# Arc length problem.

• negatifzeo

## Homework Statement

Find the arclength of the function $$y=x^2$$ when x is between 0 and 10.

## Homework Equations

Arclength here is $$\int_{0}^{10} \sqrt{1+(2x)^2} dx$$
(It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)

## The Attempt at a Solution

Trig substitution. 2x= tan (theta)

Integrate secant theta
Which is ln(sec(theta)+tan(theta))

Substituting theta back in for arcsin(2x)
ln(sec(arctan(2x))+tan(arctan(2x))) evaluated from 10 to 0. Solving for arctan(2x) my final answer is
ln(sqrt(1+4x^2)+2x) evaluated from 0 to 10. I know this is wrong as it is much to short of an arc length but I don't know where I went wrong. Any clues?

## The Attempt at a Solution

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## Homework Statement

Find the arclength of the function $$y=x^2$$ when x is between 0 and 10.

## Homework Equations

Arclength here is $$\int_{0}^{10} \sqrt{1+(2x)^2} dx$$
(It's intended to be the integral from 0 to 10 of the quare root of 1+(2x)^2. My latex skills suck.)

## The Attempt at a Solution

Trig substitution. 2x= tan (theta)

That's fine.

Integrate secant theta
Which is ln(sec(theta)+tan(theta))

That's not fine. Why would you integrate $\sec (\theta )$?

That's fine.

That's not fine. Why would you integrate $\sec (\theta )$?

Well, when I use trig substitution here I change 2x to tan(theta). This changes the integrand to sqrt(1+tan^2(theta)). The trig identity says that 1 + tangent squared equals secant squared, and since it is the quare root of that it just becomes secant.

But you've forgotten about the $dx$.

Oh, duh. Wait, why am I having a hard time remembering how to get the dx here? Does dx here equal 1/2*sec^2(theta) d(theta)?

Yes, that's right.