Question

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

Answer 1

R(1) = 1i +1j +1k; so it is comprised of unit vectors

Answer 2

R' = 2ti + 3t^2j +k

Answer 3

the slope of at t=1 should be parallel to the vector <2,3,1> i think

Answer 4

The normal plane would include the vector perpendicular to that vector if I remember right

Answer 5

dR/dt=2ti+3t^2j+k

Answer 6

what's that equation for?

Answer 7

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?

Answer 8

R' is normal to the tangent plane.

Answer 9

normal as in perpendicular?

Answer 10

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\[2x+3y+z=6\]

Answer 11

Normal plane, I meant.

Answer 12

The velocity vector is tangent to the curve, so it's perpendicular to the normal plane. That's what I mean to say above.

Answer 13

Thanks so much, what would be the tangent line then?

Answer 14

tangent line is (1,1,1)+t(2,3,1)

Answer 15

Anchor point + tangent vector x parameter...like uzma's done.

Answer 16

there are many tangent lines to the curve at that point. they make up the plane :)

Answer 17

but here normal plane is required

Answer 18

like this right?