What is the arclength of #f(t) = (-(t+3)^2,3t-4)# on #t in [0,1]#?

Question

Elijah Abram · Accepted Answer

#L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))# units.

#f(t)=(−(t+3)2,3t−4)# #f'(t)=(−2(t+3),3)# Arclength is given by: #L=int_0^1sqrt(4(t+3)^2+9)dt# Apply the substitution #2(t+3)=3tantheta#: #L=9/2intsec^3thetad theta# This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals: #L=9/4[secthetatantheta+ln|sectheta+tantheta|]# Reverse the substitution: #L=[1/2(t+3)sqrt(4(t+3)^2+9)+9/4ln|2(t+3)+sqrt(4(t+3)^2+9)|]_0^1# Insert the limits of integration: #L=2sqrt73-9/2sqrt5+9/4ln((8+sqrt73)/(6+3sqrt5))#

Elijah Abram · Answer

undefined To find the arc length of the curve described by the function $ f(t) = (-(t+3)^2, 3t-4) $ on the interval $ t $ in $[0,1]$, you can use the arc length formula:

$$ L = \int_{a}^{b} \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt $$

Where $ a $ and $ b $ are the lower and upper limits of the parameter $ t $, and $ dx/dt $ and $ dy/dt $ are the derivatives of $ x $ and $ y $ with respect to $ t $, respectively.

First, find the derivatives:

$$ dx/dt = -2(t+3) $$
$$ dy/dt = 3 $$

Next, plug these derivatives into the arc length formula and integrate over the interval $[0,1]$:

$$ L = \int_{0}^{1} \sqrt{(-2(t+3))^2 + (3)^2} \, dt $$

$$ = \int_{0}^{1} \sqrt{4(t+3)^2 + 9} \, dt $$

$$ = \int_{0}^{1} \sqrt{4t^2 + 24t + 25} \, dt $$

This integral can be solved using standard calculus techniques, such as trigonometric substitution or completing the square.

Once you have found the antiderivative and evaluated it at the upper and lower limits, you will have the arc length of the curve on the interval $[0,1]$.