Find the length of C which has parametric equations $ x=9 \cos (t)-2, \quad y=9 \sin (t)+1, \quad-\frac{\pi}{6} \leqslant t \leqslant \frac{\pi}{2} $

Question

I solved it by first finding the Cartesian equation of $C$ and its domain & range. $$ (x+2)^{2}+(y-1)^{2}=81,\;\;-2 \leq x \leq 7,\;\;-\frac{7}{2} \leq y \leq 10 $$

After sketching the curve we can see that the angle subtended by an arc $C$ at the centre is equal to $ \frac{\pi}{2}+\theta $.

I then found the distance between the two marked points on the circle, and used the cosine law to find the angle $\theta$. Of course, $$ \begin{aligned} \text {arc length C } &=\theta r \\ &=\left(\frac{\pi}{2}+\frac{\pi}{6}\right) \times 9 \\ &=6 \pi \end{aligned} $$

The answer booklet seems to suggest there is a much faster way of finding $\theta.$
If this is the case, could someone enlighten me please?

Answer 1

The parametric system defines a circle with radius $r=9.$ The parametrization is defined by the angle with vertex at the center of the circle, therefore the coordinates of the center do not influence the computation.

The interval $-\frac{\pi}{6}\leq t \leq \frac{\pi}{2}$ has a length $$l=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3},$$ which corresponds to one third of a circle. Therefore, the length of $C$ is $$l(C)=r\times l =6\pi.$$