3 years ago

Help pls it’s easy I’m just slow lol

1 answer:

4 0

Answer:

12.65

Step-by-step explanation:

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USPshnik [31]

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath+t\,\vec k$

$\implies\vec r'(t)=-2\sin t\,\vec\imath+2\cos t\,\vec\jmath+\vec k$

$\implies\|\vec r'(t)\|=\sqrt{(-2\sin t)^2+(2\cos t)^2+1^2}=\sqrt5$

Then the length of the arc is

$\displaystyle\int_0^{2\pi}\|\vec r'(t)\|\,\mathrm dt=\sqrt5\int_0^{2\pi}\mathrm dt=2\sqrt5\,\pi$

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kkurt [141]

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Ber [7]

The answer is 1/2 4

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2 years ago

Degger [83]

Answer:

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Step-by-step explanation:

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2 years ago

Ivahew [28]

Answer:

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3 years ago