The arc length of the plane curve x5t sin2ty3cos3t where...

The arc length of the plane curve x=5t-sin2t,y=3+cos3t, where 0<=t<=4pi , is given by the integral int_0^(4pi ) f(t)dt where f(t)= The arc length of the plane curve=5t-sin2t,u=3+cos3t,where0<t<4T,is given by the integral f(t)at where f(t)=

Transcript

00:01 For this problem we are to find the arc length of y that's equal to the natural log of x minus the square root of x squared minus 1.

00:09 For x values in the closed integral 1 to the square root of 257.

00:14 And we are given a hint that is to integrate with respect to y.

00:20 Now since our function is written as a function of y in terms of x, we have to rewrite this as a function of x in terms of y.

00:30 That using properties of logarithm e raise a power of y is equal to x minus the square root of x squared minus 1 so e raised to y plus e raised to negative y that is the reciprocal of x minus the square of x squared minus 1 this is equal to x minus square root of x squared minus 1 plus 1 over x minus square root of x squared minus 1 this is equal to 2x, which means that x is equal to one half of e raised to y plus e raised to negative y.

01:15 And since our interval is an interval in terms of x, we have to rewrite that as well.

01:22 So if x is in the interval 1 to the square root of 257 and y is equal to the natural log of x minus square root of x squared minus 1, then for x that's equal to 1, we have y equal to the natural log of 1 minus square root of 1 minus 1.

01:47 This is equal to 0.

01:49 And x is equal to the square root of 257.

01:51 It will give us y that's equal to the natural log of square root of 257 minus 16.

01:59 Note that this value is less than 0.

02:02 So when using this as limits for the integral, x equals square to 257 must be a lower limit...