Composition of (vector) functions

[Ted123]

Homework Statement

[PLAIN]http://img230.imageshack.us/img230/4203/vectoro.jpg

Homework Equations

The Attempt at a Solution

I know I need to find (f\circ p)'(0) which is 2-dimensional vector and then show it equals \alpha (a,1) where the number \alpha depends on v_1 and v_2 but a is a number independent of v_1 and v_2 . But how do I find (f\circ p)'(t) and hence (f\circ p)'(0) ?

 

[micromass]

I would start by calculating what (f\circ p)(t) is. This makes it easier to derive...

 

[Ted123]

I would start by calculating what (f\circ p)(t) is. This makes it easier to derive...

OK, does this check out with you?

Letting

p(t) = \begin{bmatrix} 1+v_1 t \\ 1 + v_2 t \end{bmatrix} = \begin{bmatrix} p_1(t) \\ p_2(t) \end{bmatrix}

f(x,y) = \begin{bmatrix} 5x^2 + 2xy + 2y^2 \\ 2x^2 + y^2 \end{bmatrix}

(f\circ p)(t) = f(p(t)) = f(p_1(t), p_2(t))

(f\circ p)(t) = \begin{bmatrix} 9 + 12v_1 t + 6v_2 t 2v_1 v_2 t^2 + 5v_1^2 t^2 + 2v_2^2 t^2 \\ 3 + 4v_1 t + 2v_2 t + 2v_1^2 t^2 + v_2^2 t^2 \end{bmatrix}

and

(f\circ p)'(t) = \begin{bmatrix} 12v_1 + 6v_2 + 4v_1 v_2 t + 10v_1^2 t + 4v_2^2 t \\ 4v_1 + 2v_2 + 4v_1^2 t + 2v_2^2 t \end{bmatrix}

so

(f\circ p)'(0) = \begin{bmatrix} 12v_1 + 6v_2 \\ 4v_1 + 2v_2 \end{bmatrix} = (4v_1 + 2v_2 ) \begin{bmatrix} 3 \\ 1 \end{bmatrix}

ie. \begin{bmatrix} 12v_1 + 6v_2 \\ 4v_1 + 2v_2 \end{bmatrix} \propto \begin{bmatrix} a \\ 1 \end{bmatrix}

\forall\;v_1 , v_2 where a=3

 

[micromass]

seems fine!