What is the arc length of f(t)=(t^2-4t,5-t)  over t in [3,4] ?

Mar 16, 2018

$L = \frac{1}{2} \left(2 \sqrt{17} - \sqrt{5}\right) + \frac{1}{4} \ln \left(\frac{4 + \sqrt{17}}{2 + \sqrt{5}}\right)$ units.

Explanation:

f(t)=(t^2−4t,5−t)

f'(t)=(2t−4,−1)

Arc length is given by:

$L = {\int}_{3}^{4} \sqrt{{\left(2 t - 4\right)}^{2} + 1} \mathrm{dt}$

Apply the substitution $2 t - 4 = \tan \theta$:

$L = \frac{1}{2} \int {\sec}^{3} \theta d \theta$

This is a known integral:

$L = \frac{1}{4} {\left[\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |\right]}_{3}^{4}$

Reverse the substitution:

$L = \frac{1}{4} {\left[\left(2 t - 4\right) \sqrt{{\left(2 t - 4\right)}^{2} + 1} + \ln | \left(2 t - 4\right) + \sqrt{{\left(2 t - 4\right)}^{2} + 1} |\right]}_{3}^{4}$

Insert the limits of integration:

$L = \frac{1}{2} \left(2 \sqrt{17} - \sqrt{5}\right) + \frac{1}{4} \ln \left(\frac{4 + \sqrt{17}}{2 + \sqrt{5}}\right)$