# y = ln((e^x + 1)/(e^x - 1)) , [ln2 , ln3] Find the arc length of the graph of the function over the indicated interval. The arc length of a function is the length of the described curve within some interval on the x-axis (and a corresponding interval on the y-axis). If we were to lay a piece of string over the line of the graph in this window, the graph going from corner to...

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The arc length of a function is the length of the described curve within some interval on the x-axis (and a corresponding interval on the y-axis). If we were to lay a piece of string over the line of the graph in this window, the graph going from corner to diagonally opposite corner in a curve, the arc length is the length of the piece of string used.

In this case the function to find the arc length of is

y = ln ((e^x+1)/(e^x - 1))

where ln(x)  is the natural logarithm of x  (the inverse function to e^(x) , recall).

The window over which to find the length is the rectangle given by  ln 2 <= x <= ln3   and ln2 <= y <= ln3.

The function for the arc length of a generic function y = f(x)  is

s = int_a^b sqrt(1+((dy)/(dx))^2) \quad dx

This integrates along the line of the graph, adding infinitesimal sections together where each very small section added is a straight line diagonal over the small window x_0 <=x <=x_0 + dx  , y_0 <=y <= y_0 + dy  . Though y = f(x)  may be a curve, the point is that these windows over which the length is integrated (added up) are so small that the graph is a straight line within them making it a simple thing to add the small (straight line) sections together. This concept is the basis of calculus.

In this example, since y = ln ((e^x+1)/(e^x-1)) = ln(e^x + 1) - ln(e^x-1) then its derivative (dy)/(dx)   is given by

(dy)/(dx) = (1/(e^x+1))e^x - (1/(e^x-1))e^x = e^x(1/(e^x+1) - 1/(e^x-1))  = e^x((e^x-1-e^x-1)/((e^x+1)(e^x-1)))  = -(2e^x)/((e^x+1)(e^x-1))

Now working in steps to find the integrand in the formula for the arc length s  , we have that

1+ ((dy)/(dx))^2 = 1 + (4e^(2x))/((e^x+1)^2(e^x-1)^2)   = ((e^x+1)^2(e^x-1)^2 + 4e^(2x))/((e^x+1)^2(e^x-1)^2)

and that

sqrt(1+((dy)/(dx))^2) = (sqrt((e^x+1)^2(e^x-1)^2 + 4e^(2x)))/((e^x+1)(e^x-1))

= sqrt((e^(2x)+2e^x +1)(e^(2x)-2e^(x)+1)+4e^(2x))/((e^x+1)(e^x-1))

= sqrt((e^(4x)-2e^(3x)+e^(2x)+2e^(3x) - 4e^(2x) + 2e^x + e^(2x)-2e^x + 1)+4e^(2x))/((e^x+1)(e^x-1))

Adding up that very long set of exponential terms, finding that most cancel, we finally have that

sqrt(1+((dy)/(dx))^2) = sqrt((e^(4x)-2e^(2x)+1)+4e^(2x))/((e^x+1)(e^x-1))   = sqrt(e^(4x)+2e^(2x)+1)/((e^x+1)(e^x-1))

= sqrt((e^(2x)+1)^2)/((e^x+1)(e^x-1)) = (e^(2x)+1)/((e^x+1)(e^x-1))

= (e^(2x)+1)/(e^(2x)-1)

Integrating the integrand over the required interval, we have that

s = int_(ln2)^(ln3) ((e^(2x)+1)/(e^(2x)-1)) \quad dx

  = int_(ln2)^(ln3) (2e^(2x)- e^(2x) + 1)/(e^(2x)-1) \quad dx = int_(ln2)^(ln3) (2e^(2x))/(e^(2x)-1) -1 \quad dx

= ln(e^(2x)-1)|_ln2^(ln3) - x |_(ln2)^(ln3)

= ln (e^(2ln3)-1) - ln(e^(2ln2)-1) - ln3 + ln2

= ln(9-1) - ln(4-1) - ln3 + ln2 = ln((8/3)(2/3))

= ln(16/9)  is the final answer