Integral curve of vector field $W = x^2 \frac{\partial}{\partial x}$

Question

Integral curve of vector field $W = x^2 \frac{\partial}{\partial x}$.

My textbook is saying the unique integral curve of the vector field given by $W = x^2 \frac{\partial}{\partial x}$ starting at $(1,0)$ is:

$\gamma(t) = (\frac{1}{1-t},0)$.

This confuses me a bit, clearly $W_{(1,0)}=(2,0)$ and yet $\gamma(0)=(1,0)$. What am I not understanding here?

Answer 1

If $\gamma$ is to be an integral curve, then at any point $p$ on $\gamma$, the value of $W$ at that point corresponds to the derivative of $\gamma$ at that point.

Specifically, for $t = 0$, when we have the point $p = \gamma(0) = (1,0)$, the derivative $\gamma'(0)$ is indeed equal to $W_p = (1^2, 0)$. The same will be true for any other value of $t$, which makes $\gamma$ the integral curve you're after.

Answer 2

This seems correct. $$ \gamma'(t) = \left( \frac{1}{(1-t)^2}, 0 \right) = (x^2,0) = x^2 \frac{\partial}{\partial x} + 0 \frac{\partial}{\partial y} = W \text{.} $$ $W$ is supposed to be the vector field of tangents to any integral curve, $\gamma$.