We have:
x(t)=5cost,y(t)=5sint,z(t)=3x(t)=5cost, y(t)=5sint, z(t)=3x(t)=5cost,y(t)=5sint,z(t)=3
Then:
L=∫(x′(t))2+(y′(t))2+(z′(t))2dtL=\int \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dtL=∫(x′(t))2+(y′(t))2+(z′(t))2dt
x′(t)=−5sint,y′(t)=5cost,z′(t)=0x'(t)=-5sint, y'(t)=5cost, z'(t)=0x′(t)=−5sint,y′(t)=5cost,z′(t)=0
Answer:
L=∫25sin2t+25cos2tdt=5∫dt=5t+cL=\int \sqrt{25sin^2t+25cos^2t}dt=5\int dt=5t+cL=∫25sin2t+25cos2tdt=5∫dt=5t+c
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