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Numerade Educator

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Problem 36 Hard Difficulty

(a) Find the arc length function for the curve $ y = \ln (\sin x) $, $ 0 < x < \pi $, with starting point $ ( \frac{\pi}{2}, 0) $.
(b) Graph both the curve and its arc length function on the same screen.

Answer

a) $L(x)=\ln |\csc x-\cot x|$
b) The black curve represents the curve $y=\ln (\sin x)$
The blue curve represents the length function
The red dot is the starting point $\left(\frac{\pi}{2}, 0\right)$

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Video Transcript

it's clear, so immune right here. So we have. Why is equal to Ellen of Sign So D why over the hex is equal to one over sign Oren's co sign with juice Co tangent. And so when we do one plus do you Why, over t X square is this equal to those seeking square? Just equal to the absolute value of CO Speaking of X So we're looking at the interval zero and pi and co seek it. It's bigger than or equal to zero. So we have this to be equal to coast seeking of X. We have to substitute into our art length function which gives us t is equal to pax. I'm t is equal to Pi house Cool Seaton of tea E t which is equal to l on, of course, seeking of tea minus co tangent of tea. We're finding it when t is equal to x and T is equal to Pi house. This gives us help then of co Seacon Thanks minus co tangent of tea What engine of X And since coast seeking it is bigger or equal to co tangent for access between zero and pi s of X is equal to Ellen Co Seacon of X minus co Tangent of vets. Next, we're going to graft the curve on the Ark link function. So first do our well and ah, sign of X in black. So I look something like this. This is a rush trying. And then our second function. Well, look, something like this.