What is the arc length of #f(x)=cosx# on #x in [0,pi]#?
This formula is derived by adding up infinitesimal arc lengths along the curve. Here, for example, is an online explanation: https://tutor.hix.ai
Numerically, this is approximately 3.82.
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To find the arc length of the function ( f(x) = \cos(x) ) on the interval ([0, \pi]), you can use the arc length formula:
[ L = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} , dx ]

Find the derivative of ( f(x) ). [ f'(x) = \sin(x) ]

Square the derivative. [ [f'(x)]^2 = \sin^2(x) ]

Add 1 to the squared derivative. [ 1 + [f'(x)]^2 = 1 + \sin^2(x) ]

Take the square root of ( 1 + [f'(x)]^2 ). [ \sqrt{1 + [f'(x)]^2} = \sqrt{1 + \sin^2(x)} ]

Integrate the square root from ( x = 0 ) to ( x = \pi ). [ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} , dx ]

Evaluate the integral to find the arc length.
[ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} , dx ]
[ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} , dx ]
[ L = \int_{0}^{\pi} \sqrt{1 + \sin^2(x)} , dx ]
[ L \approx 3.819 ]
Therefore, the arc length of ( f(x) = \cos(x) ) on the interval ([0, \pi]) is approximately ( 3.819 ) units.
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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