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

Question

Charles Arocha · Accepted Answer

The arc length is #14/3# units.

The arc length of a curve on the interval #[a, b]# is given by evaluating #int_a^b sqrt(1 + (dy/dx)^2)dx#. The derivative of #f'(x)#, given by the power rule, is #f'(x) = 1/2x^2 - 1/(2x^2) = (x^4 - 1)/(2x^2)# Substitute this into the above formula. #int_1^3 sqrt(1 + ((x^4 - 1)/(2x^2))^2)dx# Expand. #int_1^3 sqrt(1 + (x^8 - 2x^4 + 1)/(4x^4))dx# Put on a common denominator. #int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx# Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form #(ax)^2#. #int_1^3 sqrt((x^4 + 1)^2/(2x^2)^2)dx# Eliminate the square root using #(a^2)^(1/2) = a# #int_1^3 (x^4+ 1)/(2x^2)dx# Factor out a #1/2# and put it in front of the integral. #1/2int_1^3 (x^4 + 1)/x^2dx# Separate into different fractions. #1/2int_1^3 x^4/x^2 + 1/x^2dx# Simplify using #a^n/a^m = a^(n - m)# and #1/a^n = a^-n#. #1/2int_1^3 x^2 + x^-2dx# Integrate using #intx^ndx = x^(n + 1)/(n + 1)#, with #n in RR, n!= -1#. #1/2[1/3x^3 - 1/x]_1^3# Evaluate using the second fundamental theorem of calculus, which states that for #int_a^b F(x) = f(b) - f(a)#, if #f(x)# is continuous on #[a, b]# and where #f'(x) = F(x)#. #1/2(1/3(3)^3 - 1/3 - (1/3(1)^3 - 1/1))# Combine fractions and simplify. #1/2(9 - 1/3 - 1/3 + 1)# #1/2(10 -2/3)# #5 - 1/3# #14/3# Hopefully this helps!

Christopher Agresta · Answer

undefined To find the arc length of the curve $ f(x) = \frac{x^3}{6} + \frac{1}{2x} $ over the interval $[1,3]$, we use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

First, we find $ \frac{dy}{dx} $:

$$ f'(x) = \frac{d}{dx}\left(\frac{x^3}{6} + \frac{1}{2x}\right) = \frac{x^2}{2} - \frac{1}{2x^2} $$

Then, we plug it into the arc length formula:

$$ L = \int_{1}^{3} \sqrt{1 + \left(\frac{x^2}{2} - \frac{1}{2x^2}\right)^2} \, dx $$

Now, we calculate this integral to find the arc length.