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How do you find the arc length of the curve #y=sqrt(cosx)# over the interval [-pi/2, pi/2]?

Answer 1

William Abdalla

The arc element of the curve #y=f(x)# is given by the expression:

#dl = sqrt(dx^2+dy^2) = sqrt(dx^2+(f'(x)dx)^2) =dx sqrt(1+(f'(x))^2)#

as:

#f(x) = sqrt(cosx)#

#f'(x) = sinx/(2sqrt(cosx))#

integrating over the interval we have:

#l =int_(-pi/2)^(pi/2) sqrt (1+sin^2x/(2cosx))dx#

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Answer 2

Emily Ammerman

To find the arc length of the curve (y = \sqrt{\cos(x)}) over the interval ([- \frac{\pi}{2}, \frac{\pi}{2}]), you use the arc length formula:

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

Find the derivative of (y = \sqrt{\cos(x)}): [ \frac{dy}{dx} = -\frac{1}{2} \sqrt{\frac{1}{\cos(x)}} \cdot (-\sin(x)) = \frac{\sin(x)}{2\sqrt{\cos(x)}} ]
Square the derivative and add 1: [ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{\sin(x)}{2\sqrt{\cos(x)}}\right)^2 ]
Integrate ( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} ) from (-\frac{\pi}{2}) to (\frac{\pi}{2}): [ L = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + \left(\frac{\sin(x)}{2\sqrt{\cos(x)}}\right)^2} , dx ]
Solve the integral to find the arc length.

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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.