# How do you find the distance travelled from t=0 to t=pi by an object whose motion is x=3cos2t, y=3sin2t?

Oct 5, 2017

$6 \pi$

#### Explanation:

We seek the the distance travelled from $t = 0$ to $t = \pi$ by an object whose motion is $x = 3 \cos 2 t , y = 3 \sin 2 t$

Short Solution

We note that the parametric equations are those of a circle of radius $3$ centred on the origin and a full circle is transcribed in the interval t in $\left[0 , \pi\right]$,

Thus the distance travelled is the circumference of said circle:

$L = \left(2\right) \left(\pi\right) \left(3\right) = 6 \pi$

Long Solution

We can calculate the parametric arc length using,

$L = {\int}_{\alpha}^{\beta} \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{0}^{\pi} \sqrt{{\left(- 6 \sin 2 t\right)}^{2} + {\left(6 \cos 2 t y\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{0}^{\pi} \sqrt{{6}^{2} \left({\sin}^{2} t + {\cos}^{2} t\right)} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{0}^{\pi} \sqrt{{6}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{0}^{\pi} 6 \mathrm{dt}$
$\setminus \setminus = {\left[6 t\right]}_{0}^{\pi}$
$\setminus \setminus = 6 \pi$