EXAMPLE. Calculate the length of the arc of a curve given parametrically

$$
\left\{\begin{array}{l}
x=\left(t^{2}-2\right) \sin t+2 t \cos t, \\
y=\left(2-t^{2}\right) \cos t+2 t \sin t,
\end{array} \quad 0 \leq t \leq \pi\right.
$$
## Solution.

1. Find $x_{t}^{\prime}$ and $y_{t}^{\prime}:$

$$
x_{t}^{\prime}=t^{2} \cos t, \quad y_{t}^{\prime}=t^{2} \sin t
$$

2. Calculate the differential of the arc length:

$$
d l=\sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t=\sqrt{t^{4} \cos ^{4} t+t^{4} \sin ^{4} t} d t=t^{2} d t
$$

3. Find the arc length by evaluating the definite integral (1):

$$
l=\int_{0}^{\pi} t^{2} d t=\left.\frac{t^{3}}{3}\right|_{0} ^{\pi}=\frac{\pi^{3}}{3}
$$

Answer. $l=\pi^{3} / 3$ units of length.

Conditions of the Problems. Calculate the lengths of the arcs of the given curves.

1. $\left\{\begin{array}{l}x=t-\sin t, \\ y=1-\cos t,\end{array} \quad 0 \leq t \leq 2 \pi\right.$.
2. $\left\{\begin{array}{l}x=2 \cos t, \\ y=2 \sin t,\end{array} \quad 0 \leq t \leq \pi / 3\right.$.
3. $\left\{\begin{array}{l}x=\cos ^{3} t, \\ y=\sin ^{3} t,\end{array} \quad 0 \leq t \leq \pi\right.$.
4. $\left\{\begin{array}{l}x=2 \cos t-\cos 2 t, \\ y=2 \sin t-\sin 2 t,\end{array} \quad 0 \leq t \leq 2 \pi\right.$.
5. $\quad\left\{\begin{array}{l}x=e^{t} \cos t, \\ y=e^{t} \sin t,\end{array} \quad 0 \leq t \leq 1\right.$.
6. $\quad\left\{\begin{array}{l}x=e^{t} \sin t, \\ y=e^{t} \cos t,\end{array} \quad 0 \leq t \leq \pi\right.$.
7. $\left\{\begin{array}{l}x=3 \cos t, \\ y=3 \sin t,\end{array} \quad 0 \leq t \leq \pi / 6\right.$.
8. $\left\{\begin{array}{l}x=\cos t+t \sin t, \\ y=\sin t-t \cos t,\end{array} \quad 0 \leq t \leq \pi / 3\right.$.
9. $\left\{\begin{array}{l}x=4 \cos ^{3} t, \\ y=4 \sin ^{3} t,\end{array} \quad 0 \leq t \leq \pi / 6\right.$.
10. $\left\{\begin{array}{l}x=2(t-\sin t), \\ y=2(1-\cos t),\end{array} \quad \pi \leq t \leq \pi / 2\right.$.

Answers. 1. 8. 2. $2 \pi / 3$. 3. 3. 4. 16. 5. $\sqrt{2}(e-1) .6 . \sqrt{2}\left(e^{\pi}-1\right)$. 7. $\pi / 2$. 8. $\pi^{2} / 18$. 9. 3. $10.8-4 \sqrt{2}$.

### 8.10. Calculation of Arc Lengths $\varrho=\varrho(\varphi)$

STATEMENT OF THE PROBLEM. Calculate the length of the arc of a curve given by the equation in polar coordinates

$$
\varrho=\varrho(\varphi), \quad \alpha \leq \varphi \leq \beta
$$

SOLUTION PLAN.

If a piecewise smooth curve is given by the equation in polar coordinates $\varrho=\varrho(\varphi)$, then the arc length is

$$
l=\int_{\alpha}^{\beta} \sqrt{\varrho(\varphi)^{2}+\varrho^{\prime}(\varphi)^{2}} d \varphi
$$

where $\alpha$ and $\beta$ are the values of $\varphi$ corresponding to the boundary points of the arc.

1. Find $\varrho^{\prime}(\varphi)$.
2. Calculate the differential of the arc length

$$
d l=\sqrt{\varrho(\varphi)^{2}+\varrho^{\prime}(\varphi)^{2}} d \varphi
$$

3. Find the arc length by evaluating the definite integral (1).

Write the answer, including the units.