Question

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

$x=\cos ht$, $y=t$, $t\in \left[0,1\right]$

Short answer

The arc length is$\begin{array}{rcl}& & \frac{1}{2}\left\{e-\frac{1}{e}\right\}\end{array}$.

Step-by-step solution

Step 1. Given information.

The given parametric equations are $x=\cos ht$ and$y=t$.

Step 2. Find the derivative of the parametric equations.

$\begin{array}{rcl}x& =& \cos ht\\ f\text{'}\left(t\right)& =& \sin ht\end{array}$ and $\begin{array}{rcl}y& =& t\\ g\text{'}\left(t\right)& =& 1\end{array}$

Step 3. Find the length of the curve.

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

$\begin{array}{rcl}{\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt& =& {\int }_0^1\sqrt{{\sin }^{2}ht+1}dt\\ & =& {\int }_0^1\sqrt{{\cos }^{2}ht}dt\cdots \cdots \cdots \cdots \cdots \cdots \left\{{\cos }^{2}ht-{\sin }^{2}ht=1\right\}\\ & =& {\int }_0^1\frac{e^t+e^{-t}}{2}dt\cdots \cdots \cdots \cdots \left\{\cos ht=\frac{e^t+e^{-t}}{2}\right\}\\ & =& \frac{1}{2}{\int }_0^1{e}^{t}dt+\frac{1}{2}{\int }_0^1{e}^{-t}dt\\ & =& \frac{1}{2}{\left[e^t\right]}_0^1+\frac{1}{2}{\left[-e^{-t}\right]}_0^1\\ & =& \frac{1}{2}\left[e^1-e^0\right]+\frac{1}{2}\left[-e^{-1}+e^0\right]\\ & =& \frac{1}{2}\left\{e-1-e^{-1}+1\right\}\\ & =& \frac{1}{2}\left\{e-\frac{1}{e}\right\}\end{array}$

Step 4. Simplified answer.

Hence, the required arc length of the curve is$\begin{array}{rcl}& & \frac{1}{2}\left\{e-\frac{1}{e}\right\}\end{array}$.