Look at the image. ( Ill give brainliest to best answer.)
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Hello,
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
![l= 2*5* \int\limits^{\pi}_0{sin( \frac{t}{2} )} \, dt= 5[-cos (t/2)]_0^{\pi}\\\\ =5(0+1)=5](https://tex.z-dn.net/?f=l%3D%202%2A5%2A%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%7Bsin%28%20%5Cfrac%7Bt%7D%7B2%7D%20%29%7D%20%5C%2C%20dt%3D%205%5B-cos%20%28t%2F2%29%5D_0%5E%7B%5Cpi%7D%5C%5C%5C%5C%0A%3D5%280%2B1%29%3D5)
Here is the method but i may have make some mistakes.
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
Here is the method but i may have make some mistakes.