What is the arc length of #f(x)=1/x-1/(5-x) # in the interval #[1,5]#?

Question

Cameron Alexander · Accepted Answer

Infinite or arguably undefined, since #f(5)# is undefined.

#f(x)# has a vertical asymptote at #x=5# so the arc length in the interval #[1, 5)# is infinite. You could say that the arc length in #[1, 5]# is undefined since #f(5)# is undefined. graph{1/x-1/(5-x) [-10, 10, -5, 5]}

Charles Arocha · Answer

undefined To find the arc length of the function $f(x) = \frac{1}{x} - \frac{1}{5-x}$ in the interval $[1,5]$, you can use the arc length formula:

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

1. First, find $\frac{dy}{dx}$ by taking the derivative of $f(x)$.
2. Then, plug the derivative into the arc length formula.
3. Integrate the resulting expression over the interval $[1,5]$.

Here's how you can do it:

1. Compute the derivative $f'(x)$:
$$ f'(x) = -\frac{1}{{x^2}} + \frac{1}{{(5-x)^2}} $$

2. Compute $\left(\frac{dy}{dx}\right)^2$:
$$ \left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{{x^2}} + \frac{1}{{(5-x)^2}}\right)^2 $$

3. Add 1 to $\left(\frac{dy}{dx}\right)^2$:
$$ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\left(-\frac{1}{{x^2}} + \frac{1}{{(5-x)^2}}\right)^2\right) $$

4. Integrate $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$ from 1 to 5:
$$ L = \int_1^5 \sqrt{1 + \left(-\frac{1}{{x^2}} + \frac{1}{{(5-x)^2}}\right)^2} \, dx $$

5. Evaluate the integral numerically to find the arc length.

The resulting value of $L$ will give you the arc length of the function $f(x)$ in the interval $[1,5]$.