The area of the loop of the curve r=a sin3 θ is |...

Name: Problem 16, Chapter 6: Higher Engineering Mathematics
Uploaded: 2020-06-05T22:03:10-08:00
Duration: 16 min 15 s
Channel: Chris Trentman
Description: The area of the loop of the curve r=a sin3 θ is

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system in which each point on a plane is determined by a distance from a reference point and an angle from a reference direction. This system is particularly useful when dealing with equations and curves that are naturally expressed in terms of angles and radii, such as circular or spiral curves.

Polar Area Formula

The polar area formula, given by one half the integral of the square of the radius function over an interval of angles, is used to calculate the area enclosed by a polar curve. This formula, A = 1/2? r(?)² d?, accounts for the way area is swept out as the angle increases, making it ideal for integrating regions defined in polar coordinates.

Rose Curves

Rose curves are a family of polar graphs represented by equations of the form r = a sin(n?) or r = a cos(n?). Their distinctive petal-like shapes arise due to the periodic nature of the sine and cosine functions, and the number of petals depends on the value of n. Understanding these curves involves analyzing their symmetry and the specific intervals needed to cover one or more petals.

Symmetry in Polar Curves

Symmetry in polar curves allows for simpler area calculations by exploiting the repetitive or reflective properties of the graph. By recognizing symmetric segments, one can compute the area over a fundamental segment and multiply it appropriately to obtain the total area, thereby reducing the computational effort involved in the integration process.

Transcript

00:01 We are given a curve and we are asked to find the area contained by one loop of this curve.

00:10 The curve is r equals 2 cosine of theta minus secant of theta.

00:16 In order to find the area of a loop, let's find what this loop looks like by graphing the curve.

00:25 We're just going to graph it by hand.

00:27 Alternatively, you could use a graphing calculator.

00:32 So we see that secan of theta, which is equal to 1 over cosine theta, can be arbitrarily large.

00:49 However, we also know that sickena theta, we cannot have that theta is equal to pi over two or three pi over two, and then cina theta is undefined.

01:08 So we're going to have asymptotes along the lines.

01:12 Theta equals pi over two and three pi over two, which i'll draw in green.

01:22 Just remind us that we don't have any values there.

01:29 Now to start graphing some points, if we have the theta is equal to 0, r is going to be equal to 2 minus 1 over 1, or 2 minus 1, or 1.

01:52 If theta is equal to pi over 6, we have the r is equal to 2 times root 3 over 2 or root 3 minus 1 over root 3 over 2, or 2 over 2 or 2 over root 3, or 2 over 2 over 2, or 2 over 2 over 2, or 2 2 over 2.

02:16 2 root 3 over 3.

02:19 So root 3 minus 2 thirds root 3 or 1 third root 3.

02:25 1 3 is going to be a little bit over 1 3.

02:34 And this is at theta equals pi over 6.

02:42 So we see that our graph gets smaller.

02:48 At theta equals pi over 4, we have that r is equal to 2 times root 2 over 2 or root 2 minus 1.

02:59 Over root 2 over 2 or 2 over root 2 or 2 root 2 or 2 root 2 or just root 2.

03:10 So we have root 2 minus root 2 or 0.

03:16 So we start out at 1 and go back to the origin.

03:25 This is at pi over 4.

03:28 At pi over 3 we see that r is equal to 2 times 1 half or 1 minus 1.

03:43 Over 1 half or 2, so r is going to negative 1.

03:55 So we go through the origin into the third quadrant, and you can't have that theta is equal to pi over 2.

04:12 As theta gets closer and closer to pi over 2, we have that r is going to get closer and closer to negative 1 over some very small number close to 0, but it's positive.

04:32 So it's going to get closer and closer to negative infinity.

04:37 So you see that our graph is going to go off towards infinity as theta gets closer and closer to pi over 2.

04:55 Now if theta is going to be greater than pi over 2, say theta is equal to 2 pi over 3.

05:06 We have the r is equal to 2 times 1 1⁄2, minus 1 over 1 half, which is 2.

05:19 So again, r is equal to negative 1.

05:30 So we're coming all the way back from negative infinity to 1 or 2 close to the origin...

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