Find the length of the following polar curves. The spiral r=4 θ^2, for 0 ≤θ≤6

Name: Problem 68, Chapter 12: Calculus: Early Transcendentals
Uploaded: 2020-07-31T20:42:17-08:00
Duration: 5 min 15 s
Channel: Victoria Dollar
Description: Find the length of the following polar curves. The spiral r=4 θ^2, for 0 ≤θ≤6

Find the length of the following polar curves. The spiral...

Key Concepts

Polar Coordinates

Polar coordinates are a system for representing points in the plane using a distance and an angle rather than x and y coordinates. This system is particularly useful for curves that have radial symmetry or are defined naturally by a radius as a function of an angle, which simplifies many problems in geometry and physics.

Arc Length in Polar Coordinates

The arc length of a polar curve r(?) is found by integrating the square root of the sum of the squares of the radius and its derivative with respect to the angle. This formula, derived from the Pythagorean theorem applied to an infinitesimal segment of the curve, allows one to compute the total length of complex curves defined in polar form.

Differentiation and Integration Techniques

Calculating the arc length of a polar curve involves finding the derivative of the radius with respect to the angle and then setting up an integral that encapsulates both the radius and its derivative. Mastering differentiation and integration techniques is crucial for evaluating these integrals, especially when the integrand is complex or involves higher powers of the variable.

Transcript

00:01 Okay, so for this problem, we are asked to find the length of the arc for r equals for theta squared.

00:08 So what we need to know is that the formula for the length of the arc is the integral between a and b, square root of the function of theta squared, plus the derivative of the function squared, and then d theta.

00:28 So this here is going to be our a and b, so zero and six.

00:34 This here is our function of theta.

00:38 So the only thing we need is our derivative function, which is going to be 8 theta.

00:47 So now what we need to do is we're going to set up our equation because we have everything we need.

00:51 So the integral between 0 and 6, and then the derivative of 4 theta squared, and then we're going to square it again, plus 8 theta squared.

01:13 So now what we're going to do is simply, simplify this down the best we can.

01:18 So 4 squared is 16.

01:20 Theta squared squared again is going to be to the fourth power, plus 8 squared, which is 64, and then theta squared d theta.

01:34 Now if we notice both of these coefficients have a 16 in common.

01:41 So i'm going to go ahead and pull out that 16, and they also have a theta squared in common.

01:49 So i'm going to be left with theta squared plus 4.

01:56 So this is now going to be 4 theta.

02:06 So then the square root of theta squared plus 4 d theta.

02:12 And this is where i'm going to switch gears a little bit and i'm going to do some u substitution.

02:16 So theta squared plus 4.

02:18 So du is going to be 2 theta d theta.

02:24 So then i can change d theta to be 1 over 2 theta du.

02:31 So i can go ahead and substitute this in...

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