What is the arc length of #f(x)=(3x)/sqrt(x-1) # on #x in [2,6] #?

Question

What is the arc length of #f(x)=(3x)/sqrt(x-1)   # on #x in [2,6]     #?

Emilia Aguilar · Accepted Answer

#approx 4.514518184#

Writing

#f(x)=3x(x-1)^(-1/2)# then by the product and the chain rule we get

#f'(x)=3(x-1)^(-1/2)+3x*(-1/2)*(x-1)^(-3/2)#

which simplifies to

#f'(x)=(3(x-2))/(2*(x-1)^(3/2))# so we have to solve

#int_2^6 sqrt(1+(3(x-2)/(2*(x-1)^(3/2)))^2)dx# by a numerical method we get

#\approx4.514518184#

Charles Anctil · Answer

undefined To find the arc length of the function $f(x) = \frac{3x}{\sqrt{x - 1}}$ on the interval $[2, 6]$, we use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

First, we find $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{d}{dx} \left(\frac{3x}{\sqrt{x - 1}}\right) = \frac{3\sqrt{x - 1} - \frac{3x}{2\sqrt{x - 1}}}{x - 1} = \frac{3(2 - x)}{2(x - 1)\sqrt{x - 1}} $$

Now, we plug this into the arc length formula and integrate over the interval $[2, 6]$:

$$ L = \int_{2}^{6} \sqrt{1 + \left(\frac{3(2 - x)}{2(x - 1)\sqrt{x - 1}}\right)^2} \, dx $$

This integral is quite complex and doesn't have a simple closed-form solution. Numerical methods, such as numerical integration or software like MATLAB or Python, can be used to approximate the value of this integral and thus the arc length of the curve.