What is the arclength of $f (x) = \sqrt{4 - x^{2}}$ $f(x)=sqrt(4-x^2)$ in the interval $[- 2, 2]$ $[-2,2]$ ?

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What is the arclength of $f (x) = \sqrt{4 - x^{2}}$ $f(x)=sqrt(4-x^2)$ in the interval $[- 2, 2]$ $[-2,2]$ ?

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Answer 1 · 2018-07-10T16:06:08

The arc length is $2 π$ $2pi$ units.

Explanation:

This is a semi circle of radius $2$ $2$ . We simply find the perimeter of the full circle of radius $2$ $2$ and divide it by $2$ $2$ to get our arc length.

$P = 2 (2) π = 4 π$ $P = 2(2)pi = 4pi$

$\frac{P}{2} = \frac{4 π}{2} = 2 π$ $P/2 = (4pi)/2 = 2pi$

Therefore the arc length is $2 π$ $2pi$ units.

Hopefully this helps!

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What is the arclength of $f (x) = \sqrt{4 - x^{2}}$ $f(x)=sqrt(4-x^2)$ in the interval $[- 2, 2]$ $[-2,2]$ ?

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Explanation:

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What is the arclength of f(x)=sqrt(4-x^2) f(x)=√4−x2f(x)=sqrt(4-x^2) in the interval [-2,2][−2,2][-2,2]?

1 Answer

Explanation:

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What is the arclength of $f (x) = \sqrt{4 - x^{2}}$ $f(x)=sqrt(4-x^2)$ in the interval $[- 2, 2]$ $[-2,2]$ ?