r(t)=[cos(3t),sin(3t),3t]r(t)=[{\cos(3t),\sin(3t),3t}]r(t)=[cos(3t),sin(3t),3t]
r′(t)=[−3sin(3t),3cos(3t),3]r'(t)=[{-3\sin(3t),3\cos(3t),3}]r′(t)=[−3sin(3t),3cos(3t),3]
∣r′(t)∣=(−3sin(3t)2+(3cos(3t))2+32==32\begin{vmatrix} r'(t) \end{vmatrix}=\sqrt{(-3\sin(3t)^2+(3\cos(3t))^2+3^2}=\\=3\sqrt{2}∣∣r′(t)∣∣=(−3sin(3t)2+(3cos(3t))2+32==32
The length of the curve =∫0π32dt=32π\int_0^\pi3\sqrt{2}dt=3\sqrt{2}\pi∫0π32dt=32π
Answer: a.32π3\sqrt2\pi32π .
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments