How do you find the arc length of the curve #y=x^5/6+1/(10x^3)# over the interval [1,2]?

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Answer 1 · 2018-03-21T08:46:01

#1219/240#

#y = x^5/6+1/{10x^3} implies#

#dy/dx= (5x^4)/6-3/(10x^4) implies#

#1+(dy/dx)^2 = 1+( (5x^4)/6-3/(10x^4))^2#

Now, since #1 = 4 times (5x^4)/6 times 3/(10x^4)#, the expression on the right above simplifies !

#1+(dy/dx)^2 =( (5x^4)/6+3/(10x^4))^2#

Thus, the required arc length

#L = int_a^b sqrt{1+(dy/dx)^2}dx = int_1^2( (5x^4)/6+3/(10x^4)) dx = (x^5/6-1/(10x^3))_1^2 =(2^5/6-1/(10times 2^3))-(1/6-1/10) = 1219/240 #