What is the arc length of f(x) = ln(x) f(x)=ln(x) on x in [1,3] x[1,3]?

1 Answer

The arc length color(red)(s=2.3019875345832)" "s=2.3019875345832 units

Explanation:

Given function f(x)=y=ln xf(x)=y=lnx with x from 1 to 3

The formula for arc length ss

s=int_1^3 sqrt(1+(dy/dx)^2) dxs=311+(dydx)2dx

dy/dx=(1/x)dx/dx=1/xdydx=(1x)dxdx=1x

s=int_1^3 sqrt(1+(1/x)^2) dxs=311+(1x)2dx

s=int_1^3 (sqrt(x^2+1))/x dxs=31x2+1xdx

Using Algebraic Substitution

Let z=sqrt(x^2+1)z=x2+1
and z^2=x^2+1z2=x2+1
Differentiating both sides
2z *dz=2x*dx+02zdz=2xdx+0
2z *dz=2x*dx2zdz=2xdx

dividing both sides by x^2x2
2z *dz=2x*dx2zdz=2xdx

(2z)/x^2 *dz=(2x)/x^2*dx=(2dx)/x 2zx2dz=2xx2dx=2dxx

(2z)/x^2 *dz=(2dx)/x 2zx2dz=2dxx

(2z*dz)/(z^2-1) =(2dx)/x 2zdzz21=2dxx

and
(z*dz)/(z^2-1) =(dx)/x zdzz21=dxx

Let us go back to

s=int_1^3 (sqrt(x^2+1))/x dx=int_1^3 (sqrt(x^2+1))(dx/x)s=31x2+1xdx=31(x2+1)(dxx)

Replacing now with variable zz

s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=31(x2+1)(dxx)=102(z)(zdzz21)

Because z=sqrt(x^2+1)" "z=x2+1
If xx is from x=1x=1 to x=3x=3, then, zz is from z=sqrt2z=2 to z=sqrt10z=10

Now we integrate

s=int_1^3 (sqrt(x^2+1))(dx/x)=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=31(x2+1)(dxx)=102(z)(zdzz21)

s=int_sqrt2^sqrt10 (z)((z*dz)/(z^2-1))s=102(z)(zdzz21)
s=int_sqrt2^sqrt10 ((z^2*dz)/(z^2-1))s=102(z2dzz21)
s=int_sqrt2^sqrt10 ((z^2-1+1)*dz)/(z^2-1)s=102(z21+1)dzz21

s=int_sqrt2^sqrt10 ((z^2-1)/(z^2-1)+1/(z^2-1))*dzs=102(z21z21+1z21)dz

s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dzs=102(1+1z21)dz

s=int_sqrt2^sqrt10 (1+1/(z^2-1))*dz=[z+1/2*ln((z-1)/(z+1))]_sqrt2^sqrt10s=102(1+1z21)dz=[z+12ln(z1z+1)]102

s=sqrt10+1/2ln((sqrt10-1)/(sqrt10+1))-(sqrt2+1/2ln((sqrt2-1)/(sqrt2+1)))s=10+12ln(10110+1)(2+12ln(212+1))

color(red)(s=2.3019875345832)" "s=2.3019875345832 units

God bless....I hope the explanation is useful.