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

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beckden | High School Teacher | (Level 1) Educator

Posted on

I made a mistake above the second to last line should be

`=1/2 ln((3+sqrt(2)/2)/(1/2))=1/2ln(3+2sqrt(2))`

And the last line should be

So our answer is `1/2ln(3+2sqrt(2))~~0.8814`

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beckden | High School Teacher | (Level 1) Educator

Posted on

Curve length from a to b `= int_a^bsqrt(1+(dy/dx)^2)dx` ` `

First find dy/dx where `sinx=e^y`

`cos(x)dx=e^y dy`

`dy/dx = cos(x)/e^y = cos(x)/sin(x) = cot(x)`

Curve length `= int_(pi/4)^(pi/2) sqrt(1+(cot(x))^2) dx`

Since `1+cot^2(x)=csc^2(x)` , `sqrt(1+cot^2(x))=sqrt(csc^2(x))=csc(x)`

`=int_(pi/4)^(pi/2) csc(x) dx=int_(pi/4)^(pi/2) 1/sin(x) dx `

We can integrate

`int(1/sin(x) dx)=int(sin(x)/(sin^2(x))) dx=int(sin(x)/(1-cos^2(x))) dx`

Using `u=cos(x)` `du=-sin(x) dx`

`int sin(x)/(1-cos^2(x))dx=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

`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 `

 

`=1/2ln((1+u)/(1-u))+C `

Substituting back in we get

`int csc(x) dx = -1/2ln((1+cos(x))/(1-cos(x)))+C=1/2ln((1-cos(x))/(1+cos(x)))+C`

So finally

Length`=int_(pi/4)^(pi/2) csc(x) dx = 1/2 ln((1-cosx)/(1+cos(x)))|_(pi/4)^(pi/2)`

=`1/2(ln((1-0)/(1+0)))-1/2ln((1-sqrt(2)/2)/(1+sqrt(2)/2))=1/2ln(1)+1/2ln((1+sqrt(2)/2)/(1-sqrt(2)/2))`

`=1/2(0)+1/2ln((1+sqrt(2)+1/2)/(1-1/2))`

`=1/2(ln(3/2-sqrt(2)/(1/2))=1/2(ln(3-2sqrt(2))`

So our answer is `1/2ln(3-2sqrt(2))~~0.8814`

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