This "Using the arc length formula below, set up..." has been explained

Secondary
Calculus and Analysis
Calculus 2
Applications of integrals

PEEWSRIGWETRYqx Answered2021-12-16

Using the arc length formula below, set up a correct formula for the arc length of a semi circle with its radius = 5 and then find its arc length.

\int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2} d x}

Answer & Explanation

Buck Henry

Expert

2021-12-17Added 33 answers

Explanation:

peterpan7117i

Expert

2021-12-18Added 39 answers

I can't make out the answer. Can you type it please?

nick1337

Expert

2021-12-28Added 573 answers

Sure,
Equation of circle $x^{2} + y^{2} = 2$
$\begin{matrix} \frac{d y}{d x} = - \frac{x}{y} \\ ∴ A r c l e n g h t = \int_{5}^{- 5} \sqrt{1 + (- \frac{x}{y})^{2} d x} \\ = \int_{5}^{- 5} \sqrt{(\frac{x^{2} + y^{2}}{y^{2}}) d x} = 5 \int_{5}^{- 5} \frac{d x}{y} = 2 \times 5 \int_{5}^{0} \frac{d x}{\sqrt{25 - x^{2}}} \\ 5 [\sin^{- 1} (\frac{x}{5})]^{5} = 10 [\sin^{- 1} \frac{5}{5} - \sin^{- 1} 0] \\ 10 \times \frac{x}{2} = 5 x \end{matrix}$