# Calculating the length of a curve

Tags: curve, length
P: 232
 Quote by arildno You write it as: $$L=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$ You may look at the code by making a right hand click on it. The command begins with "tex" within rectangular brackets, and ends with "\tex" within rectangular brackets, without apostrophes.
Thanks a lot, I edited it the exact way that you had it with the tex and \tex, but it still didn't come out...
 Sci Advisor HW Helper PF Gold P: 12,016 Sorry about that. It should be "/tex" in the last bracket, not "\tex"
P: 232
 Quote by arildno Sorry about that. It should be "/tex" in the last bracket, not "\tex"
What's wrong with it now? Haha, this is more complicated than it should be.

Edit: nevermind! got it.

By the way, back to the problem. We will have:

$$L=\int_{-2}^{2}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$

Our function is $$y=x^2$$, so the derivative is 2x (OP if you want to see why, it's because of the power rule).

$$L=\int_{-2}^{2}\sqrt{1+(2x)^{2}}dx$$
$$L=\int_{-2}^{2}\sqrt{1+4x^{2}}dx$$

Here, would the best substitution be $$4x=tan(t)$$? I did it out that way and I got the answer:

$$ln(4x+\frac{4x}{sin(arctan(4x))})$$

by drawing the triangle with those trig functions, it simplifies to:

$$ln(4x+\frac{4x}{\frac{4x}{\sqrt{1+4x}}}$$
$$ln(4x+\sqrt{1+4x})$$

plugging in the limits of integration:

$$ln(8+\sqrt{1+8})-ln(-8+\sqrt{1-8})$$

OK I'm getting complex numbers, what did I do wrong??
 Sci Advisor HW Helper PF Gold P: 12,016 You should use tan(t)=2x, not 4x. Besides, your area element is probably wrong. furthermore, it is much simpler to use a hyperbolic substitution, 2x=Sinh(u), rather than a trigonometric substitution.
 P: 232 Here's another way that you can do it, since the curve $$y=x^2$$ is symmetrical. $$\int_-2^2 \sqrt{1+(4x)^2}dx$$ $$2\int_0^2 \sqrt{1+4x^2}dx$$ Substitute 2x=tan(x): $$2\int_0^2 \sqrt{1+tan^2(x)}$$ $$2\int_0^2 sec(x)dx$$ $$2ln(tan(x)+sec(x))-ln(tan(0)+sec(0)$$ $$2ln(tan(x)+sec(x)-ln(1)$$ $$2ln(2x+\sqrt{1+4x^2})$$ $$2ln(4+\sqrt{65})$$

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