+y^2+z^2)-2(x+y+z)=0) with the constraint that x^2+y^2+z^2=x^3+y^3+z^3=3 we see that the constraint on the derivatives becomes 3(3)-2(x+y+z)=0 which is just the planar equation 2(x+y+z)=9. This feels wrong to me; these curves should not intersect at a plane. Am I right?
 
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Those aren't curves, they are surfaces. Tangent lines aren't usually what you talk about with surfaces although I guess there's no law against it.:rolleyes:

If it's any help to you, here's a picture of the two surfaces:

surfaces.jpg
 
I believe this is not quite finished, though. You have found that your points must lie in the plane x + y + z = 9/2 , but you must still introduce a constraint, since plainly this equation alone permits "solutions" which are far from the sphere. You could, say, write z in terms of x and y using the equation for the sphere.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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