63–74. Arc length of polar curves Find the length of the follo...

Question

63–74. Arc length of polar curves Find the length of the following polar curves.

The complete cardioid r = 4 + 4 sin θ

Accepted Answer

Hello. In this video, we are going to be computing the total length of the polar curve RF theta equal to 5 + 5 sine theta traced once for theta within 0 to 2 pi. Now, in order to do this, we are going to use the length formula for the polar curve. That is going to be defined as the length is equal. To the integral from 0 to 2 of the square root of R squared plus the derivative of R with respect to theta squared d theta. So there are a few things that we need in order to integrate or find the length of the polar curve. Now, the first thing we want to do is want to go ahead and solve for R2d. Now, R2d is just the given curve, but quantity squared. So Rsquared is going to equal to the quantity 5 plus 5 sin of theta quantity squared. And by folding out this quantity, we will be left with 25 plus 50 sin of theta. Plus 25 sine squared of theta. So this is going to be the quantity R squared. Now, the next thing we need in the integral is the derivative of R with respect to theta squared. So the first thing we will do is we will go ahead and take the derivative with respect to theta. That is going to leave us with 5 cosine of theta. And by squaring this quantity, we get DRD theta squared is equal to 25 cosine squared of theta. Now, for the integral, we are going to take the sum of both of these quantities. So R squared plus DR D theta squared is going to equal to 25 plus 50 of sine theta. Plus 25 sin squared theta. Plus 25 cosine square theta. Notice that we have a sine square plus cosine squared quantity located in the sum. If we were to factor out 25, we will have 25 multiplied by sine squared plus cosine squared. And by using the Pythagorean identity, this is going to simplify to be 25 divided by 1, which is just 25. So, these last two terms will simplified to be 25, and the overall sum is going to be 25 plus 50 sin of theta plus 25, which further simplifies to be 50 plus 50 sin of theta. And if we were to factor out of 50, we can further simplify the sum to be 50 multiplied by 1 plus sign of theta. OK, now what we need to do for the integral is we need to take the square root of this quantity. So by taking the square root, we get the square root of R2 plus DRD theta squared, and that is going to equal to 50, multiplied by 1, plus sine of theta, all within a square root. Now, removing the 50 from the square root can be rewritten as 5 square root of 2 multiplied by the square root of 1 plus sine of theta. Now there's an identity that we can apply to the square root of one plus sine theta. The square root of 1 plus sine of theta is equivalent to the absolute value of sin of theta divided by 2, plus cosine of theta divided by 2. And in fact, this also has its own identity, where the sum within the absolute value is equivalent to the square root of 2, sine of theta divided by 2, plus theta divided by 4. And so what this means is that we can simplify the sum to be 50 or 5 square root of 2, multiplied by square root of 2, multiplied by the absolute value of sine of theta divided by 2, plus the divided by 4. And this is going to simplify to leave us with 10. Multiplied by the absolute value of sine of theta divided by 2, plus theta divided by 4. And now we can go ahead and integrate. So now if we were to set up the integral, we will have the integral, so the length is equal to the integral from 0 to 2 pi, of the coefficient of 10, which will move as a constant outside of the integral, of sine, the absolute value of sine of the divided by 2, plus pi divided by 4, D theta. Now, here we can go ahead and simplify the integral by using a substitution. If we allow U to equal to theta divided by 2, plus pi divided by 4, then DU is equal to 12 D theta, which can also be rewritten as 2DU is equal to D theta. And so by applying the substitution, we can further simplify the integral to be 20 multiplied by the integral, from pi divided by 4 to 5 pi divided by 4. A sign of you. To you Now here there are two things to note. On the interval from pi divided by 4 to i, sine is positive, and on the interval from pi to 5 divided by 4, sine is negative. And so what this means is that we can rewrite the integral to be the sum of integrals. And so that's going to be 20 multiplied by the integral, from pi divided by 4, to pi, a positive sign of you. Minus the integral from pi to 5 pi divided by 4 of negatives sign of you. Do you And if we were to integrate both of the integrals, we will be left with 20 multiplied by the quantity, 1 plus square root of 2 divided by 2, plus 1 minus the square root 2 divided by 2. And this is going to give us a final solution of 40. So the total length of the arc, the total length of the polar equation is going to be 40, and this is going to be the solution to the problem. So I hope this video helps you in understanding how to approach this type of problem, and we will go ahead and see you all in the next video.

63–74. Arc length of polar curves Find the length of the following polar curves.

The complete cardioid r = 4 + 4 sin θ

Key Concepts

Polar Coordinates and Polar Curves

Arc Length Formula for Polar Curves

Differentiation of Polar Functions

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