How do you find the arc length of the curve #y=(5sqrt7)/3x^(3/2)-9# over the interval [0,5]?

1 Answer
mason m
Jun 16, 2018

#s=int_0^5sqrt(1+175/4x)color(white).dx=1/525(879^(3/2)-8)#

Explanation:

The arc length of a function #y# on the interval #[a,b]# is given by:

#s=int_a^bsqrt(1+(dy/dx)^2)color(white).dx#

Here:

#y=(5sqrt7)/3x^(3/2)-9#

#dy/dx=(5sqrt7)/3(3/2x^(1/2))=(5sqrt7)/2x^(1/2)#

Then the arc length desired is:

#s=int_0^5sqrt(1+((5sqrt7)/2x^(1/2))^2)color(white).dx#

#s=int_0^5sqrt(1+175/4x)color(white).dx#

Let #u=1+175/4x#, which implies that #du=175/4dx#. Moreover, note the change of bounds of the integral under this substitution: #x=0=>u=1# and #x=5=>u=879/4#. The integral is then:

#s=4/175int_1^(879//4)u^(1/2)color(white).du#

Integrating using the power rule for integration:

#s=4/175(2/3u^(3/2))|_1^(879//4)#

#s=8/525((879/4)^(3/2)-1^(3/2))#

#s=8/525(879^(3/2)/8-1)#

#s=1/525(879^(3/2)-8)#

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