"Do the following for the given curve on..." solved and explained

pancha3

Answered question

2021-09-11

Do the following for the given curve on the given interval.
a. Set up an integral for the length of the curve.
b. Graph the curve to see what it looks like.
c. Use your graphers

Answer & Explanation

Macsen Nixon

Skilled2021-09-12Added 117 answers

$\frac{d y}{d x} = \frac{d}{d x} (\sin {x} - \cos {x})$
$= \cos {x} - [x (- \sin {x}) + \cos {x}]$
$= \cos {x} + 2 \sin {x} + \cos {x}$
$= 2 \cos {x} + x \sin {x}$
a) set up integral
$L = \int_{0}^{2 π} {\sqrt{(1 + \frac{d x}{d y}}}}^{2} d x$
$= \int_{0}^{2 π} \sqrt{{(1 + 2 \cos {x} + {x} \sin {x})}^{2}} d x$
$= \int_{0}^{2 π} \sqrt{1 + 4 \cos {x} + x^{2} \sin^{2} x + 4 x \cos {x} \sin {x}}$
b)

c) $λ = \int_{0}^{2 π} \sqrt{1 + {(\frac{d y}{d x})}^{2}}$
$= \int_{0}^{2 π} \sqrt{1 + {(2 \cos {x} + x \sin {x})}^{2}} d x$
$= \int_{0}^{2 π} \sqrt{1 + 4 \cos^{2} {x} + x^{2} \sin^{2} {x} + 4 x \cos {x} \sin {x}}$
$[\frac{1}{2 \sqrt{1 + 4 \cos^{2} {x} + x^{2} <}}$

Do you have a similar question?Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?