Find the lengths of the curves. The curve r=a sin^2(θ/ 2), 0 ≤θ≤π, a>0

Name: Problem 24, Chapter 10: University Calculus: Early Transcendentals
Uploaded: 2020-04-19T20:42:55-08:00
Duration: 9 min 3 s
Channel: Carlos Pinilla
Description: Find the lengths of the curves. The curve r=a sin^2(θ/ 2), 0 ≤θ≤π, a>0

Find the lengths of the curves. The curve r=a sin^2(θ/ 2),...

Key Concepts

Polar Coordinates

Polar coordinates offer a way to represent points in the plane by specifying a radius and an angle. This coordinate system is particularly useful for describing curves that have rotational symmetry or are better expressed in terms of an angle rather than Cartesian coordinates.

Arc Length in Polar Coordinates

The arc length of a curve defined in polar coordinates is determined by the integral L = ?[?? to ??] ?(r² + (dr/d?)²) d?. This formula is crucial as it combines the contributions from both the radial distance and its change with respect to the angle, providing a method to measure the actual distance along the curve.

Integration Techniques

Solving the arc length integral often requires the application of integration techniques such as substitution and the use of trigonometric identities. These methods help simplify the integrand, making it possible to evaluate the integral and obtain the length of the curve.

Transcript

00:01 So we want to find what is the length? length of the curve given by r is equal to a squared sine of sine squared of theta halves is not a score just a times sine squared of theta halves for going from zero up to and we need some a positive.

00:42 So the length of the curve is given by the integral from 0 to pi of the square root of r squared plus the r d theta squared.

01:04 So for this r d r d theta we'll be given by a by the channel we get 2, sine of theta squared on the derivative of times the derivative of sine of theta square which is cosine of theta squared and then by the inner derivative one half this part is the by the triangle that's derivative of sine of theta and so we'll go something nice happens here and the two cancel there and the r d theta is that so that completing this we're gonna have r squared plus the r d theta squared and the squared and the squared of that would be equal to so a sine of sine of s squared of theta halves squared that is this first part and then the derivative is equal to a times sine of therohps times cosine of therhavs so then it would be plus a times sine of therah halves times cosine of therahabes that squared so that we take in a square root of that we can factor the a squared and here we will get a sign to the fourth times sine square times cosine square so that we can factor a sign square in there because this this this will give us sign to the fourth and this will be sine a square times cosine plus times cosine square so that we can factor that as sine squared count to both times sine squared x squared third a half and what is left there is cosine squared of theta halves so we'll be doing square root of that so that this would be just a and then times sine of theta halves times or square root of this is just one so that's this interval from 0 up to pi it would be the intral of this from 0 up to pi so the entail deeta so that length will be equal to a times the intro from 0 up to pi of sine x theta halves d theta so here we need to be careful with one thing and is that sine, sine goes like that.

05:06 So we have a theta halves...

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