Question

Find the arc lengths of the curves defined by the parametric equations on the specified intervals.

$x=3t+1$,$y=-2t+3$,$t\in \left[0,1\right]$

Short answer

The arc length is$\sqrt{13}$units.

Step-by-step solution

Step 1. Given information.

The given parametric equation is$x=3t+1$,$y=-2t+3$.

Step 2. Find the derivative of the parametric equations.

$\begin{array}{rcl}x& =& 3t+1\\ f\text{'}\left(t\right)& =& 3\end{array}$ and $\begin{array}{rcl}y& =& -2t+3\\ g\text{'}\left(t\right)& =& -2\end{array}$

Step 3. Substitute the value of f't and g'(t) in arc length formula.

The arc length formula is:

${\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt$, where $\left[a,b\right]=\left[0,1\right]$.

${\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt={\int }_0^1\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}dt$

Step 4. Find the arc length.

The required length of the curve is:

$\begin{array}{rcl}{\int }_0^1\sqrt{{\left(3\right)}^2+{\left(-2\right)}^2}dt& =& {\int }_0^1\sqrt{13}dt\\ & =& \sqrt{13}{\int }_0^{1}dt\\ & =& \sqrt{13}{\left[t\right]}_0^1\\ & =& \sqrt{13}\left[1-0\right]\\ & =& \sqrt{13}\end{array}$

Step 5. Simplified answer.

Hence, the required arc length of the curve is$\sqrt{13}$units.