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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
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R(1) = 1i +1j +1k; so it is comprised of unit vectors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the slope of at t=1 should be parallel to the vector <2,3,1> i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The normal plane would include the vector perpendicular to that vector if I remember right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what's that equation for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0R' 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
 5 years ago
Best ResponseYou've already chosen the best response.0R' is normal to the tangent plane.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0normal as in perpendicular?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0Normal plane, I meant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The velocity vector is tangent to the curve, so it's perpendicular to the normal plane. That's what I mean to say above.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks so much, what would be the tangent line then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tangent line is (1,1,1)+t(2,3,1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Anchor point + tangent vector x parameter...like uzma's done.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are many tangent lines to the curve at that point. they make up the plane :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but here normal plane is required
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