anonymous
  • anonymous
Calculus: Find the equations of the tangent line and normal plane to the curve R(t) = t^2i + t^3j + tk at t=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
R(1) = 1i +1j +1k; so it is comprised of unit vectors
anonymous
  • anonymous
R' = 2ti + 3t^2j +k
anonymous
  • anonymous
the slope of at t=1 should be parallel to the vector <2,3,1> i think

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anonymous
  • anonymous
The normal plane would include the vector perpendicular to that vector if I remember right
anonymous
  • anonymous
dR/dt=2ti+3t^2j+k
anonymous
  • anonymous
what's that equation for?
anonymous
  • anonymous
R' is the slope of the tangemt to the curve; to get the equation of it you have to include it as the slope of the line right?
anonymous
  • anonymous
R' is normal to the tangent plane.
anonymous
  • anonymous
normal as in perpendicular?
anonymous
  • anonymous
So the form of the tangent plane will be\[2x+3y+z=d\]and since the point (1,1,1) lies in it (where t=1), you have d=6, so\[2x+3y+z=6\]
anonymous
  • anonymous
Normal plane, I meant.
anonymous
  • anonymous
The velocity vector is tangent to the curve, so it's perpendicular to the normal plane. That's what I mean to say above.
anonymous
  • anonymous
Thanks so much, what would be the tangent line then?
anonymous
  • anonymous
tangent line is (1,1,1)+t(2,3,1)
anonymous
  • anonymous
Anchor point + tangent vector x parameter...like uzma's done.
anonymous
  • anonymous
there are many tangent lines to the curve at that point. they make up the plane :)
anonymous
  • anonymous
but here normal plane is required
anonymous
  • anonymous
like this right?
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