What is the arc length of #f(x)= 1/sqrt(x-1) # on #x in [2,4] #?

Question

What is the arc length of #f(x)=    1/sqrt(x-1) # on #x in [2,4]  #?

Christopher Agresta · Accepted Answer

#L ~~ 2.05411#

#f'(x) = -1/{2 sqrt{ (x-1)^{3} } }#

The length is given by

#L = int_2^4 sqrt{1 + f'(x)^2} "d"x#

#= int_2^4 sqrt{1 + (-1/{2 sqrt{ (x-1)^{3} } })^2} "d"x#

#= int_2^4 sqrt{1 + 1/{4 (x-1)^3 } } "d"x#

#~~ 2.05411#

Ezra Adams · Answer

undefined To find the arc length of $ f(x) = \frac{1}{\sqrt{x-1}} $ on the interval $ x \in [2,4] $:

1. Use the formula for arc length: 
$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

2. Calculate $ \frac{dy}{dx} $ which is the derivative of $ f(x) $.

3. Substitute $ \frac{dy}{dx} $ into the formula.

4. Integrate the expression over the interval $ x \in [2,4] $.

$$ L = \int_{2}^{4} \sqrt{1 + \left(\frac{d}{dx} \left(\frac{1}{\sqrt{x-1}}\right)\right)^2} \, dx $$

$$ = \int_{2}^{4} \sqrt{1 + \left(-\frac{1}{2(x-1)^{3/2}}\right)^2} \, dx $$

$$ = \int_{2}^{4} \sqrt{1 + \frac{1}{4(x-1)^3}} \, dx $$

This integral needs to be evaluated, which might involve some techniques like trigonometric substitution or other methods depending on the complexity of the function. After integration, you'll get the arc length $ L $ of the function $ f(x) $ over the interval $ x \in [2,4] $.