I am trying to solve this problem:
Show that the curve $t \to (t, t^2, t^3)$ embeds $\mathbb{R}^1$ into $\mathbb{R}^3$. Find two independent functions that globally define the image. Are your functions independent on all of $\mathbb{R}^3$, or just on an open neighborhood of the image?
I have problem when I try to show independence.
The two functions I am considering are Now we consider the functions $g_1, g_2: \mathbb{R}^3 \to \mathbb{R}$. \begin{eqnarray*} g_1(x_1, x_2,x_3) & = & x_1^2 - x_2\\ g_2(x_1, x_2,x_3) & = & x_1^3 - x_3\\ \end{eqnarray*}
Then how can I show that the gradient $\nabla g_1 = (2x_1, -1, 0)$ and $\nabla g_2 = (3x_1^2, 0, -1)$ are linearly independent?
Thanks,
The vector fields $\nabla g_1 = (2x_1, -1, 0)$ and $\nabla g_2 = (3x_1^2, 0, -1)$ are clearly linearly independent at each point $(x_1, x_2, x_3)$. If not, there exist real $a, b \ne 0$ such that $a(2x_1, -1, 0) + b(3x_1^2, 0, -1) = 0$; but $a(2x_1, -1, 0) + b(3x_1^2, 0, -1) = (2ax_1 + 3bx_1^2, -a, -b)$. If $(2ax_1 + 3bx_1^2, -a, -b) = 0$, then we must have $a = b = 0$, a conclusion which is incidentally independent of the value of the coordiante $x_1$ or even of the point $(x_1, x_2, x_3)$! $\nabla g_1$ and $\nabla g_2$ are linearly independent both as vector fields on $R^3$ and as simple vectors at each individual point. Hope this clarifies. Cheers.