Arc length: Review the discussion of arc length in Sections 6.3 and 11.4. Compute the length of the following curves.

The plane curve $\left(t,t^2\right)$from $t=4\mathrm{to}t=6$

We have to compute the length of the following curves.

The plane curve $\left(t,t^2\right)$ from $t=4\mathrm{to}t=6$

Before finding the arc length we have to find the $\mathbf{r}\text{'}\left(t\right)$

$$\begin{gathered} \mathrm{r}\left(t\right)=\left(t,t^2\right) \\ \frac{d\mathrm{r}\left(t\right)}{dt}=\left(\frac{d}{dt}t,\frac{d}{dt}{t}^2\right) \\ \mathrm{r}\text{'}\left(t\right)=\left(1,2t\right) \end{gathered}$$

$$\begin{gathered} {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt={\int }_4^6||(1,2t)||dt \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt={\int }_4^6\sqrt{{\left(1\right)}^2+{\left(2t\right)}^2}dt \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt={\int }_4^6\sqrt{1+{\left(2t\right)}^2}dt \end{gathered}$$

$$\begin{gathered} \mathrm{Let}u=2t \\ \frac{du}{dt}=2 \\ \frac{1}{2}du=dt \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}{\int }_{t=4}^{t=6}\sqrt{1+u^2}du \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}{\left[\frac{1}{2}u\sqrt{1+u^2}+\frac{1}{2}{\log }_e\left|u+\sqrt{1+u^2}\right|\right]}_{t=4}^{t=6} \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}{\left[\frac{1}{2}2t\sqrt{1+2t^2}+\frac{1}{2}{\log }_e\left|2t+\sqrt{1+2t^2}\right|\right]}_{t=4}^{t=6} \end{gathered}$$

$$\begin{gathered} {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}{\left[\frac{1}{2}2t\sqrt{1+2t^2}+\frac{1}{2}{\log }_e\left|2t+\sqrt{1+2t^2}\right|\right]}_{t=4}^{t=6} \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}\left[\left(\frac{1}{2}·2·6\sqrt{1+{(2·6)}^2}+\frac{1}{2}{\log }_e\left|2·6+\sqrt{1+{(2·6)}^2}\right|\right) \\ -\left(\frac{1}{2}·2·4\sqrt{1+{(2·4)}^2}+\frac{1}{2}{\log }_e\left|2·4+\sqrt{1+{(2·4)}^2}\right|\right)\right] \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\left[\frac{1}{2}\left(6\sqrt{1+{\left(12\right)}^2}+\frac{1}{2}{\log }_e\left|12+\sqrt{1+{\left(12\right)}^2}\right|\right) \\ -\left(4\sqrt{1+{\left(8\right)}^2}+\frac{1}{2}{\log }_e\left|8+\sqrt{1+{\left(8\right)}^2}\right|\right)\right] \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}\left[\left(6\sqrt{1+144}+\frac{1}{2}{\log }_e\left|12+\sqrt{1+144}\right|\right) \\ -\left(4\sqrt{1+64}+\frac{1}{2}{\log }_e\left|8+\sqrt{1+64}\right|\right)\right] \end{gathered}$$

$$\begin{gathered} {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt=\frac{1}{2}\left[\left(6\sqrt{145}+\frac{1}{2}{\log }_e\left|12+\sqrt{145}\right|\right) \\ -\left(4\sqrt{65}+\frac{1}{2}{\log }_e\left|8+\sqrt{65}\right|\right)\right] \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt\approx \frac{1}{2}\left[\left(72.25+0.69\right)-\left(32.25+0.60\right)\right] \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt\approx \frac{1}{2}\left[72.94-32.85\right] \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt\approx \frac{1}{2}\times 40.09 \\ {\int }_4^6||\mathbf{r}\text{'}\left(t\right)||dt\approx 20.045 \end{gathered}$$