Difference between a point of a curve $\gamma(t_0)$ and the vector $\gamma'(t_0)$ tangent to $\gamma$ in $t_0$

Question

Consider a curve $\gamma: \mathbb{R} \rightarrow \mathbb{R^n}$.

I don't understand why if $\gamma(t_0)$ is a point of the curve then $\gamma'(t_0)$ must be a vector (precisely the tangent vector to the curve in the point $t_0$).

I know that the two ideas of point and vector are completely different but then their representation becomes similar in many situations.

Nevertheless in this case I'm afraid I am confused about this difference. What exactly makes $\gamma(t_0)$ different from $\gamma'(t_0)$?

Answer 1

I think your resource's point is that $\gamma(t_0)$ represents a point in $\mathbb{R}^n$ while $\gamma'(t_0)$ is a vector representing the direction and speed of the curve.

For example, if $t_0$ represents time, then maybe $\gamma(t_0)$ represents the position of an object in $\mathbb{R}^n$ at time $t_0$. However, $\gamma'(t_0)$ is the instantaneous direction in which $t_0$ is going. $\gamma'(t_0)$ is still represented by an element in $\mathbb{R}^n$, but that element represents a vector, or the direction in which the curve is traveling at $t_0$.

For example, if $\gamma(t_0)=(1, 2)$ and $\gamma'(t_0)=(2, 4)$, then we know that $t_0$ is in Quadrant I in $\mathbb{R}^2$ at $t_0$ and we know that the curve is going to keep going further into Quadrant I because both the x and y coordinates of $\gamma'(t_0)$ are positive. We also have an idea of how fast $\gamma(t_0)$ is traveling in the direction its traveling. If $\gamma(t_0)$ stopped on its regular path and simply kept going according to its speed at time $t_0$, which is $\gamma'(t_0)$, then after $n$ units of time, the curve would be at $(1+2n, 2+4n)$ because $\gamma'(t_0)=(2, 4)$. This is different from if $\gamma'(t_0)=(1, 2)$: Although $(1, 2)$ and $(2, 4)$ are vectors pointing in the same direction, they have different magnitudes and the latter tells us that the curve is traveling faster in that direction that if it had the former derivative.

Thus, $\gamma(t_0)$ represents a point in $\mathbb{R}^n$ while $\gamma'(t_0)$ represents a vector that tells what direction the curve is traveling and how fast it's traveling in that direction. That's why the former is a point and the latter is a vector.

Answer 2

Consider this situation: you're driving a car through some region (say, $\Bbb R^2$) now, $\gamma(t)$ will denote your position at time $t$, while $\gamma'(t)$ will where you are going and how fast: that's your speed, the direction of this vector is where your steering wheel is pointed to, and the length will be what's on the speedometer.

If $\gamma'(t)$ wasn't a vector, we'd lose some information: From what I've said, it's easy to see that if I give you only the initial position, but you know all the values of $\gamma'(t)$, you can reconstruct the whole $\gamma(t)$ trajectory (just go to the initial position and steer the wheel, and press the accelerator as much as $\gamma'(t)$ dictates)!

I hope this gives you some intuition.