Difference between revisions of "009C Sample Final 1, Problem 9"
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!Foundations: | !Foundations: | ||
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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: | !Step 1: | ||
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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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!Step 2: | !Step 2: | ||
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
Revision as of 14:34, 4 February 2016
A curve is given in polar coordinates by
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\theta}
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\leq \theta \leq 2\pi}
Find the length of the curve.
| Foundations: |
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| The formula for the arc length Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} of a polar curve Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=f(\theta)} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha_1\leq \theta \leq \alpha_2} is |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta} . |
Solution:
| Step 1: |
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| First, we need to calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dr}{d\theta}} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\theta,~\frac{dr}{d\theta}=1} . |
| Using the formula in Foundations, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int_0^{2\pi}\sqrt{\theta^2+1}d\theta} . |
| Step 2: |
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| Now, we proceed using trig substitution. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\tan x} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d\theta=\sec^2xdx} |
| Final Answer: |
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