Question

In Exercises 28–30 show that the given vector-valued functions are not arc length parametrizations. Then find an arc length parametrization for the curves defined by those functions.

$\mathbf{r}\left(t\right)=(3\cos t,3\sin t)$

Short answer

The arc length parametrization of the given curve is

Step-by-step solution

Step 1. Given information.

The given function is$\mathbf{r}\left(t\right)=⟨3\cos t,3\sin t⟩$.

Step 2. Required answer.

Now,

$$\begin{gathered} ||r^{\text{'}}\left(t\right)||=\sqrt{(-3\sin t{)}^2+(3\cos t{)}^2} \\ =\sqrt{9\left({\sin }^{2}t+{\cos }^{2}t\right)}=3 \\ Consider{\left.{\int }_c^d||r^{\text{'}}\left(t\right)||dt={\int }_c^{d}3dt=3t\right]}_c^d \\ =3(d-c)\ne d-c \end{gathered}$$

Hence the given vector-valued functions are not arc length parameterizations on the interval $[0,1]$the arc length of the curve is $3(1-0)=3$