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I believe I've addressed (1) in another question of yours (24529718).
For (2), the arc length of the curve parameterized by x(t) = 3 cos(t ) and y(t) = 3 sin(t ) over 0 ≤ t ≤ π is
[tex]\displaystyle \int_0^\pi \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2 + \left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]
We have
[tex]\dfrac{\mathrm dx}{\mathrm dt} = -3\sin(t) \text{ and }\dfrac{\mathrm dy}{\mathrm dt} = 3\cos(t)[/tex]
so that the integral reduces to
[tex]\displaystyle \int_0^\pi \sqrt{9\sin^2(t) + 9\cos^2(t)}\,\mathrm dt = 3\int_0^\pi\mathrm dt[/tex]
since [tex]\cos^2(t)+\sin^2(t)=1[/tex] for all t. The remaining integral is trivial:
[tex]\displaystyle 3\int_0^\pi\mathrm dt = 3t\bigg|_0^\pi = 3(\pi-0) = \boxed{3\pi}[/tex]