Calculating the length of a curve

arildno

You write it as:
[tex]L=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^{2}}dx[/tex]
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.

piercebeatz

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...

arildno

Sorry about that. It should be "/tex" in the last bracket, not "\tex"

piercebeatz

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:

[tex]L=\int_{-2}^{2}\sqrt{1+(\frac{dy}{dx})^{2}}dx[/tex]

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

[tex]L=\int_{-2}^{2}\sqrt{1+(2x)^{2}}dx[/tex]
[tex]L=\int_{-2}^{2}\sqrt{1+4x^{2}}dx[/tex]

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

[tex]ln(4x+\frac{4x}{sin(arctan(4x))})[/tex]

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

[tex]ln(4x+\frac{4x}{\frac{4x}{\sqrt{1+4x}}}[/tex]
[tex]ln(4x+\sqrt{1+4x})[/tex]

plugging in the limits of integration:

[tex]ln(8+\sqrt{1+8})-ln(-8+\sqrt{1-8})[/tex]

OK I'm getting complex numbers, what did I do wrong??

arildno

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.

piercebeatz

Here's another way that you can do it, since the curve [tex]y=x^2[/tex] is symmetrical.

[tex]\int_-2^2 \sqrt{1+(4x)^2}dx[/tex]
[tex]2\int_0^2 \sqrt{1+4x^2}dx[/tex]

Substitute 2x=tan(x):

[tex]2\int_0^2 \sqrt{1+tan^2(x)}[/tex]
[tex]2\int_0^2 sec(x)dx[/tex]
[tex]2ln(tan(x)+sec(x))-ln(tan(0)+sec(0)[/tex]
[tex]2ln(tan(x)+sec(x)-ln(1)[/tex]
[tex]2ln(2x+\sqrt{1+4x^2})[/tex]
[tex]2ln(4+\sqrt{65})[/tex]