Question

Given a differentiable vector-valued function $r\left(t\right),$explain why $r\text{'}\left(t_0\right)$ is tangent to the curve defined by $r\left(t\right)$ when the initial point of $r\text{'}\left(t_0\right)$ is placed at the terminal point of$r\text{'}\left(t_0\right)$

Short answer

$r\text{'}\left(t_0\right)$is a tangent to the curve defined by $r\left(t\right)$when the initial point of $r\text{'}\left(t_0\right)$is placed at the terminal of $r\left(t_0\right)$

Step-by-step solution

Step 1. Given information

The given a differentiable vector - valued function $r\left(t\right)$

Step 2. The objective is to explain why r'(t0) is a tangent to the curve defined by r(t) when the initial point of r'(t0) is placed at the terminal point of r(t) .

For this the definition of the derivative of the vector - valued function should be considered first:
The Derivative of a vector - valued function:
Let$r\left(t\right)$be a differentiable vector function. Then the derivative of$r\left(t\right)$is

$$\begin{gathered} r\text{'}\left(t\right)=\underset{h\to 0}{\lim }\frac{r(t+h)-r\left(t\right)}{h} \\ r\text{'}\left(t\right)=\underset{h\to 0}{\lim }\frac{r(t_0+h)-r\left(t\right)}{h} \end{gathered}$$

Derivative of$r\left(t\right)$consists of$r(t+h)-r\left(t\right)$in the numerator, the difference of vector point from the terminal point of$r\left(t\right)$. If$r\text{'}\left(t_0\right)$starts from the terminal point of$r\left(t\right)$it becomes tangent to the curve.
Thus $r\text{'}\left(t_0\right)$is a tangent to the curve defined by $r\left(t\right)$when the initial point of $r\text{'}\left(t_0\right)$is placed at the terminal of$r\left(t_0\right)$