# What is the arclength of #(sint/(t+cos2t),cost/(2t))# on #t in [pi/12,pi/3]#?

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Derivative by WolframAlpha

Arc length with parametric equations

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To find the arc length of the curve ((\frac{\sin t}{t+\cos 2t}, \frac{\cos t}{2t})) on the interval (t \in [\frac{\pi}{12}, \frac{\pi}{3}]), we use the formula for arc length:

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

where (a) and (b) are the endpoints of the interval.

Given that (x(t) = \frac{\sin t}{t+\cos 2t}) and (y(t) = \frac{\cos t}{2t}), we differentiate them with respect to (t):

[ \frac{dx}{dt} = \frac{(\cos t)(t+\cos 2t) - (\sin t)(1 - 2t\sin 2t)}{(t+\cos 2t)^2} ] [ \frac{dy}{dt} = \frac{-\sin t}{2t} - \frac{\cos t}{2t^2} ]

Now, we plug these derivatives into the formula for arc length and integrate over the given interval:

[ L = \int_{\frac{\pi}{12}}^{\frac{\pi}{3}} \sqrt{\left(\frac{(\cos t)(t+\cos 2t) - (\sin t)(1 - 2t\sin 2t)}{(t+\cos 2t)^2}\right)^2 + \left(\frac{-\sin t}{2t} - \frac{\cos t}{2t^2}\right)^2} dt ]

Once the integrand is simplified, you can numerically evaluate this integral to find the arc length.

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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.

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.

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.

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