How do you find the exact length of the polar curve $r = e^{θ}$ $r=e^theta$ ?

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How do you find the exact length of the polar curve $r = e^{θ}$ $r=e^theta$ ?

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Answer 1 · 2014-09-07T18:33:52

If $θ$ $theta$ goes from $θ_{1}$ $theta_1$ to $θ_{2}$ $theta_2$ , then the arc length is $\sqrt{2} (e_{2}^{θ_{2}} - e_{1}^{θ_{1}})$ $sqrt{2}(e^{theta_2}-e^{theta_1})$ .

Let us look at some details.
$L = \int_{θ_{1}}^{θ_{2}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $L=int_{theta_1}^{theta_2}sqrt{r^2+({dr}/{d theta})^2}d theta$
since $r = e^{θ}$ $r=e^{theta}$ and $\frac{d r}{d θ} = e^{θ}$ ${dr}/{d theta}=e^{theta}$ ,
$= \int_{θ_{1}}^{θ_{2}} \sqrt{{(e^{θ})}^{2} + {(e^{θ})}^{2}} d θ$ $=int_{theta_1}^{theta_2}sqrt{(e^{theta})^2+(e^{theta})^2}d theta$
by pulling $e^{θ}$ $e^{theta}$ out of the square-root,
$= \int_{θ_{1}}^{θ_{2}} e^{θ} \sqrt{2} d θ = \sqrt{2} \int_{θ_{1}}^{θ_{2}} e^{θ} d θ$ $=int_{theta_1}^{theta_2}e^{theta}sqrt{2} d theta=sqrt{2}int_{theta_1}^{theta_2}e^{theta} d theta$
by evaluating the integral,
$= \sqrt{2} {[e^{θ}]}_{θ_{1}}^{θ} = \sqrt{2} (e_{2}^{θ_{2}} - e_{1}^{θ_{1}})$ $=sqrt{2}[e^{theta}]_{theta_1}^{theta_2}=sqrt{2}(e^{theta_2}-e^{theta_1})$

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How do you find the exact length of the polar curve $r = e^{θ}$ $r=e^theta$ ?

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