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Calculating the length of a curve 
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#19
Nov2512, 02:34 PM

P: 232




#20
Nov2512, 03:45 PM

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



#21
Nov2512, 05:22 PM

P: 232

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{18})[/tex] OK I'm getting complex numbers, what did I do wrong?? 


#22
Nov2612, 07:36 AM

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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. 


#23
Nov2612, 07:40 AM

P: 232

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] 


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