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Vectors

Find the eqn of normal plane and osculating plane to the curve x=3t-t^3, y=3t^2 , z=3t+t^2 at t=1

Profile image of prakash p
16 Years agoGrade
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1 Answer

Profile image of Vikas TU
9 Years ago
put t^2 = y/3 in z eqn.
that is:
z = 3t + y/3
and 
Put 
t = (z – y/3 )/3
in x = 3t – t^3.
to get the paramertic relation in xyz.