/sci/ - Science & Math

[Anonymous No.2916665]

hey /sci/ its been a while, how do i calculate the arc length of a curve from 0 to 2 with the derivative being (x^2-2x)^1/2? i no the formula but i keep getting 0. help would be nice

[Anonymous No.2916735]

wow...i completely forgot how to do that...damn

[Anonymous No.2916815]

<div class="math">\int _{0}^{2}\! \left| \sqrt{1+((x^2-2x)^{\frac{1}{2}})^2} \right| {dx}=1</div>

[Anonymous No.2916835]

hehehehe....

[Anonymous No.2916908]

>>2916815
well shit i mean its not like thats what i have the answer key for...i just need to no how you get that

[Anonymous No.2916945]

[\
\int^0 _2 \sqrt{1+ f'(x)^2}dx
\]

[Anonymous No.2917036]

Use the distance formula... <span class="math"> d = \sqrt{x^{2} + y^{2}} [/spoiler] thus arc length <span class="math"> = \int^{2}_{0}\sqrt{dx^2+dy^2}[/spoiler]

[Anonymous No.2917070]

ok well i found out how to do it no thanks to the fucktards on /sci/. seriously arent atleast some of you supposed to be reasonably intelligent?

[Anonymous No.2917089]

>>2917070
You're just so stupid few of us bothered.

[Anonymous No.2917096]

>>2917070

You expect kings to do the work of peasants?

[Anonymous No.2917270]

>>2917096
>>2917089
well arent you two badasses
you could try and back your statements up with actual proof or you could sit there and pretend to be a master race of superior intellectuals

[Anonymous No.2917981]

>>2916815

nope.

[Anonymous No.2917984]

>>2917036

nope.

[Anonymous No.2918032]

${ds}^2 = {dx}^2 + {dy}^2$

$\dfrac{{ds}^2}{{dx}^2} = 1 + \dfrac{{dy}^2}{{dx}^2}$

${\dfrac{ds}{dx}}^2 = 1 + {\dfrac{dy}{dx}}^2$

$\dfrac{ds}{dx} = \sqrt{1 + {\dfrac{dy}{dx}}^2}$

$s = \int^b_a \sqrt{1 + {\dfrac{dy}{dx}}^2}\,dx$

So in your case this would be:

$s = \int^2_0 \sqrt{1 + {{{x}^{2} - {2x}}^{\frac{1}{2}}^2}\,dx$

$s = \int^2_0 \sqrt{1 + {{x}^{2} + {2x}}\,dx$

[Anonymous No.2918037]

{ds}^2 = {dx}^2 + {dy}^2

\dfrac{{ds}^2}{{dx}^2} = 1 + \dfrac{{dy}^2}{{dx}^2}

{\dfrac{ds}{dx}}^2 = 1 + {\dfrac{dy}{dx}}^2

\dfrac{ds}{dx} = \sqrt{1 + {\dfrac{dy}{dx}}^2}

s = \int^b_a \sqrt{1 + {\dfrac{dy}{dx}}^2}\,dx

So in your case this would be:

s = \int^2_0 \sqrt{1 + {{{x}^{2} - {2x}}^{\frac{1}{2}}^2}\,dx

s = \int^2_0 \sqrt{1 + {{x}^{2} + {2x}}\,dx

[Anonymous No.2918039]

>>2918037

The answer's 4, by the way.

http://www.wolframalpha.com/input/?i=integral+%280%2C2%29+sqrt%281+%2B+x%5E2+%2B+2x%29

[Anonymous No.2918052]

>>2918037
>writes out LaTeX code
>doesn't add tags
why.jpeg