3. (12 points) A particle is moving along a curve C in 3-dimensional space with

Question

3. (12 points) A particle is moving along a curve C in 3-dimensional space with position vector r(t) = <t, f(t), g(t)>, t ? 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 h0, 0, 0i;

• the velocity vector of the particle is orthogonal to the vector h2, 1, 0i for any t ? 0; and

• the particle is moving on the surface z = 2x ^2 + y^ 2 (this surface is an elliptic paraboloid). That is, the entire curve C is on this surface.

(a) Find the functions f(t) and g(t).

(b) Find the tangential and normal components of the acceleration for any t ? 0.

(c) Find the curvature ?(t) of r(t) at any t ? 0.

Explanation / Answer

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 ..

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