Find the exact arc length of the curve
x = cos 3t, y = sin 3t ;(0 ≤ t ≤π)
L=∫(x′)2+(y′)2dtL=\int \sqrt{(x')^2+(y')^2}dtL=∫(x′)2+(y′)2dt
x′=−3sin3t,y′=3cos3tx'=-3sin3t, y'=3cos3tx′=−3sin3t,y′=3cos3t
L=∫0π(−3sin3t)2+(3cos3t)2dt=3∫0πdt=3πL=\int^{\pi}_0 \sqrt{(-3sin3t)^2+(3cos3t)^2}dt=3\int^{\pi}_0 dt=3\piL=∫0π(−3sin3t)2+(3cos3t)2dt=3∫0πdt=3π
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments