give the integral formula for the length of the curve sinx=e^y from x=pie/4 to x=pie/2

Tuesday, August 4, 2015

give the integral formula for the length of the curve sinx=e^y from x=pie/4 to x=pie/2

Curve length from a to b $\displaystyle={\int_{{a}}^{{b}}}\sqrt{{{1}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{x}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}$

First find dy/dx where
$\displaystyle{\sin{{x}}}={{e}}^{{y}}$

$\displaystyle{\cos{{\left({x}\right)}}}{\left.{d}{x}\right.}={{e}}^{{y}}$
dy

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\cos{{\left({x}\right)}}}}{{{e}}^{{y}}}=\frac{{\cos{{\left({x}\right)}}}}{{\sin{{\left({x}\right)}}}}=$
cot(x)

Curve length $\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}$
sqrt(1+(cot(x))^2) dx

Since $\displaystyle{1}+{{\cot}}^{{2}}{\left({x}\right)}={{\csc}}^{{2}}{\left({x}\right)}$ ,
$\displaystyle\sqrt{{{1}+{{\cot}}^{{2}}{\left({x}\right)}}}=\sqrt{{{{\csc}}^{{2}}{\left({x}\right)}}}={\csc{{\left({x}\right)}}}$

$\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}$
csc(x) dx=int_(pi/4)^(pi/2) 1/sin(x) dx

We can
integrate

$\displaystyle\int{\left(\frac{{1}}{{\sin{{\left({x}\right)}}}}{\left.{d}{x}\right.}\right)}=\int{\left(\frac{{\sin{{\left({x}\right)}}}}{{{{\sin}}^{{2}}{\left({x}\right)}}}\right)}$
dx=int(sin(x)/(1-cos^2(x))) dx

Using $\displaystyle{u}={\cos{{\left({x}\right)}}}$
$\displaystyle{d}{u}=-{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{\sin{{\left({x}\right)}}}}{{{1}-{{\cos}}^{{2}}{\left({x}\right)}}}{\left.{d}{x}\right.}=\int$
-(du)/(1-u^2)

Using partial
fractions

1/(1-u^2)=A/(1-u)+B(1+u)
gives

A(1+u)+B(1-u) = 1 A=1/2,
B=1/2

And

$\displaystyle\int$
(du)/(1-u^2)=int 1/2(1/(1-u))+1/2(1/(1+u)) du=1/2(-ln(1-u)+ln(1+u))+C

$\displaystyle=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{u}}}{{{1}-{u}}}\right)}}}+{C}$

Substituting back in we
get

$\displaystyle\int{\csc{{\left({x}\right)}}}{\left.{d}{x}\right.}=$
-1/2ln((1+cos(x))/(1-cos(x)))+C=1/2ln((1-cos(x))/(1+cos(x)))+C

So
finally

Length $\displaystyle={\int_{{\frac{\pi}{{4}}}}^{{\frac{\pi}{{2}}}}}{\csc{{\left({x}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{2}}$
ln((1-cosx)/(1+cos(x)))|_(pi/4)^(pi/2)

= $\displaystyle\frac{{1}}{{2}}{\left({\ln{{\left(\frac{{{1}-{0}}}{{{1}+{0}}}\right)}}}\right)}-\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}-\frac{\sqrt{{{2}}}}{{2}}}}{{{1}+\frac{\sqrt{{{2}}}}{{2}}}}\right)}}}=\frac{{1}}{{2}}{\ln{{\left({1}\right)}}}+\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+\frac{\sqrt{{{2}}}}{{2}}}}{{{1}-\frac{\sqrt{{{2}}}}{{2}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left({0}\right)}+\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+\sqrt{{{2}}}+\frac{{1}}{{2}}}}{{{1}-\frac{{1}}{{2}}}}\right)}}}$

$\displaystyle=\frac{{1}}{{2}}{\left({\ln{{\left(\frac{{3}}{{2}}-\frac{\sqrt{{{2}}}}{{\frac{{1}}{{2}}}}\right)}}}=\frac{{1}}{{2}}{\left({\ln{{\left({3}-{2}\sqrt{{{2}}}\right)}}}\right.}\right.}$

So
our answer is $\displaystyle\frac{{1}}{{2}}{\ln{{\left({3}-{2}\sqrt{{{2}}}\right)}}}\approx{0.8814}$

Help Notes

Tuesday, August 4, 2015