Question

Question #1a66a

Answer 1

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...`