Centroids

Find the coordinates of the centroi...

Question

Centroids

Find the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.

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 coordinates of the centroid parentheses of x, y bar of the arc of the cycloid defined by the parametric equations x is equal to t minus sin of t and y is equal to 1 minus cosine of t. Over 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 problem we are ultimately asked to determine or to find what the coordinates are for this specific centroid. Of the arc of the centroid, which is defined by the following parametric equations and the specific interval that are provided to us by the prom itself. So, now that we know that we're ultimately trying to determine what the coordinates of the centroid is, we're trying to determine parentheses of X, Y bar. Our first step that we need to take is based on our prior knowledge, in order to find the centroid, which is X, Y bar in parentheses, we need to note that in order to find the centroid of a parametric curve. Specifically, we will need to recall and use the following formulas. Firstly, x bar will be equal to 1 divided by l, multiplied by the integral, evaluated from T1 to T2 of X of T. Multiplied by the square root of DX by DT to the 2 plus DY by DT to the power of 2. And we need to note that we will have DT at the end of our expression, and we need to note that Y bar will be equal to 1 divided by L multiplied by the integral evaluated from T1 to T2 of Y of T multiplied by the square root of. DX by DT to the 2 plus DY by DT to the power of 2 DT. And we need to note that L, specifically capital L, is the total arc length of the curve. And we need to note, based on our prior knowledge, that the process involves 3 main phases for solving for this specific type of prompt. We will first need to calculate the arc length element, which is denoted as DS, and then we will need to determine the total arc length, which is denoted as L. And then we will need to evaluate the weight, weighted, specifically, we need to evaluate the weighted integrals for the coordinates. For this specific prom, we will need to recall a note that DS will be equal to the square root of DX by DT to the power of 2, plus DY by DT to the power of 2. And we need to note that L will be equal to the integral evaluated from T1 to T2 of the square root of DX by DT to the power of 2 plus DY by DT to the power of 2. DT Fantastic. So our first step that we need to take at this point is we need to calculate the arc length element, so we need to solve for DS first. So, in order to do that, we will need to find the derivatives of X and Y with respect to T. So, solving for DX by DT, which will be equal to 1 minus cosine of T, that will mean that DY by DT will be equal to sin of T. Now, at this point, we need to simplify the term that is inside of our square root. So, once again, focusing on our square root, which will be the square root of dx by dT to the 2 plus Dy by DT to the 2. So, we're simplifying the term inside of our square root, so we need to take parentheses of DX by DT to the 2, plus parentheses of DY by DT to the 2, which will be equal to parentheses of 1 minus cosine of T to the 2, plus. Sin of T to the power of 2, which will be equal to 1 minus 2 multiplied by cosine of T plus cosine square of T plus sin square of T. And we need to note and recall the following identity that states that cosine squat plus sin squat is equal to 1. That will mean we can simplify our expression +22 minus 2 multiplied by cosine of t, which will be equal to 2 multiplied by parentheses of 1 minus cosine of t. Fantastic. We are on a roll. So with that said, we now need to recall and use the identity that states that 1 minus cosine of t will be equal to 2 multiplied by sin to the power of 2 t divided by 2. Fantastic. So using that identity, we will note that we can go ahead and write 2 multiplied by square brackets of 2 multiplied by sin to the power of 2 of T divided by 2. Will be equal to or multiplied by sin to the power of 2 of T divided by 2. Fantastic. Moving right along here, we now need to take the square root, and we need to also note that T divided by 2 is greater than or equal to 0 on our closed interval square brackets of 0, pi. So that will mean we can safely write that DS is equal to 2 multiplied by sin of T divided by 2 DT. Fantastic. So, moving right along here, our second step, step 2, is we need to determine what the total arc length L is. So, in order to do that, based on our prior knowledge, we will need to integrate the arc length from 0 to pi. So we'll take that L is equal to the integral, evaluate from 0 to pi of 2 multiplied by sin. Of T divided by 2, DT, fantastic. Moving right along here, we now need to recall and use the antiderivative, the integral of sin of KT DT is equal to -1 divided by k multiplied by cosine of KT. So when we use our antiderivative, we will be able to go ahead and write that L will be equal to square brackets of -4 multiplied by cosine of t divided by 2. Which will be evaluated from 0 to pi. Which will be equal to -4 multiplied by cosine of pi divided by 2. -4 multiplied by cosine of 0. Which when we evaluate this expression, we will find that L will be simply equal to 4. Fantastic. So, our third step now is we need to calculate the Y coordinate, which will be Y bar. So Y bar is equal to 1/4 multiplied by the integral evaluated from 0 to pi. Of parentheses of 1 minus cosine of t multiplied by square brackets of 2 multiplied by sin of t divided by 2. DT Fantastic. So now, at this point, we need to substitute, so we're going to be substituting 1 minus cosine of T is equal to 2 multiplied by sin to the power of 2 of T divided by 2 back into our expression, and when we do, we will find that Y bar is equal to. 1/4 multiplied by the integral evaluated from 0 to pi of 4 multiplied by sin to the power of 3 of t divided by 2. DT, which will be equal to the