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4 singlthe arc Iength formula inithe parametric form find the arc length of theicurve IgiMenibyuxl= cos(t) + t sin(t)lland v = singt) ticos(t) in the interval 0 =...

Question

4 singlthe arc Iength formula inithe parametric form find the arc length of theicurve IgiMenibyuxl= cos(t) + t sin(t)lland v = singt) ticos(t) in the interval 0 = t1< 22 #Please be careful with your algebra, Trig land calculus!

4 singlthe arc Iength formula inithe parametric form find the arc length of theicurve IgiMenibyuxl= cos(t) + t sin(t)lland v = singt) ticos(t) in the interval 0 = t1< 22 #Please be careful with your algebra, Trig land calculus!



Answers

Arc Length write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ \begin{array}{ll} \underline{\text { Parametric Equations }} & \underline{\text { Interval }} \\ x=t+\sin t, \quad y=t-\cos t &\quad 0 \leq t \leq \pi \end{array} $$

You were given the parametric equations, takes scratch. X is equal to E. T plus two. And why is he going to to T plus one and the domain by not too two, two. And we had to find the at length which given by the integral from alpha to better often tough dx DT squared plus T, Y t t squared T T. And here first we have to change their mind the X T. T. He is the derivative of E. To the poverty that just gives us each party and the people to zero. So we just have to the T the first derivative of Y with respect to T. Here we have to therefore our back length is given by the integral from minus two 22 of the root of T X. T. T squared, which is E. To the power of T squared plus 2 to the power of two Gt. And this can be further simplified to the root of he, to the power of to T plus four T. T. And this is the solution.

In this question we are given the parametric equations X. Is equal to two T minus T squared. And why is he going to to T. People 3/2 and the domain one to to inclusive. Now we have to find the at length. You know that at length is equal to the integral from Peter hopeful move to better of the root of TX 80 squared plus do I. T. T. It's good T. T. And here dx did to you differentiating X. We get to my name's to T. D. Y. T. Here we get three over to play by two T. To the power off. Uh huh. And In this counselor doing three teeth pour half and our strength will be given by the integral from two From 1 to 2 and of the hood. Uh huh. 2 -2 T squared plus this way off 32 to the power house squared. You hear T. T. And we can simply leave it like this is this is the integral and evaluated. Is it?

We're gonna find the arc length of the given vector. So looking at the equation um for arc length we are going to be taking the derivative of R. F. T. First and then finding its magnitude. So the derivative of R f T. Is going to be a to co sign of two T. In the I. Direction and then minus two sign of two T. In the J direction plus one. Kay. To find our magnitude. We're going to square each of those components, add them together and then take the square root. So we are going to have a four co sign square to T and a four sine squared to T. So because of pythagorean we can go ahead and our identity tells us that sine squared plus cosine squared is just an equal one. So we have four plus one And we find our magnitude is the square root of five. So we can go ahead and integrate that in terms of T. So we're going to get square to five times t. And we'll first place our pie in and then subtract putting zero in. Of course putting zero in. Well zero it out. And so are arc length is pi times the square to five

Okay, we're going to find the arc length of the given vector. So looking at our formula, we are going to need the magnitude of the derivative of our empty. So first let's take the derivative. Notice that you have a co sign T. All to the third power. So great to use chain roll. The three comes down, we have a cosine squared and then we multiply by the derivative. The inside which is a negative sign T. That is all in the I direction. Our next derivative is similar, but we have signed inside so we have three and then we go to sine squared and the derivative is co sign T. And that is in the J direction and then the derivative of one is just zero. And so we don't have to write RK component. So now we're ready do our magnitude. And so we're going to square each of those. So for I component we have benign co signed to the fourth in a sine squared and RJ component, we will have nine signed to the fourth in a cosine squared. So looking at all the things in common with these two terms, they all have a nine or they both have a nine. They both have a co sine squared and they both have a sine squared. So each of those can be factored out and we can take the square root. So the square root of nine is three. The square root of cosine squared is just a co sign in the square root of sine squared is just a sign. That's going to leave a cosine squared plus sine squared inside the square root. Well, we know from our Pythagorean identities that equals one. So we're just integrating three scientific society. So in order to integrate that we need to use U substitution. Let's go ahead and make Ur sign T. Then our D. You will be arcosanti DT. So that means we are just integrating three. You do you Now considering putting a zero into sign? We're going to get a zero out. So even in terms of zero, that lower bound value is zero. But when we put a pi over two into sign we get a one out. So we'll be integrating from 0-1. So we can go up a power reciprocal and then we can put our value of one in subtracting putting the value of zero in. And that gives us three halves as the arc length of our vector


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