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

$x=e^t\cos t,y=e^t\sin t,t\in [0,1]$

The parametric curve$x=e^t\cos t,y=e^t\sin t,t\in [0,1]$

Consider the parametric equations $x=e^{\text{'}}\cos t,y=e^{\text{'}}\sin t,t\in [0,1]$.

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 $x=e^{\text{'}}\cos t,y=e^{\text{'}}\sin t$.

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That is $f\left(t\right)=e^t\cos t$

The derivative with respect to t is written as follows,

On simplifying.

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

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

The derivative with respect to t is written as follows,

Thus,

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

On further simplification,

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

Thus,

Length of the curve $={\sqrt{2}}^1{\int }^{\text{'}}dt$

By substituting the limits,

Length of the curve $=\sqrt{2}\left[e^3-e^0\right]$

Length of the curve $=\sqrt{2}[e-1]=2.43$

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

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

When $t=0$.

When t=1.

Then,

The graphical representation of the points $(1,0)(2.71,0.01)$ is as follows,

Therefore, the length of the curve is equals to $\sqrt{2}[e-1]or2.43.$