Q 11. - Calculus

Question

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

$x=\sin \left(kt\right)$, $y=\cos \left(kt\right)$, $t\in \left[0,2\mathrm{\pi }\right]$, where is $k$ is constant.

Step-by-Step Solution

Answer

The arc length is $\begin{array}{rcl}& & 2k\mathrm{\pi }\end{array}$.

Step 1. Given information.

The given parametric equations are $x=\sin \left(kt\right)$ and $y=\cos \left(kt\right)$, where $k$ is a constant.

Step 2. Find the derivative of the parametric equations.

$\begin{array}{rcl}x& =& \sin \left(kt\right)\\ f\text{'}\left(t\right)& =& k \cos \left(kt\right)\end{array}$       and        $\begin{array}{rcl}y& =& \cos \left(kt\right)\\ g\text{'}\left(t\right)& =& -k \sin \left(kt\right)\end{array}$

Step 3. Substitute the value of f ' ( t ) and g ' ( t ) in the 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 $\left[a,b\right]=\left[0,2\mathrm{\pi }\right]$.

$\begin{array}{rcl}{\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt& =& {\int }_0^{2\mathrm{\pi }}\sqrt{{\left(k \cos \left(kt\right)\right)}^2+{\left(-k \sin \left(kt\right)\right)}^2}dt\\ & =& {\int }_0^{2\mathrm{\pi }}\sqrt{k^2{\cos }^{2}kt+k^2{\sin }^{2}kt}dt\\ & =& k{\int }_0^{2\mathrm{\pi }}\sqrt{{\cos }^{2}kt+{\sin }^{2}kt}dt\\ & =& k{\int }_0^{2\mathrm{\pi }}\sqrt{1}dt\\ & =& k{\left[t\right]}_0^{2\mathrm{\pi }}\\ & =& 2k\mathrm{\pi }\end{array}$

Step 4. Simplified answer.

Hence, the required arc length is $\begin{array}{rcl}& & 2k\mathrm{\pi }\end{array}$.