Lengths of Curves

Find the lengths of the cur...

Question

Lengths of Curves

Find the lengths of the curves in Exercises 13–19.

x = 5 cos t − cos 5t, y = 5 sin t − sin 5t, 0 ≤ t ≤ π/2

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Find the length of the curve defined by the following parametric equations x is equal to 3 multiplied by cosine of t plus cosine of 3 t. Y is equal to 3 multiplied by sin of t minus sine of 3 of t. On the interval, 0 is less than or equal to t, and t is less than or equal to pi. OK, so it appears for this particular prompt, we're asked to take all the information that is provided to us by the prom itself, and we're ultimately asked to determine what the length of the curve is. So what is the length of the curve? So, based on our prior knowledge, we need to recall a note that in order to find the length of a parametric curve, we need to recall and use the arc length formula that states that capital L, the arc length, is equal to the integral evaluated from A to B of the square root of DX by DT to the 2. Plus DY by DT to the power of 2 DT. And we will need to differentiate both X and Y with respect to T to find DX by DT and DY by DT, and we will need to square and sum these derivatives to get DX by DT to the 2 plus DY by DT to the 2. And we will need to simplify the expression by recalling and using trigometric identities to remove the square root, and we will need to therefore integrate the resulting expression over the given interval, which is square brackets of 0 comma pi. Fantastic. So with that in mind, keeping that overview of how we're going to solve this problem in mind, our first step is we are going to find the derivatives. So when we recall and use the chain rule, we will be able to write that dx by dt will be equal to -3 multiplied by sin oft minus 3 multiplied by sin of 3t. And DY by DT will be equal to 3 multiplied by cosine of T minus 3 multiplied by cosine of 3T. Fantastic. Our second step now is we need to square and sum our derivatives. So let us square the derivatives. So we'll take DX by DT, so the power of 2 will be equal to 9 multiplied by sin to the power of 2 of T, plus 18 multiplied by sin of t multiplied by sin of 3 t. Plus 9 multiplied by sin to the power 2 of 3 T. And similarly, DY by DT 2 will be equal to 9 multiplied by cosine to the 2 of T minus 18 multiplied by cosine of T multiplied by cosine of 3T. Plus 9 multiplied by cosine to the power of 2 of 3 T. Fantastic. So now we need to sum the squares and simplify, and we need to be able to recall and recognize that when we add these values together, we will notice the following identity which states that sine to the power 2 of theta plus cosine to the power 2 of theta will be equal to 1. So using this trigometric identity. We need to go ahead and write that parentheses of dx by dt to the 2 plus parentheses of Dy by dt to the 2 will be equal to 9 multiplied by sin to the 2 oft plus cosine to the 2 oft plus 9 multiplied by parentheses of sin to the power of 2 of 3 t plus. Cosine to the power 2 of 3T. Plus 18 multiplied by sin of t multiplied by sin of drum roll, sin of. 3 T. And we need to subtract. 18 multiplied by cosine of t. Multiplied by cosine of 3 T. Fantastic, we are on a roll. So, with that in mind, we need to note that this will be equal to 9 multiplied by 1 + 9 multiplied by 1 minus 18 multiplied by cosine of t multiplied by cosine of 3T plus 18 multiplied by sin oft multiplied by sin of 3T. Fantastic, moving right along here, and we need to note that this will be equal to 18. So essentially, here, I'm going to take a moment here to pause, let's rewind for a second. So ultimately, what we need to do is we're going to be trying to simplify this expression down to its most simple form. But to do that, let us take a moment to note that this will be equal to 18 minus 18 multiplied by square brackets of cosine of 3t multiplied by cosine of t minus sin of 3t multiplied by sin of t. And we need to take a moment now to pause briefly and recall the following trigometric identity. So we need to recall and note that cosine of A plus B is equal to cosine of A multiplied by cosine of B. Minus sin of a. Multiplied by sin of b. Fantastic. So when we apply this identity, we can go ahead and write that this will be equal to 18 minus 18 multiplied by cosine. Of 3 T plus T. And when we write this expression down in its most simplified form, we will find that it's equal to 18 multiplied by a parentheses of 1 minus cosine of 4 T. Fantastic. Our 3rd step now is we need to set up the integral and apply the substitution method, so we'll need to ultimately recall and apply the substitution method. So, we will need to note that our arc length formula will become L is equal to the integral evaluated from 0 to pi of the square root of 18 multiplied by parentheses of 1 minus cosine of 4t. DT fantastic. And we need to recall and use the identity that states that, so let's make a star. So that will state that 1 minus cosine of 2 theta is equal to 2 multiplied by sin square of theta, and that will mean that we will then get 1 minus cosine of 4 T. Will be equal to 2 multiplied by sin to the power 2 of 2 t. So that will mean when we apply this identity, our interval as a result will become L is equal to the integral, evaluate from 0 to pi of the square root of 36 multiplied by sin to the power of 2 of 2T. DT Which will be equal to the integral evaluated from 0 to pi of 6 multiplied by the absolute value of sin of 2 t. DT and in order to solve for the integral we will need to recall and use the u substitution method. So you will state that u will be equal to 2 t. Then that will mean that d u by dt will be equal to 2 and that dt will be equal to 1/2 d u. So we also need to change our limits, so we need to note that when T is equal to 0, that will mean that U is equal to 0. And when T is equal to pi, that will mean that U is equal to 2 pi. Fantastic. So that will mean that L will be equal to the interval evaluated from 0 to 2 of 6 multiplied by the absolute value of sin of u multiplied by 1/2 duu, which will be equal to the integral evaluated from 0 to 2 pi 3 multiplied by the absolute value of sin of u. Do you Fantastic. So our 4th and final step is we will need to evaluate the definite integral and as we should recall and note since the sine function is positive on square brackets of 0, pi and because it is positive on this interval. And we'll make a note with a plus sign and quotation marks since it's positive on this interval. And it's going to be negative on the interval square brackets of pi, 2 pi. We need to note that as a result, the absolute value signs will make both of these areas positive, so we can recall and use the symmetry of a sine wave, so we can state that l is equal to 3 multiplied by parentheses of the integral evaluated from 0 to pi of sin of u duu plus the integral evaluated from pi to 2 pi. Of negative sign of you, do you? Which will mean that l is equal to 3 multiplied by parentheses of square brackets of negative cosine of u evaluated from 0 to pi. Plus Square brackets of cosine of u, and don't forget that this will be evaluated from pi to 2 pi. Fantastic. So that will mean that L will be equal to 3 multiplied by negative cosine of pi minus of negative cosine of 0. Plus cosine of, and I'm gonna get rid of the parentheses so we don't get confused. So it's gonna be. Once again, L is going to be equal to 3 multiplied by parentheses of negative cosine of pi minus parentheses of negative cosine of 0, plus cosine of 2 pi minus cosine of pi. And when we plug this expression into our calculator, we will find that it's simply equal to 12, and we can't forget that our units are units, and that's it, that is our final answer. So the length of the curve is simply 12 units. So going back to the top to recap what we've learned, we can safely say without a doubt that our final answer for the length of the curve will be 12 units, and that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Lengths of CurvesFind the lengths of the curves in Exercises 13–19.x = 5 cos t − cos 5t, y = 5 sin t − sin 5t, 0 ≤ t ≤ π/2

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Lengths of Curves

Find the lengths of the curves in Exercises 13–19.

x = 5 cos t − cos 5t, y = 5 sin t − sin 5t, 0 ≤ t ≤ π/2