1 answer

Find the arc length of the function below?

Question:

#y=\ln(\secx)#, with parameters #0\lex\le\pi/4#?


Answers

The arc has a length of #ln(sqrt2+1).#

Explanation:

For a function in the form #y=f(x),# the arc length from #[a,b]# is given by

#L=int_a^bsqrt(1+(dy/dx)^2)dx#.

The given function #y=lnsecx# is in the form #y=f(x)#, so we'll use the above formula. Furthermore, #[a,b]=[0,pi/4]#, so these are our integral's bounds.

Differentiate using the Chain Rule:

#y=lnsecx#
#(dy)/dx=1/secx*d/dxsecx#

#dy/dx=(cancelsecxtanx)/(cancelsecx)#

#dy/dx=tanx#

Then,

#L=int_0^(pi/4)sqrt(1+tan^2x)dx#

Recall the identity #1+tan^2x=sec^2x#. Then we get

#L=int_0^(pi/4)sqrt(sec^2x)dx#

#L=int_0^(pi/4)secxdx#

This is a common integral and worth memorizing. In general,

#intsecxdx=ln|secx+tanx|+C#

Then,

#int_0^(pi/4)secxdx=ln(secx+tanx)|_0^(pi/4)#

Absolute value bars are dropped because secant and tangent are positive on #[0, pi/4]#.

#=ln(sec(pi/4)+tan(pi/4))-ln(sec(0)+tan(0))#

#=ln(sqrt2+1)-ln(1)#

#=ln(sqrt2+1)#

The arc has a length of #ln(sqrt2+1).#

.

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