# Math Help - Length of a curve

1. ## Length of a curve

Can someone help with these two questions.

Determine the length of the given curve

$y=x^2$ from 0 to $\frac {\sqrt 3} {2}$

$y=\ln x$ from 1 to $\sqrt 3$

2. Originally Posted by chengbin
Can someone help with these two questions.

Determine the length of the given curve

$y=x^2$ from 0 to $\frac {\sqrt 3} {2}$

$y=\ln x$ from 1 to $\sqrt 3$
Hi

The elementary displacement along a curve is

$dl = \sqrt{dx^2 + dy^2}$

If $y = f(x)$ then $dy = f'(x) \:dx$

Therefore $dl = \sqrt{1 + f'^2(x)}\:dx$

The length of the curve is $L = \int_{x_0}^{x_1}\sqrt{1 + f'^2(x)}\:dx$

3. Thanks, I know that, but I just can't get the answer in my solution book. So I was hoping someone will write a step by step instruction of solving it.

4. If you apply the formula for the first one

$L = \int_{0}^{\frac{\sqrt{3}}{2}}\sqrt{1 + 4x^2}\:dx$

which can be solved by parts

5. And I don't know how to integrate $\int \frac {dx} {cos^3x}$

6. Originally Posted by chengbin
And I don't know how to integrate $\int \frac {dx} {cos^3x}$
try to do so

$\int {\frac{{dx}}{{{{\cos }^3}x}}} = \int {\frac{{\cos x}}
{{{{\cos }^4}x}}dx} = \int {\frac{{\cos x}}{{{{\left( {1 - {{\sin }^2}x} \right)}^2}}}dx} = \left\{ \begin{gathered}\sin x = t, \hfill \\\cos xdx = dt \hfill \\ \end{gathered} \right\} =$

$= \int {\frac{{dt}}{{{{\left( {1 - {t^2}} \right)}^2}}}} = \int {\frac{{dt}}{{{{\left( {{t^2} - 1} \right)}^2}}}} = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}}}.$

$\frac{1}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}} = \frac{A}{{{{\left( {t + 1} \right)}^2}}} + \frac{B}{{t + 1}} + \frac{C}{{{{\left( {t - 1} \right)}^2}}} + \frac{D}{{t - 1}} \Leftrightarrow$

$\Leftrightarrow 1 = A{\left( {t - 1} \right)^2} + B\left( {t + 1} \right){\left( {t - 1} \right)^2} + C{\left( {t + 1} \right)^2} + D\left( {t - 1} \right){\left( {t + 1} \right)^2} \Rightarrow$

$\Rightarrow \left. {\begin{array}{*{20}{c}}{t = - 1} \hfill \\{t = 1} \hfill \\{t = 2} \hfill \\{t = 0} \hfill \\\end{array} } \right|{\text{ }}\left\{ \begin{gathered}4A = 1, \hfill \\4C = 1, \hfill \\A + 3B + 9C + 9D = 1, \hfill \\A + B + C - D = 1; \hfill \\ \end{gathered} \right.$ ${\text{ }} \Leftrightarrow \left\{ \begin{gathered}A = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\B = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\D = - {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}, \hfill \\C = {1 \mathord{\left/{\vphantom {1 4}} \right.\kern-\nulldelimiterspace} 4}. \hfill \\ \end{gathered} \right.$

7. $= \int {\frac{{dt}}{{{{\left( {{t^2} - 1} \right)}^2}}}} = \int {\frac{{dt}}{{{{\left( {t + 1} \right)}^2}{{\left( {t - 1} \right)}^2}}}}.$

$(t^2-1)^2=(t-1)^2(t+1)^2$

???

8. $\begin{gathered}\boxed{{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)} \hfill \\{\left( {{t^2} - 1} \right)^2} = {\left( {\left( {t + 1} \right)\left( {t - 1} \right)} \right)^2} = {\left( {t + 1} \right)^2}{\left( {t - 1} \right)^2} \hfill \\ \end{gathered}$

Do you understand it??

9. OMG, that's the dumbest mistake I've done in a long time.

Oh I found out how to integrate it using integration by parts.

I'm not sure how your way of doing it works.

