Question

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

Answer 1

`1219/240`

Explanation:

`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`