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Tangent Plane for a torus

  1. Nov 14, 2006 #1
    I know I am making a stupid mistake but I am not sure what it is...

    Find an equation for the plane tangent to the torus X(s,t)=((5+2cost)coss, (5+2cost)sins, 2sint) at the point ((5-(3)^1/2)/(2)^1/2, (5-(3)^1/2/(2)^1/2, 1).

    First I have to find what s and t are in order to sub them in for dT/ds and dT/dt. So first I solved for t using 2sint=1 and get t=pi/6. However, when I attempt to sub it into one of the other equations to solve for s I get 2 different answers for s when using each of the different equations. What am I doing wrong?
  2. jcsd
  3. Nov 15, 2006 #2


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    sin(t)= 1/2 means either t= [itex]\pi/6[/itex], in which case cos(t)= [itex]\sqrt{3}/2[/itex] or t= [itex]5\pi/6[/itex], in which case cos(t)= [itex]-\sqrt{3}/2[/itex]. Obviously, since x= (5+ 2cos(t))cos(s)) and y= (5+ 2cos(t))sin(s), if x= y, as you have here, then sin(s)= cos(s)= [itex]\pm1/\sqrt{2}[/itex]. Then 5+ 2cos(t)= 5-sqrt(3) so that cos(t)= [itex]-\frac{\sqrt{3}}{2}[/itex]. From sin2(t)+ cos2(t)= 1 that gives immediately sin(t)= 1/2 as needed. The given point has t= [itex]5\pi/6[/itex], s= [itex]\pi/4[/itex].
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