integral evaluated from 0 to pi of sin to the power of 3 of t divided by 2 dt. So now we need to rewrite sin to the power of 3 of T divided by 2. And we're rewriting once again. Let's make a note. We're rewriting sign to the power of 3. Of T divided by 2, as we're rewriting this as sin of T divided by 2 multiplied by 1 minus cosine to the power 2 of T divided by 2. And when we do that, we will find that Y bar is equal to the integral evaluated from 0 to pi of, so it's going to be 0 to pi. Of sin of t divided by 2 minus cosine to the power of 2 oft divided by 2. And we need to multiply this by sin of t divided by 2. DT, fantastic. So with that said. We could therefore simplify a little bit more and we can write that y bar will be equal to the integral evaluated from 0 to pi of sin of t divided by 2. DT Minus the integral evaluated from 0 to pi of cosine square of t divided by 2 multiplied by sin of t. Divided by 2 DT. Fantastic. So for the 2nd integral, you will need to recall and use the substitution method. So you will state that u is equal to cosine of t divided by 2. Which will mean that DU by DT will be equal to 1/2 multiplied by sin of T divided by 2, which will mean that sin of T divided by 2 DT will be equal to -2DU. Fantastic. And we need to note that when T is equal to 0, that will mean that you will be equal to cosine of 0 divided by 2, which will be equal to 1. And when T is equal to pi, that will mean that you will be equal to cosine of, once again, cosine of pi divided by 2, which will be equal to, when we plug that into our calculator, will be simply equal to 0. So that will mean We could safely write that Y bar will be equal to the integral. Evaluate from 0 to pi of sin of t divided by 2 dt plus the integral evaluated from 0 to 1 of 2 multiplied by u to the power 2DU. And we need to integrate term by term, which will mean that Y bar will be equal to square, so it's gonna be Y bar will be equal to square brackets of -2 multiplied by cosine of T divided by 2. And we need to note that this will be evaluated from 0 to pi plus square brackets of 2 multiplied by u to the power of 3 divided by 3, evaluated from 0, evaluated from 1 to 0. Fantastic. So that will mean that Y bar will be equal to 0 minus 2 + 2 multiplied by 0 to 3 divided by 3 minus 2 multiplied by 1 to 3 divided by 3, which will be simply equal to 4 divided by 3. Fantastic. Our 4th and final step, so this is step 4, is we need to calculate the X coordinate X bar, and X bar will be equal to 1 divided by 4, multiplied by the integral, evaluated from 0 to pi of T minus sin of T. Multiplied by square brackets of 2 multiplied by sin of t divided by 2. DT, which will be equal to 1/2 multiplied by the integral evaluated from 0 to pi oft multiplied by sin of t divided by 2 dt. -1/2 multiplied by the interval evaluated from 0 to pi of sin of t multiplied by sin of t divided by 2. DT Fantastic. So keeping that in mind, what we need to do next is we need to note this all for the first part. We will need to recall and use integration by parts. So we will state that U is equal to T, which will mean that DU by DT will be equal to 1, and that will mean that DU will be equal to DT. And we need to note that DV will be equal to sin of T divided by 2, DT, which will mean That the integral of DV will be equal to the integral of sin of T divided by 2. DT, which will mean that V will be simply equal to -2 multiplied by cosine of T divided by 2. Fantastic. So it's gonna be cosine of T divided by 2. Moving right along here so we can go ahead and write that 1/2 multiplied by square brackets of -2 multiplied by cosine of t divided by 2 plus 4 multiplied by sin of t divided by 2. And we need to note that this will be evaluated from, and don't forget your square brackets, so this will be evaluated from 0 to pi. And we need to note that this will be simply equal to when we perform our evaluation, it will be equal to 2, and solving for the second part, we will need to note that we will use sin of t is equal to 2 multiplied by sin oft divided by 2 multiplied by cosine of t divided by 2. Which will mean we can go ahead and write that 1/2 multiplied by the integral evaluate from 0 to 2 multiplied by sin oft divided by 2 multiplied by cosine of t divided by 2, multiplied by sin oft divided by 2. DT fantastic. And this will be equal to the integral evaluated from 0 to pi of sin to the 2 of T divided by 2, multiplied by cosine of T divided by 2 DT. And we need to use, so once again, we will need to use our substitution method again. So you will be equal to sin of T divided by 2, meaning that DU by DT will be equal to 1/2 multiplied by cosine of T divided by 2, which will mean that DU will be equal to 1/2 multiplied by cosine of T divided by 2 DT. So that will mean we can go ahead and write that square brackets of 2 divided by 3. Multiplied by sin to the power of 3. Of T divided by 2. Will be evaluated from 0 to pi, which will be equal to when we perform that evaluation, 2 divided by 3. So now we need to combine our parts. So x bar will be equal to 2 minus 2 divided by 3, which will be equal to 4 divided by 3. Which will mean our final result and our final answer will be parentheses of xy bar will be equal to parentheses of 4 divided by 34 divided by 3, and that's it. That is our final answer. Hooray, we did it. We solve for our final answer so. Going back to the top, we could safely say that our final answer without a doubt will be as follows. So once again what we did is we had to use a couple different integrations, but we've ultimately found our answer for the coordinates of the centroid, which are once again parentheses. Of 4 divided by 34 divided by 3, 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.

CentroidsFind the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.

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Centroids

Find the coordinates of the centroid of the curve x = cos t, y = t + sin t, 0 ≤ t ≤ π.