Cycloid

a. Find the length of one arch of the...

Question

Cycloid

a. Find the length of one arch of the cycloid x = a(t − sin t), y = a(1 − cos t).

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. Consider the asteroid given by the parametric equations. X is equal to a multiplied by cosine to the power of 3 of t. Y is equal to a multiplied by sin to the 3 of t. With a is greater than 0. Find the length of the curve for one quarter corresponding to t from 0 to pi divided by 2. OK, so it appears for this particular prong we're asked to take all the information that is provided to us by the prom itself, and we're asked to ultimately find what the length of the curve is for one quarter corresponding to T from 0 to pi divided by 2. So now that we know that we're ultimately trying to find the specific length of the curve, so we're determining what the length of the curve is, our first step that we need to take is we need to recall a note that in order to find the length of one quarter of this particular asteroid, we will need to recall and use the parametric arc length formula, which will denote as lower. Case S is equal to the integral evaluated from T1 to T2 of the square root of parentheses of DX divided by DT to the power of 2. Plus parentheses of Dy divided by dt to the power of 2. DT So based on our prior knowledge, we need to recall and note that the core idea is to find the rates of change for both x and y relative to the parameter t, and by squaring these derivatives and summing them, we will essentially apply Pyagorin's theorem. Which means A2 + B2 equals C2, in order to find the infinitesimal distance along the curve. Then we will need to integrate this particular expression over the specific interval, square brackets of 0, pi divided by 2, which indicates a closed interval, in order to accumulate these small segments into the total length of the first quadrant arc. OK. So, uh, with that in mind, our first step that we need to take is we now need to begin to differentiate the parametric equations. So, we need to start solving this problem by finding the derivatives of the given equations with respect to T, and we will do this by recalling and using the chain rule. So, for our first expression, which is X is equal to, so our first expression or first equation for our parametric equations will be X is equal to A multiplied by cosine to the power of 3 of T. And for this specific equation, we will write that DX by DT will be equal to 3 multiplied by A multiplied by cosine to the 2 of T multiplied by negatives of T, which will be equal to -3 multiplied by A multiplied by cosine square of T multiplied by sin of T. And 4 y is equal to a multiplied by sin to the power 3 of T. For our second parametric equation, we will be able to write that Dy by DT will be equal to 3 multiplied by a multiplied by sin to the power of 2 of T multiplied by cosine of t. Which will be equal to 3 multiplied by a multiplied by sine to the power of 2 oft multiplied by cosine oft fantastic. Our second step now is we need to recall and apply the arc length formula. So in order to do that, we will need to substitute these derivatives into the integrand, which, as we should recall, our integrand is the square root of parentheses of dx by DT to the power of 2 plus parentheses of Dy by DT to the power of 2. So, keeping that in mind, we can go ahead and write that parentheses of DX by DT2 plus parentheses of DY. By DT, so Dy by DT to the power of 2 will be equal to parentheses of 3 multiplied by a multiplied by cosine to the power 2 oft multiplied by sin of t. To the power 2 plus parentheses of 3 multiplied by a multiplied by sin to the power of 2 oft multiplied by cosine of t to the power of 2 and this will be equal to 9 multiplied by a to the power 2 multiplied by cosine to the power of 4 oft multiplied by sine to the power of 2 of t. Plus 9 multiplied by a to the power 2 multiplied by sin to the power of 4 oft multiplied by cosine to the power of 2 of t. And in order to simplify this expression, we will need to factor out the common term which for this particular case the common term is 9 multiplied by a 2 multiplied by sin to the 2 oft multiplied by cosine to the power of 2 of t. And when we factor out this term, we will find that it will be equal to 9 multiplied by a to the power of 2 multiplied by sin to the 2 oft multiplied by cosine to the 2 oft multiplied by parentheses of cosine to the power of 2 t plus sin to the power of 2 of t. And at this stage, we need to recall and use the identity that states that cosine to the power of 2 of T plus sin to the power of 2 of T is simply equal to 1. Which will mean that we could write that this will be equal to simply 9 multiplied by a to the power of 2 multiplied by sin to the 2 oft multiplied by cosine to the power of 2 of t. And we need to note at this point that by taking the square root and noting that sin of t and cosine oft are non-negative, fort is an element of the closed interval square brackets of 0, pi divided by 2, we can go ahead and write once again. At this point, to reiterate, we are taking the square root, and when we take the square root, and we also note that sin of t and cosine oft are non-negative, fort is an element of the closed interval 0, pi divided by 2, we can go ahead and safely write the following. We can write that the square root of parentheses of dx by dt to the 2 plus Dy by dt to the 2 will be simply equal to 3 multiplied by a multiplied by sin of t multiplied by cosin a t. Fantastic. So now for our 3rd step, which is our 3rd and final step, we now need to evaluate our integral. So we need to set up the integral for the specified interval, which once again is square brackets of 0, pi divided by 2. Which will mean that S is equal to the integral evaluated from 0 to pi divided by 2 of 3 multiplied by a multiplied by sin of t multiplied by cosine of t dt. And we need to recall and use the substitution method, so we will state that u will be equal to sin of t, which will mean that duu by dt will be equal to cosine of t, which will mean that duu will be equal to cosine of t dt. And 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 divided by 2, so once again, when t. Is equal to pi divided by 2. That will mean that u is simply equal to 1. So that will mean that S is equal to 3 multiplied by a multiplied by the integral evaluated from 0 to 1 of U. D u is equal to 3 multiplied by a multiplied by square brackets of u to the 2 divided by 2 evaluated from 0 to 1, so that will mean that s is equal to 3 multiplied by a multiplied by parentheses of 1/2 minus 0, which will be simply equal to 3a or 3 multiplied by a divided by 2. And don't, don't forget that we need to write units, because we're trying to find, once again, we're asked to find the length of the curve, and we need to make sure we always have our units, which in this case, since they're not specified, they'll just simply be units. And that's it. That is our final answer. Therefore, without a doubt, the length of one quarter of this specific asteroid will be 3A divided by 2. So looking at the problem itself to recap once again what we've learned, we could safely say without a doubt once again that our final answer without a doubt will be as follows. So 3A divided by 2 units will be the length of one quarter of this specific asteroid, 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.

Cycloida. Find the length of one arch of the cycloid x = a(t − sin t), y = a(1 − cos t).

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Cycloid

a. Find the length of one arch of the cycloid x = a(t − sin t), y = a(1 − cos t).