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(1 point) Calculate the length of the path over the given interval:(sin 6t, cos 6t), 0 < t < T...

Question

(1 point) Calculate the length of the path over the given interval:(sin 6t, cos 6t), 0 < t < T

(1 point) Calculate the length of the path over the given interval: (sin 6t, cos 6t), 0 < t < T



Answers

In Exercises $1-2,$ use Eq. (3) to find the length of the path over the given interval, and verify your answer using geometry. $$ (3 t-1,2-2 t), \quad 0 \leq t \leq 5 $$

Here. We're trying to find the length of the path of sign of three t co sign of three T from zero to pi. So we need to use the length formula of the integral from Alfa beta of the square root of the derivative of the X coordinate squared, plus the derivative of the Y coordinate square DT. So first we need to find the derivative of the X coordinate which, using the chain rule, will give us three co sign of three T and the derivative the white coordinate. Also using the chain rule will give us negative three sign of three t. So then we can substitute these on our limits of integration into our equation. And we get the lakes is equal to the integral from zero to pi of this square root of three co sign of three t quantity squared plus negative three sign of three t quantity squared Bt. So when we simplify this lead the integral from zero to pi of the square root of nine co sign squared of three T plus nine sine squared of three t Did you see me? And we can factor out the nine I take it the integral from zero to pi of the square root of nine times the square root uh, co sine squared of three t What sine squared of three t gt. So now the square root of nine, this thing turns into the integral from zero to pi of the square to nine, which is just three. Then we can use the Pythagorean identity that says pro sine squared. Plus sine squared is just one. So you get three times a squared of one DT, which which ends up just being three times the integral from zero to pi day t. And when we integrate, we get three times team evaluated from zero high. And when we evaluate that, we get three times pi minus zero, which gives us a length a green high.

It's important that we're able to determine the arc length given a parametric equation. So there's a really important formula for this. Um And it's going to use um some kind of a deviation of pythagorean theorem um and some differentials and derivative terms. So we want to find the arc length of this path. It's going to be the integral mhm. From A to B or alpha to beta. We'll call this L arc length and it's going to be times square root of dx DT. So the derivative of the extreme with respect to t squared plus dy DT script. So often times were given this parametric equation that's in the form um t squared minus six and then She plus four or something like that. And when we take the derivative of this is the X. Term with respect to T. And then we take the derivative of the Y. With respect to T. We square those take the square root and integrate it from our 0 to 5, perhaps whatever it is. And that will end up giving us our arc length

It's important that we're able to determine the arc length given a parametric equation. So there's a really important formula for this. Um And it's going to use um some kind of a deviation of pythagorean theorem um and some differentials and derivative terms. So we want to find the arc length of this path. It's going to be the integral mhm. From A to B or alpha to beta. We'll call this L arc length and it's going to be times square root of dx DT. So the derivative of the extreme with respect to t squared plus dy DT script. So often times were given this parametric equation that's in the form um t squared minus six and then She plus four or something like that. And when we take the derivative of this is the X. Term with respect to T. And then we take the derivative of the Y. With respect to T. We square those take the square root and integrate it from our 0 to 5, perhaps whatever it is. And that will end up giving us our arc length

All right. So this question wants us to compute the arc length of another Parametric curve from 01 So we use the same formula that we've been using, which is Theis. A girl from a to B of all under the square root ex prime squared, close y prime squared. So we need to compute X prime. And why prime To do this ex private e equals the derivative of the first part, which is just the derivative of t cubed plus one, which ends up to be three t squared. And then why prime of tea equals the derivative of the second equation T squared minus three, which is to t. So now we can plug into our arc length equation. Length equals the integral from 0 to 1 of the square root of ex prime squared. Plus why Prime Squared DT, which becomes the integral from 0 to 1 of square root nine t to the fourth plus for T squared DT, which equals inside girl from 01 of tea times square root of nine T squared plus four dt. And this is approximately 1.44 which you can use your calculator for because almost every arc length question you to CNN Exam is on the calculator section or it's just gonna make you set up the integral. But if you really wanted to do it by hand, you just use a U sub and set the middle set. You let you equal to this part and you consult from there, but


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