A particle is moving along a curve C in 3-dimcnsional space with position vector

Question

A particle is moving along a curve C in 3-dimcnsional space with position vector r(t) = t, f(t), g(t) , t ge 0, for some differentiable functions f(t) and g(t). It is indicated by some experimental data that the initial position of the particle is at the origin 0, 0, 0 ; the velocity vector of the particle is orthogonal to the vector (2,1,0) for any t ge 0; and the particle is moving on the surface z = 2x2 + y2 (this surface is an elliptic paraboloid). That is, the entire curve C is on this surface. Find the functions f(t) and g(t). Find the tangential and normal components of the acceleration for any t ge 0. Find the curvature K(t) of r(f) at any t ge 0.

Explanation / Answer

DROPPING (T) AS UNDERSTOOD.

R=[T,F,G].............................................1

R[0]=[0,F(0) , G(0) ] =[0,0,0]

SO WE GET

F[0]=0................................................2

G[0]=0..............................3

V = R' = [1,F',G']

THIS IS ORTHOGONAL TO [2,1,0] ....HENCE

[1,F',G'] . [2,1,0] = 1* 2+F ' * 1+G ' * 0=0

F ' = - 2

INTEGRATING

F = - 2 T + K

BUT FROM 2

F[0]=-2*0+K=0

K=0

SO WE GET

F = - 2 T

PUTTING IN 1 WE GET

R = [ T , - 2 T , G ]

THIS LIES ON

Z = 2 X^2 + Y^2

SO WE GET

G = 2[ T^2] + [ - 2 T ] ^ 2 = 6 T^2

SO WE GET

F[T]= - 2 T .....AND .....G[T] = 6 T^2 ........ANSWER

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