Let $f :\mathbb R^2 → \mathbb R^3$ be defined by $f (u, v) = (u, v, u\sin(uv))$ . Let $ g(t)$ be a path with $g(0) = (1,π)$ and $g'(0) = (1,2)$. What is the velocity of the path $f ◦ g $ at $t = 0$?
I have no idea how to start or solve this problem. I know that I need to take the derivative to find the velocity. But I don't know how to do it when path's are involved. Please help.
On
But I don't know how to do it when path's are involved.
The Chain Rule for Jacobian matrices is this: $\mathsf J[f\circ g](t)=\mathsf J[f](g(t))~\mathsf J[g](t)$
$\qquad\qquad\begin{align}\dfrac{\mathrm d [f\circ g]}{\mathrm d t}(t)&=\left.\dfrac{\partial f(u,v)}{\partial [u,v]}\right\rvert_{[u,v]^\top=g(t)}\cdot \dfrac{\mathrm d g(t)}{\mathrm d t}\\[1ex]&=\left.{\begin{bmatrix}\dfrac{\partial f_1(u,v)}{\partial u}&\dfrac{\partial f_1(u,v)}{\partial v}\\\dfrac{\partial f_2(u,v)}{\partial u}&\dfrac{\partial f_2(u,v)}{\partial v}\\\dfrac{\partial f_3(u,v)}{\partial u}&\dfrac{\partial f_3(u,v)}{\partial v}\end{bmatrix}\\}\right\vert_{\raise{2ex}{\begin{bmatrix}u\\v\end{bmatrix}=g(t)}} \cdot \begin{bmatrix}u'(t)\\v'(t)\end{bmatrix}\end{align}$
Let $g(t)=(x(t), y(t))$, then $g(0)=(x(0), y(0))=(1, \pi)$ and $g'(0)=(x'(0), y'(0))=(1, 2)$.
Sub. $(u,v)=(x(t), y(t))$ in to $f(u,v)$:
$$ \begin{aligned}~ [f\circ g](t) &= f(x(t),y(t))\\ &= (x,y, x\sin(xy))\\ \end{aligned} $$
Differentiate w.r.t. $t$:
$$ \begin{aligned}~ [f\circ g]'(t)&= (x',y',x'\sin(xy)+x\cos(xy)(x'y+xy'))\\ \end{aligned} $$
Sub. $t=0$: $$ \begin{aligned}~ [f\circ g]'(0)&= (1, 2, 1\sin(\pi)+1\cos(\pi)(1\cdot\pi+1\cdot2))\\ &= (1, 2, -(\pi+2)) \end{aligned} $$