Question #1a66a

1 Answer
Steve M
Jan 6, 2017

Arc Length #= ln (sqrt(2)+1 )) = 0.8813137... #

Explanation:

The Arc Length for a Curve #y=f(x)# is given by

# L=int_alpha^beta sqrt(1+(dy/dx)^2) \ dx =int_alpha^beta sqrt(1+(f'(x))^2) \ dx#

So in this problem we have

# y=ln(secx) => dy/dx=tanx#

So the Arc Length is;

# L=int_0^(pi/4) sqrt(1+tan^2x) \ dx #
# \ \ \=int_0^(pi/4) sqrt(sec^2x) \ dx #
# \ \ \=int_0^(pi/4) secx \ dx #
# \ \ \=[ ln | secx+tanx | ]_0^(pi/4) #
# \ \ \=ln | sec(pi/4)+tan(pi/4) | - ln | sec(0)+tan(0) | #
# \ \ \=ln | sqrt(2)+1 | - ln | 1+0 | #
# \ \ \=ln (sqrt(2)+1 ) - ln 1 #
# \ \ \=ln (sqrt(2)+1 )) #
# \ \ \=0.8813137... #

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