Difference between revisions of "009C Sample Final 1, Problem 9"

Revision as of 15:34, 4 February 2016

A curve is given in polar coordinates by

r=\theta

0\leq \theta \leq 2\pi

Find the length of the curve.

Foundations:
The formula for the arc length $L$ of a polar curve $r=f(\theta )$ with $\alpha _{1}\leq \theta \leq \alpha _{2}$ is
$L=\int _{\alpha _{1}}^{\alpha _{2}}{\sqrt {r^{2}+{\bigg (}{\frac {dr}{d\theta }}{\bigg )}^{2}}}d\theta$ .

Solution:

Step 1:
First, we need to calculate ${\frac {dr}{d\theta }}$ . Since $r=\theta ,~{\frac {dr}{d\theta }}=1$ .
Using the formula in Foundations, we have
$L=\int _{0}^{2\pi }{\sqrt {\theta ^{2}+1}}d\theta$ .

Step 2:
Now, we proceed using trig substitution. Let $\theta =\tan x$ . Then, $d\theta =\sec ^{2}xdx$

Final Answer:

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 !Foundations: &nbsp;
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+|The formula for the arc length <math>L</math> of a polar curve <math>r=f(\theta)</math> with <math>\alpha_1\leq \theta \leq \alpha_2</math> is
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+|<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta</math>.
 |}
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 !Step 1: &nbsp;
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+|First, we need to calculate <math>\frac{dr}{d\theta}</math>. Since <math>r=\theta,~\frac{dr}{d\theta}=1</math>.
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+|Using the formula in Foundations, we have
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+|<math>L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta</math>.
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 |}
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 !Step 2: &nbsp;
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+|Now, we proceed using trig substitution. Let <math>\theta=\tan x</math>. Then, <math>d\theta=\sec^2xdx</math>
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