Video: Expressing the Length of a Parametric Curve as an Integral

Express the length of the curve with parametric equations 𝑥 = 𝑡 − 2 sin 𝑡 and 𝑦 = 1 − 2 cos 𝑡, where 0 ≤ 𝑡 ≤ 4𝜋, as an integral.

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Video Transcript

Express the length of the curve with parametric equations 𝑥 equals 𝑡 minus two sin 𝑡 and 𝑦 equals one minus two cos 𝑡, where 𝑡 is greater than or equal to zero but less than or equal to four 𝜋, as an integral.

So when we’re looking at a problem like this and we want to find the length of a curve with parametric equations, we have this general formula. And what the formula is, is that if we want to find the length of the curve, then it’s equal to the definite integral between the limits of 𝛼 and 𝛽. And that’s the limits, in our case, of 𝑡 of the square root of d𝑥 d𝑡 all squared plus d𝑦 d𝑡 all squared.

So we can see that the first thing we need to do is differentiate 𝑥 and differentiate 𝑦 because we want to find d𝑥 d𝑡 and d𝑦 d𝑡. So first of all, we want to differentiate 𝑥. And we’ve got 𝑥 equals 𝑡 minus two sin 𝑡. And if we differentiate 𝑡 minus two sin 𝑡, we’re gonna get one, because if you differentiate 𝑡 you get one, then minus two cos 𝑡. And we get that because our two will remain the same cause it’s just a constant and doesn’t get affected by differentiation. But then if we think about sin 𝑡- well, sin 𝑥, if we differentiate sin 𝑥, we get cos 𝑥. So we get one minus two cos 𝑡.

So now, what we’re gonna do is we’re gonna differentiate 𝑦. And that’s to find d𝑦 d𝑡. So we’re differentiating them with respect to 𝑡. So when we do that what we’re gonna get is two sin 𝑡. And that’s cause if we differentiate one minus two cos 𝑡 or if we differentiate one that disappears, becomes zero, and then if we differentiate cos 𝑡, so in our case, we’d look down here at cos 𝑥, then we get negative sin 𝑥.

Well, if we differentiate negative two cos 𝑡, it’s gonna be positive two sin 𝑡. And that’s cause, as we said, we had a negative and then negative sin 𝑥 which gives us positive sin 𝑥, in this case, positive sin 𝑡. Okay, great, so we now have d𝑥 d𝑡 and d𝑦 d𝑡. And we also have our limits because our limits are zero and four 𝜋. And we know that because if we look at our interval, it tells us that 𝑡 is greater than or equal to zero. So that means it doesn’t include zero. So zero is gonna be our lowest limit. And then it’s less than or equal to four 𝜋. And again, it’s less than or equal to, so that means that four 𝜋 is going to be our upper limit.

So then if we put this all into our general formula for the length of a curve, we’re gonna get the definite integral between the limits four 𝜋 and zero of the square root of one minus two cos 𝑡 all squared plus two sin 𝑡 all squared. So we’re now going to simplify this. And to simplify it, what we’ve got to do is expand the parenthesis cause we got one minus two cos 𝑡 all squared. So first of all, we can get one. So we got one multiplied by one which is just one. Then we’ve got minus two cos 𝑡. And then we’ve got minus another two cos 𝑡. And then finally, we’ve got negative two cos 𝑡 multiplied by negative two cos 𝑡 which would give us four cos squared 𝑡.

So, therefore, we’re gonna have the definite integral between the limits four 𝜋 over zero of the square root of, then we’ve got one minus four cos 𝑡 plus four cos squared 𝑡 and then plus four sin squared 𝑡. And that’s because we had two sin 𝑡 all squared. And then we can take four out as a factor. So we can have the square root of one minus four cos 𝑡 plus four. And then we’ve got cos squared 𝑡 plus sin squared 𝑡. And we do that because cos squared 𝑡 plus sin squared 𝑡 is one of our trig identities. And we know it’s equal to one.

So, therefore, if we use this trig identity and simplify them further, we’re gonna get that the length of the curve, with parametric equations 𝑥 equals 𝑡 minus two sin 𝑡 and 𝑦 equals one minus two cos 𝑡 where 𝑡 is greater than or equal to zero but less than or equal to four 𝜋, as an integral is the definite integral between the limits four 𝜋 and zero of the square root of five minus four cos 𝑡. And this is the answer, as we said, as an integral. And it’s fully simplified.

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