What is the arclength of the polar curve $f (θ) = sin 4 θ - cos 3 θ$ $f(theta) = sin4theta-cos3theta$ over $θ \in [\frac{π}{3}, \frac{π}{2}]$ $theta in [pi/3,pi/2]$ ?

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What is the arclength of the polar curve $f (θ) = sin 4 θ - cos 3 θ$ $f(theta) = sin4theta-cos3theta$ over $θ \in [\frac{π}{3}, \frac{π}{2}]$ $theta in [pi/3,pi/2]$ ?

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Answer 1 · 2017-03-20T00:36:30

The arclength is approximately $0.46371$ $0.46371$

Explanation:

The equation for arclength of a polar curve is:

$\int_{a}^{b} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $int_a^b sqrt(r^2 + ((dr)/(d theta))^2) d theta$

In this case,

$r^{2} = {(sin 4 θ - cos 3 θ)}^{2}$ $r^2 = (sin4theta-cos3theta)^2$

$\frac{d r}{d θ} = \frac{d}{d θ} (sin 4 θ - cos 3 θ)$ $(dr)/(d theta) = d/(d theta) (sin4theta-cos3theta)$

${(\frac{d r}{d θ})}^{2} = {(4 cos 4 θ + 3 sin 3 θ)}^{2}$ $((dr)/(d theta))^2 = (4cos4theta + 3sin3theta)^2$

So, the arclength can be found by:

$\int_{\frac{π}{3}}^{\frac{π}{2}} \sqrt{{(sin (4 θ) - cos (3 θ))}^{2} + {(4 cos 4 θ + 3 sin 3 θ)}^{2}} d θ$ $int_(pi/3)^(pi/2) sqrt( (sin(4theta)-cos(3theta))^2 + (4cos4theta + 3sin3theta)^2) d theta$

$\approx 0.46371$ $~~ 0.46371$

Final Answer

I am not entirely sure how to solve this integral manually; if anyone else has a valid method of integration, feel free to extend this answer, but as far as I can see, the only way to solve this problem is with a graphing calculator or similar interface.

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What is the arclength of the polar curve $f (θ) = sin 4 θ - cos 3 θ$ $f(theta) = sin4theta-cos3theta$ over $θ \in [\frac{π}{3}, \frac{π}{2}]$ $theta in [pi/3,pi/2]$ ?

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Explanation:

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Explanation:

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What is the arclength of the polar curve $f (θ) = sin 4 θ - cos 3 θ$ $f(theta) = sin4theta-cos3theta$ over $θ \in [\frac{π}{3}, \frac{π}{2}]$ $theta in [pi/3,pi/2]$ ?