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

Question

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

Aurora Alvarado · Accepted Answer

#L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))# units.

#f(t)=(t^2−4t,5−t)# #f'(t)=(2t−4,−1)# Arc length is given by: #L=int_3^4sqrt((2t-4)^2+1)dt# Apply the substitution #2t-4=tantheta#: #L=1/2intsec^3thetad theta# This is a known integral: #L=1/4[secthetatantheta+ln|sectheta+tantheta|]_3^4# Reverse the substitution: #L=1/4[(2t-4)sqrt((2t-4)^2+1)+ln|(2t-4)+sqrt((2t-4)^2+1)|]_3^4# Insert the limits of integration: #L=1/2(2sqrt17-sqrt5)+1/4ln((4+sqrt17)/(2+sqrt5))#

Emilia Aguilar · Answer

undefined To find the arc length of the curve represented by the parametric equations $ f(t) = (t^2 - 4t, 5 - t) $ over the interval $ t $ in $[3,4]$, we use the formula for arc length:

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

We first find the derivatives $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:

$$ \frac{dx}{dt} = \frac{d}{dt}(t^2 - 4t) = 2t - 4 $$
$$ \frac{dy}{dt} = \frac{d}{dt}(5 - t) = -1 $$

Substituting these derivatives into the arc length formula:

$$ L = \int_{3}^{4} \sqrt{(2t - 4)^2 + (-1)^2} dt $$

$$ L = \int_{3}^{4} \sqrt{4t^2 - 16t + 17} dt $$

$$ L = \int_{3}^{4} \sqrt{4(t^2 - 4t + \frac{17}{4})} dt $$

$$ L = \int_{3}^{4} 2\sqrt{t^2 - 4t + \frac{17}{4}} dt $$

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

$$ L = 2\left[\frac{1}{2} \ln\left| 2t - 4 + 2\sqrt{t^2 - 4t + \frac{17}{4}}\right|\right]_{3}^{4} $$

$$ L = \ln\left| 2\sqrt{t^2 - 4t + \frac{17}{4}} + 2t - 4\right| \Bigg|_{3}^{4} $$

$$ L = \ln\left| 2\sqrt{4 - 4(4) + \frac{17}{4}} + 2(4) - 4\right| - \ln\left| 2\sqrt{9 - 4(3) + \frac{17}{4}} + 2(3) - 4\right| $$

$$ L = \ln\left| 2\sqrt{1} + 4\right| - \ln\left| 2\sqrt{\frac{1}{4}} + 2\right| $$

$$ L = \ln\left| 2 + 4\right| - \ln\left| 2 + 2\right| $$

$$ L = \ln(6) - \ln(4) $$

$$ L = \ln\left(\frac{3}{2}\right) $$

Therefore, the arc length of the curve over the interval $ t $ in $[3,4]$ is $ \ln\left(\frac{3}{2}\right) $.