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

Question

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

Ezra Adams · Accepted Answer

Approximately #1.48# units.

The arc length of a curve on #x in [a, b]# is given by #A = int_a^b sqrt(1 + (dy/dx)^2)dx# Taking the derivative, we get #f'(x) = (x/sqrt(x^2 - 1)(x) - sqrt(x^2 - 1)(1))/x^2# #f'(x) = (x^2/sqrt(x^2 - 1) - sqrt(x^2 - 1))/x^2# #f'(x) = ((x^2 - x^2 + 1)/sqrt(x^2 - 1))/x^2# #f'(x) = 1/(sqrt(x^2 - 1)x^2)# So now applying the formula, we get: #A = int_-2^-1 sqrt(1 + (1/(sqrt(x^2 - 1)x^2))^2) dx# #A = int_-2^-1 sqrt(1 + 1/(x^6 - x^4)) dx# #A = int_-2^-1 sqrt((x^6 - x^4 + 1)/(x^6 - x^4)) dx# This integral doesn't have an elementary solution. According to Wolfram Alpha, the approximation for this integral would be #1.48# units. Hopefully this helps!

Elijah Abram · Answer

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

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

where $a = -2$ and $b = -1$. First, we find $ \frac{dy}{dx} $ by differentiating $ f(x) $:

$$ f'(x) = \frac{d}{dx} \left( \frac{\sqrt{x^2 - 1}}{x} \right) $$
$$ = \frac{d}{dx} \left( \frac{(x^2 - 1)^{1/2}}{x} \right) $$
$$ = \frac{1}{x} \cdot \frac{1}{2} (x^2 - 1)^{-1/2} \cdot 2x - (x^2 - 1)^{1/2} \cdot \frac{1}{x^2} $$
$$ = \frac{1}{x} \cdot \frac{x}{\sqrt{x^2 - 1}} - \frac{\sqrt{x^2 - 1}}{x^2} $$
$$ = \frac{1}{\sqrt{x^2 - 1}} - \frac{\sqrt{x^2 - 1}}{x^2} $$

Now, we can plug this into the arc length formula and integrate:

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

This integral can be challenging to solve analytically, but you can approximate it using numerical methods.