# Find the arc length(in exact value) of the curve defined by `x(t)=sin^(-1)t` ; `y(t)=ln sqrt(1-t^2)` on the interval   0 ≤ t ≤ 1/2.

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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In parametric equations the formula for arc length is:

`int sqrt( ((dx)/(dt))^2 + ((dy)/(dt))^2 ) dt `

Here:

`x="Sin"^(-1) x` , so: `(dx)/(dt) = (1)/(sqrt(1-t^2))`

`y= "ln" sqrt(1-t^2) = (1)/(2) "ln" (1-t^2)` , so: `(dy)/(dt) = (1)/(2) *(1)/(1-t^2) * -2t = -(t)/(1-t^2) `

So:

`int _0 ^(1/2) sqrt( (1)/(1-t^2) + (t^2)/((1-t^2)^2) ) dt `

`=int _0 ^(1/2) sqrt( (1-t^2)/((1-t^2)^2) + (t^2)/((1-t^2)^2) ) dt `

`=int _0 ^(1/2) sqrt( (1)/((1-t^2)^2) ) dt `

`=int _0 ^(1/2) (1)/(1-t^2) dt `

From here you can do partial fractions:

`(1)/(1-t^2) = (A)/(1-t)+(B)/(1+t) `

`1 = A(1+t)+B(1-t)`

Plug in `t=-1` to find that `1=2B`, so `B=(1)/(2)`

Plug in `t=1` to find that `1=2A`, so `A=(1)/(2)`

Thus:

`int _0 ^(1/2) (1)/(1-t^2) dt = (1)/(2) int _0^(1/2) (1)/(1+t) + (1)/(1-t) dt`
`= (1)/(2) ( "ln"|1+t| - "ln"|1-t|) |_0^(1/2) = (1)/(2) ( "ln"|(1+t)/(1-t)| ) |_0^(1/2)`
` = (1)/(2)"ln" 3 -0`

So the arc length is `(1)/(2) "ln" 3`

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

The formula of an arclength is:

`S = int_a^b sqrt[1+((dy)/(dx))^2] dx `         where S is the arc length

Note that in the above formula, there is no `t` variable.Hence, an equation in a form of `y=f(x)` must be determined.

To do so, use substitution method.

`x = sin^(-1)t`

`sin x = t`

Then, substitute `t=sinx` to the second equation `y=ln sqrt(1-t^2)`

`y = ln sqrt(1-t^2) = ln sqrt(1-sin^2x)`

Then, replace `1-sin^2` with `cos^2x` . This is based on the Pythagorean identity `sin^2 + cos^2x = 1` .

`y = ln sqrt(cos^2x)`

`y = ln cos x`

Take the derivative of y with respect to x. Use the formula `d/(dx) ln u = 1/u * u'`

`(dy)/(dx) = 1/(cos x) (-sin x) = -(sinx/cosx) = -tan x`

Then, determine the limits of the integral. Substitute the boundaries of the given interval  `0lt=tlt=1/2`  to  ` x=sin^(-1)t` .

`t = 0 `   ===>   `x = sin(-1)(0)`

`x = 0 and x = pi`

`t= 1/2`    ===>   `x = sin^(-1)1/2`

`x= pi/6 and x=(5pi)/6`

Do not consider `pi` and `(5pi)/6 ` as values of x, because if substituted to `y = ln cosx` it results to undefined function.

` y=ln cos pi = ln -1`

`y = ln (5pi)/6 = ln (-sqrt3/2)`

(Note that the argument of a logarithm should always be positive.)

Hence, the limit of the integral is `x=0` and `x=pi/6` .

Then, susbtitute `(dy)/(dx)= - tanx` and the limits of the integral to the formula of arclength.

`S= int_0^(pi/6) sqrt[1+(-tanx)^2] dx = int_0^(pi/6) sqrt (1+tan^2x) dx`

Base on the Pythagorean identity `1+tan^2x = sec^2x` .

`S=int_0^(pi/6) sqrt(sec^2x)dx = int_0^(pi/6) secx dx`

`S = ln| sec x + tan x |` `|_0^(pi/6)`

`S = ln |sec(pi/6) + tan (pi/6)| - ln |sec0 + tan 0|= ln |(2sqrt3)/3 + sqrt3/3| - ln|1 - 0|`

`S = lnsqrt3 - ln1 = ln sqrt3 - 0 = ln sqrt 3`

Express the square root as fractional exponent.

`S =ln 3^(1/2)`

To simplify further, apply the property of logarithm which is `ln a^m = m ln a` .

`S= 1/2 ln 3`

Hence, arc length is `1/2 ln3` .

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