Anybody help me with the other one? y=lnx

10. For the first one there is a more simpler way by parts

$L = \int\:\sqrt{1 + 4x^2}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int \frac{4x^2}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int \frac{1+4x^2-1}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int\:\sqrt{1 + 4x^2}\:dx + \int \frac{1}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - L + \frac12\:\int \frac{1}{\sqrt{1 + u^2}}\:du$ where u = 2x

$2L = x \:\sqrt{1 + 4x^2} + \frac12\:sinh^{-1}u$

$L = \frac12\:x \:\sqrt{1 + 4x^2} + \frac14\: sinh^{-1}(2x)$

11. Originally Posted by running-gag
For the first one there is a more simpler way by parts

$L = \int\:\sqrt{1 + 4x^2}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int \frac{4x^2}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int \frac{1+4x^2-1}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - \int\:\sqrt{1 + 4x^2}\:dx + \int \frac{1}{\sqrt{1 + 4x^2}}\:dx$

$L = x \:\sqrt{1 + 4x^2} - L + \frac12\:\int \frac{1}{\sqrt{1 + u^2}}\:du$ where u = 2x

$2L = x \:\sqrt{1 + 4x^2} + \frac12\:sinh^{-1}u$

$L = \frac12\:x \:\sqrt{1 + 4x^2} + \frac14\: sinh^{-1}(2x)$
Actually over here in my school, I was advised to

remember these formulas

$\int{\sqrt{x^2 + a^2}}dx = \frac{x\times \sqrt{x^2 +a^2} }{2} + \frac{a^2(log|x+\sqrt{x^2+a^2}|)}{2} + C
$

$
\int{\sqrt{x^2 - a^2}}dx = \frac{x\times \sqrt{x^2 -a^2} }{2} - \frac{a^2(log|x+\sqrt{x^2-a^2}|)}{2}+ C
$

$
\int{\sqrt{a^2-x^2}}dx = \frac{x\times \sqrt{a^2 -x^2} }{2} - \frac{a^2(sin^{-1}(~\frac{x}{a})~)}{2} + C
$

Though we were asked to remember it, I never did it, alwas

liked the proving stuff, its better , Better to prove it

A

12. Originally Posted by chengbin
can anyone help with y=lnx?
Even if it takes many posts tell us, No spoon feeding now, learn from first

1) What is f'(x) in this case ?
2) What was written(Formula) in RG's first post
3)How to put
4) how to integrate
Show these steps and the place you hanged this time!

13. I got y=x^2 now. It was a pain the in butt to do it. I'll just write my steps here to ask if there is a better way.

$y=x^2$ From 0 to $\frac {\sqrt3} {2}$

$\frac {dy} {dx}=2x$

$L= \int^\frac {\sqrt3} {2}_0 \sqrt{1+4x^2}dx$

Let $x=\frac {1} {2} tan\theta$

$dx=\frac {d\theta} {2cos^\theta}$

When x=0, $\theta=0$

When $x=\frac {\sqrt3} {2}$, $\theta=\frac {\pi} {3}$

$L=\int^\frac {\pi} {3}_0 \sqrt{tan\theta+1}dx$

$L=\int^\frac {\pi} {3}_0 (\frac {1} {cos\theta}) (\frac {1} {2cos^2\theta})$

$L=\frac {1} {2} \int^\frac {\pi} {3}_0 \frac {d\theta} {cos^3\theta}$

$\frac {d\theta} {cos^3\theta}=\frac {d\theta} {cos\theta cos^2\theta}$

$=\bigg [ \frac {tan\theta} {cos\theta} \bigg ]^\frac {\pi} {3}_0 - \int^\frac {\pi} {3} \frac {sin\theta cos\theta} {cos^2\theta} d\theta$

$=2\sqrt3 - \int^\frac {\pi} {3}_0 \frac {sin^2\theta} {cos^3\theta}d\theta$

$2\sqrt3 - \int^\frac {\pi} {3}_0 \frac {d\theta} {cos^3\theta} + \int^\frac {\pi} {3}_0 \frac {d\theta} {cos\theta}$

$2\int^\frac {\pi} {3}_0=2\sqrt3+\int^\frac {\pi} {3}_0 \frac {d\theta} {cos\theta}$

$\int \frac {d\theta} {cos\theta}=\int \frac {cos\theta} {cos^2\theta}d\theta$

$\int \frac{cos\theta} {1-sin^2\theta} d\theta$

Let $u=sin\theta du=cos\theta d\theta$

$=-\int \frac {du} {u^2-1}$

$\frac {1} {2} \int (\frac {1} {u+1} - \frac {1} {u-1})du$

$=1/2(\ln|u+1|-\ln|u-1|$

$=\frac {1} {2} \ln \frac {|sinx+1|} {|sinx-1|}$

$=2\sqrt3+\bigg [ \frac {1} {2} \ln \frac {|sinx+1|} {|sinx-1|}\bigg ]^\frac {\pi} {3}_0$

$=2\sqrt3+\frac {1} {2} \ln \frac {|\sqrt3+2|} {|\sqrt3-2|}$

$=2\sqrt3+\ln |\sqrt3+2|$

$L=\frac {\sqrt3} {2} + \frac {1} {4} \ln (\sqrt3+2)$

This is one ugly integral.

14. Originally Posted by chengbin
[snip]
Anybody help me with the other one? y=lnx
Make the substitution $u^2 = 1 + x^2$.