An ellipse with semimajor axis length 2 and semiminor axis...

An ellipse with semimajor axis length 2 and semiminor axis length 1 is traced by the parametric equations x(t)=2cost,y(t)=sint,0<=t<=2pi . A sector of the ellipse is taken by restricting t to 0,(pi )/(3). The arc length of this sector equals int_0^((pi )/(3)) sqrt(x^(')(t)^(2)+y^(')(t)^(2))dt. Compute this integral on a piecewise-uniform grid with m=2. Increase n to achieve numerical convergence. Jo f/39 m=2.Increase n to achieve numerical convergence. x+ydt.Compute this integral on a piecewise-uniform grid with the parametric equations x=2cost,yt=sin,0i2n.A sector of 13. An ellipse with semimajor axis length 2 and semiminor axis length 1 is traced by

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00:01 Hi there, so to solve this problem we know that the area for this problem is given and that is 0 to 2 times pi of 1 divided by 2 the radius to the square integrated over theta from this we can use this expression that we are given so this will give us l squared divided by 2 from 0 to 2 times pi of the differential in theta this divided by 1 minus epsilon times the cosine of theta this to the square okay, now if we set seed equals to the exponential of i times theta and the differential seed is just i times the exponential of i times theta times the differential in theta then we can write then as i times c times the differential in theta the integral becomes a contoured integral around the unit circle c so the area integral is given by so that will be 2 times a this divided by the length square is equal to the integral over the contouring c of minus i times c to the minus 1 integrated over c this divided by 1 minus 1 divided by 2 times epsilon times seed plus c to the minus 1 and this to the square let's keep solving this integral so that will give us 4 divided by epsilon to the square the integral over c of minus i times seed integrated over seed this divided by seed a square minus 2 times epsilon to the minus 1 times seed plus 1 and then this to the square so in here a times epsilon to the square divided by 2 times l a square is equal to the integral from over c of seed minus c sub 0 this to the square this time seed minus seed 1 this to the square and in the numerator we will have minus i times seed integrated over seed where? seed of 1 and 0 is equal to 1 divided by epsilon this plus or minus the square root of 1 divided by epsilon to the square minus 1 now since we know that epsilon is greater than 0 well is greater than 0 but less than 1 and then epsilon to the minus 1 will be greater than 1 and only the pole at seed 0 equals the exponential of minus 1 minus the exponential of minus 2 minus 1 elevated to 1 divided by 2 lies inside the circle seed the magnitude of seed equals to 1 clearly it is a double pole of the integral so to determine the root the root issue at seed equals to c sub 0 we see that seed is equal to c sub 0 plus another value.

03:09 Let's label this with this symbol okay so we know that minus i times seed this divided by seed minus c sub 0 to the square times seed minus seed 1 that to the square could be written as minus i times c sub 0 plus this value then this divided by this value to the square just like this...

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