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anonymous

  • 5 years ago

Calculus: Find the equations of the tangent line and normal plane to the curve R(t) = t^2i + t^3j + tk at t=1

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  1. anonymous
    • 5 years ago
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    R(1) = 1i +1j +1k; so it is comprised of unit vectors

  2. anonymous
    • 5 years ago
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    R' = 2ti + 3t^2j +k

  3. anonymous
    • 5 years ago
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    the slope of at t=1 should be parallel to the vector <2,3,1> i think

  4. anonymous
    • 5 years ago
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    The normal plane would include the vector perpendicular to that vector if I remember right

  5. anonymous
    • 5 years ago
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    dR/dt=2ti+3t^2j+k

  6. anonymous
    • 5 years ago
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    what's that equation for?

  7. anonymous
    • 5 years ago
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    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?

  8. anonymous
    • 5 years ago
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    R' is normal to the tangent plane.

  9. anonymous
    • 5 years ago
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    normal as in perpendicular?

  10. anonymous
    • 5 years ago
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    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\[2x+3y+z=6\]

  11. anonymous
    • 5 years ago
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    Normal plane, I meant.

  12. anonymous
    • 5 years ago
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    The velocity vector is tangent to the curve, so it's perpendicular to the normal plane. That's what I mean to say above.

  13. anonymous
    • 5 years ago
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    Thanks so much, what would be the tangent line then?

  14. anonymous
    • 5 years ago
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    tangent line is (1,1,1)+t(2,3,1)

  15. anonymous
    • 5 years ago
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    Anchor point + tangent vector x parameter...like uzma's done.

  16. anonymous
    • 5 years ago
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    there are many tangent lines to the curve at that point. they make up the plane :)

  17. anonymous
    • 5 years ago
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    but here normal plane is required

  18. anonymous
    • 5 years ago
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    like this right?

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