Q. 53 - Calculus

Question

In Exercises 49-53 sketch the parametric curve and find its length.

$x=5+2t,y=e^t+e^{-t},t\in [0,1]$

Step-by-Step Solution

Answer

The graphical representation of the points $(5,2)(7,3,07)$ is as follows,

The length of the curve is equal to $\left(\left(e-\frac{1}{e}\right)\right.$

Step 1: Given information

The parametric curve  is $x=5+2t,y=e^t+e^{-t},t\in [0,1]$

Step 2: Calculation

Consider the parametric equations $x=5+2t,y=e^{\text{'}}+e^{-t},t\in [0,1]$.

The objective is to draw the parametric curve and find the arc length of the curve.

The formula to find the arc length of the curve is,

Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$

First, find the derivative of the parametric equations $x=5+2t,y=e^{\text{'}}+e^{-t}$.

Take$x=5+2 t$

That is $f\left(t\right)=5+2 t$

The derivative with respect to t is written as follows,

$f^{\text{'}}\left(t\right)=5+2t$

On simplifying.

$$\begin{gathered} f^{\text{'}}\left(t\right)=\frac{d}{dt}(5+2t)[sinced(uv)=udv+vdu] \\ {f}^{\text{'}}\left(t\right)=\frac{d}{dt}5+\frac{d}{dt}2t \\ {f}^{\text{'}}\left(t\right)=0+2 \\ {f}^{\text{'}}\left(t\right)=2 \end{gathered}$$

Take $y=e^{\text{'}}+e^{-t}$

That is$g\left(t\right)=e^{\text{'}}+e^{-t}$

The derivative with respect to t is written as follows,

$$\begin{gathered} \frac{d}{dt}g\left(t\right)=\frac{d}{dt}\left(e^{\text{'}}+e^{-t}\right) \\ \frac{d}{dt}g\left(t\right)=\frac{d}{dt}{e}^{\text{'}}+\frac{d}{dt}{e}^{-t} \end{gathered}$$

Thus,

$g^{\text{'}}\left(t\right)=e^t-e^{-t}$

Now by using the Length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$.

Length of the curve $={\int }_0^1\sqrt{(2{)}^2+{\left(e^{\text{'}}-e^{-t}\right)}^2}dt$

$={\int }_0^1\sqrt{4+e^{2t}+e^{-2t}-2e^{2t}·e^{-2t}}dt$

On further simplification,

Length of the curve $=\left[e^1-e^{-1}-e^1+e^{\text{'}}\right]$

$$\begin{gathered} ={\int }_0^1\sqrt{{\left(e^{\text{'}}+e^{-t}\right)}^2}dt \\ ={\int }_0^1\left(e^{\text{'}}+e^{-t}\right)dt \end{gathered}$$

Thus,

Length of the curve$={\int }_0^1{e}^{\text{'}}dt+{\int }_0^1{e}^{-t}dt$

Length of the curve $={\left[e^{\text{'}}-e^{-1}\right]}_0^1$

By substituting the limits,

Length of the curve $=\left[e^1-e^{-1}-e^0+e^0\right]$

Then

Length of the curve $=\left[e^1-e^{-1}-e^1+e^{\text{'}}\right]$

$=\left[e^1-e^{-1}-e^1+e^{\text{'}}\right]$

$$\begin{gathered} =\left[e^1-e^{-1}\right] \\ =\left[e-\frac{1}{e}\right] \end{gathered}$$

To draw the curve for the parametric equations, first, find the points when $t\in [0,1]$.

Given that $t\in [0,1]$

When t=0,

$$\begin{gathered} (x,y)=\left(5+2t,e^{\text{'}}+e^{-t}\right) \\ (x,y)=\left(5+2·0,e^0+e^{-0}\right) \\ (x,y)=(5,2) \end{gathered}$$

When t=1,

Then

$$\begin{gathered} (x,y)=\left(5+2t,e^{\text{'}}+e^{-t}\right) \\ (x,y)=\left(5+2·1,e^{\text{'}}+e^{-1}\right) \\ (x,y)=(7,3.07) \end{gathered}$$

The graphical representation of the points $(5,2)(7,3,07)$ is as follows,

Therefore, the length of the curve is equal to $\left(\left(e-\frac{1}{e}\right)\right.$