22–23. Arc length Find the length of the following curves.
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Question

22–23. Arc length Find the length of the following curves.

x = cos 2t, y = 2t - sin 2t; 0 ≤ t ≤ π/4

Accepted Answer

Find the length of the curve given by X equals cosine 3T, Y equals 3T minus 3T for T between 0 and pi divided by 6. Now, we know the arc length formula for a parametric curve is given by L equals the integral from A to B. Of the square root, D X D T square. Plus D Y D T squared. This is in terms of DT. Now, we know our bounds are from 0 to pi divided by 6. Let's find DXDT and DYDT. The XDT is just a derivative of cosine 3T, which is -3, 3 3T. DYDT is a derivative. Of why Which is 3 -3, cosine 3 T. So now, we can substitute this back into our equation. L equals the integral from 0 to pi divided by 6. Of the square root 33 T squared. Plus 3 minus 3 cosine 3 T squared DT. From here we can simplify underneath the radical. We can go ahead and square everything to get 9. Signed squared, VT. Plus 9 multiplied by 1 minus cosine, 3 T squared. DT We can square root the nines. To get 3 Integral 0 of the pi divided by 6 of the square root. Sin squared 3 T. Plus 1 minus cosine 3 T squared DT. We can then simplify our expression inside the root even further. We can expand. Our 1 minus cosine 3 T squared. And use a Pythagorean identity to give us 2 multiplied by 1 minus cosine. 3 T. No. We have the integral A 3 0 of the pi divided by 6. Of The square root 2 minus 2 cosine 3TDT. Now, we can use our half angle identity to rewrite this. This is going to be rewritten. As 4 squared of 3 T divided by 2 underneath the square roots. Now, from here, we can just take the square root to get 3 and equal 0 to pi divided by 6 of 23 T divided by 2 DT. From here we have a use substitution. We have U equals 3 T divided by 2. TU equals 3 halves DT. This means DT equals 2/3 DU. Meaning we have 3. Inequal With our new you bounce. 0 toi divided by 4. Of 2/3 multiplied. By sign of you, do you. Now, we can go ahead and rewrite this install. Also note that by our half angle, we also have the 2 multiplying in the front. So we will move a 2 out in the front. We end up getting 4 integral 0 to pi divided by 4, sine U D U. Which is just 4 multiplied by negative cosine of U from 0 to pi divided by 4. We end up getting 4 multiplied by negative cosine pi divided by 4. Plus cosine of zero. This gives us 4 multiplied. By 1 minus square of 2 divided by 2. Which just gives us 4 minus 2 square roots of 2. Since this is a leak, this is in units as well, meaning our answer is 4 minus 2 square roots of 2. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

22–23. Arc length Find the length of the following curves.
x = cos 2t, y = 2t - sin 2t; 0 ≤ t ≤ π/4

Key Concepts

Parametric Equations

Arc Length Formula for Parametric Curves

Differentiation of Trigonometric Functions

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