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What is the arc length of #f(x)=cosx-sin^2x# on #x in [0,pi]#?

Answer 1

Ezra Adams

#S = 4.28154#

Arc length #S = int_0^pi sqrt(1+(y')^2) dx#

#y' = -sin x - 2 sin x cos x#

#S = int_0^pi sqrt(1+(-sin x - 2 sin x cos x)^2) dx#

#= int_0^pi sqrt(1+sin^2 x + 4 sin^2 x cos^2 x + 4 sin^2 x cos x) dx#

#= int_0^pi sqrt(1+sin^2 x + sin^2 2x + 2 sin2x sin x ) dx#

That doesn't simplify easily so computer solution is:

#S = 4.28154#

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Answer 2

Emilia Aguilar

To find the arc length of a function ( f(x) ) on an interval ([a, b]), the formula is:

[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx ]

For ( f(x) = \cos(x) - \sin^2(x) ) on ( x \in [0, \pi] ):

[ \frac{dy}{dx} = -\sin(x) - 2\sin(x)\cos(x) ]

[ L = \int_{0}^{\pi} \sqrt{1 + \left(-\sin(x) - 2\sin(x)\cos(x)\right)^2} , dx ]

[ L \approx 4.525 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.