What is the arc-length integral for the curve y = x^2 - ln(x), 1 ≤ x ≤ 2?

[rhothi]

Find the length of the curve
y = x^2 − ln(x), 1 ≤ x ≤ 2.

I have have to find the arc-length of the given equation. I have checked my work several times and i ultimately come up with the integral of ((4x^4 -3x^2 +1)/(x^2))^(1/2), which I just cannot seem to solve. Any help would be appreciated.

 

[SammyS]

Homework Statement

Find the length of the curve
y = x^2 − ln(x), 1 ≤ x ≤ 2.

Homework Equations

The Attempt at a Solution

I have have to find the arc-length of the given equation. I have checked my work several times and i ultimately come up with the integral of ((4x^4 -3x^2 +1)/(x^2))^(1/2), which I just cannot seem to solve. Any help would be appreciated.

Hello rhothi. Welcome to PF !

Please give details as to how you got that integral, so we can help you.

 

[rhothi]

I used the formula for calculating arc length. The formula is ((1 + (f'(x))^(2))^(1/2) from a to b. As a result, I got f'(x) to be 2x - 1/x. When I squared the function i received ((4x^4 -4x^2 +1). From there I plugged f'(x)^2 into the equation to get the integral that I mentioned.

 

[Karnage1993]

$\displaystyle \left(2x - \frac{1}{x}\right)^2$ is not $4x^4 -4x^2 +1$.

Note that $(a+b)^2 = a^2 + 2ab + b^2$.

Try it again.

 

[rhothi]

I meant that it is (4x^4−4x^2+1)/(x^2). After adding 1 underneath the radical I received ((4x^4 -3x^2 +1)/(x^2))^((1/2)).

 

[SammyS]

I used the formula for calculating arc length. The formula is ((1 + (f'(x))^(2))^(1/2) from a to b. As a result, I got f'(x) to be 2x - 1/x. When I squared the function i received ((4x^4 -4x^2 +1). From there I plugged f'(x)^2 into the equation to get the integral that I mentioned.

It looks like you have a typo.

Are you saying that $\displaystyle \ \ 1+\left(2x-\frac{1}{x}\right)^2=\frac{4x^4 -3x^2 +1}{x^2}\ ?$

That's correct.

The result for the integral, $\displaystyle \ \ \int\sqrt{\frac{4x^4 -3x^2 +1}{x^2}}\,dx\ \$ looks very messy.

You can solve for x as a function of y for x > 0 . That result doesn't look much more promising.

Off hand, I can't think of any parametrization which would give a nicer result.

 

[rhothi]

Yes that is what I meant. How would I solve the equation in terms of y? I have no idea how to compute the integral in terms of x.

 

[SammyS]

Yes that is what I meant. How would I solve the equation in terms of y? I have no idea how to compute the integral in terms of x.

Here is a parametrization which I thought looked promising, but not surprisingly, it gives a similarly messy result.

Let x = e^t, then y = (e^t) - ln(e^t) = e^2t - t .

 

[Dick]

Here is a parametrization which I thought looked promising, but not surprisingly, it gives a similarly messy result.

Let x = e^t, then y = (e^t) - ln(e^t) = e^2t - t .

If you put the integral into Wolfram Alpha, it does give you an indefinite integral in terms of elementary functions. I didn't think it would. That probably means you could do it by completing a square, thinking of a clever substitution, maybe doing an integration by parts or two and voila. But the complexity of the answer makes it look like a really tough road for mere mortals. I'd guess if this was an exercise in computing arclength, you want to make it a lot easier than that. I'm going to guess there is probably a typo in the problem statement.