What is the arc length of #f(x)=sqrt(1+64x^2)# on #x in [1,5]#?

1 Answer
Jun 1, 2018

A first-order approximation gives #L=4sqrt65-4/sqrt65tan^(-1)(32/321)# units.

Explanation:

#f(x)=sqrt(1+64x^2)#

#f'(x)=(64x)/sqrt(1+64x^2)#

Arc length is given by:

#L=int_1^5sqrt(1+(4096x^2)/(1+64x^2))dx#

Apply the substitution #8x=u# and rearrange:

#L=sqrt65/8int_8^40sqrt(1-64/65 1/(1+u^2))du#

For #u in [8,40]#, #64/65 1/(1+u^2)<1#. Take the series expansion of the square root:

#L=sqrt65/8int_8^40sum_(n=0)^oo((1/2),(n))(-64/65 1/(1+u^2))^ndu#

Isolate the #n=0# term and simplify:

#L=sqrt65/8int_8^40du+sqrt65/8sum_(n=1)^oo((1/2),(n))(-64/65)^nint_8^40(1/(1+u^2))^ndu#

Apply the substitution #u=tantheta#:

#L=4sqrt65+sqrt65/8sum_(n=1)^oo((1/2),(n))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n-2)thetad theta#

Isolate the #n=1# term:

#L=4sqrt65-4/sqrt65int_(tan^(-1)(8))^(tan^(-1)(40))d theta+sqrt65/8sum_(n=2)^oo((1/2),(n))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n-2)thetad theta#

Rescale #n#:

#L=4sqrt65-4/sqrt65(tan^(-1)(40)-tan^(-1)(8))+sqrt65/8sum_(n=1)^oo((1/2),(n+1))(-64/65)^(n+1)int_(tan^(-1)(8))^(tan^(-1)(40))cos^(2n)thetad theta#

Apply the Trigonometric power-reduction formula:

#L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-64/65)^nint_(tan^(-1)(8))^(tan^(-1)(40)){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((2n),(k))cos((2n-2k)theta)}d theta#

Integrate term by term:

#L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))theta+sum_(k=0)^(n-1)((2n),(k))sin((2n-2k)theta)/(n-k)]_(tan^(-1)(8))^(tan^(-1)(40))#

Hence:

#L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)+sum_(k=0)^(n-1)((2n),(k))(sin((2n-2k)tan^(-1)40)-sin((2n-2k)tan^(-1)8))/(n-k)]#

Apply the sum-to-product formula for the sine function:

#L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)+2sum_(k=0)^(n-1)((2n),(k))(sin{(n-k)(tan^(-1)40-tan^(-1)8)}cos{(n-k)(tan^(-1)40+tan^(-1)8)})/(n-k)]#

Apply the angle-sum identities for the tangent function:

#L=4sqrt65-4/sqrt65tan^(-1)(32/321)-8/sqrt65sum_(n=1)^oo((1/2),(n+1))(-16/65)^n[((2n),(n))tan^(-1)(32/321)-2sum_(k=0)^(n-1)((2n),(k))(sin{(n-k)tan^(-1)(32/321)}cos{(n-k)tan^(-1)(48/319)})/(n-k)]#