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

Question

Christopher Agresta · Accepted Answer

Approximate arc length via a numerical method is:

# 146.64 # (2dp) The arc length of a curve: # vec(r) (t) = << f(t), g(t)) >> # Over an interval #[a,b]# is given by: # L = int_a^b \ || vec(r) (t) || \ dt # # \ \ = int_a^b \ sqrt(f'(t)^2 +g'(t)^2) \ dt # So, for the given curve: # vec(r)(t) = (t-2t^3, 4t+1) \ \ \ t in [-2,4] # the arc length is given by: # L = int_(-2)^4 \ sqrt( (d/dt(t-2t^3))^2 + (d/dt(4t+1))^2 ) \ dt # # \ \ = int_(-2)^4 \ sqrt( (1-6t^2)^2 + (4)^2 ) \ dt # # \ \ = int_(-2)^4 \ sqrt( 1-12t^2+36t^4 + 16 ) \ dt # # \ \ = int_(-2)^4 \ sqrt( 36t^4-12t^2 + 17 ) \ dt # The integral does not have an elementary antiderivative,and so we evaluate the definite integral al numerically: # L = 146.64637228 ... #

Christopher Allers · Answer

undefined To find the arc length of the curve $ (t - 2t^3, 4t + 1) $ on the interval $ t \in [-2, 4] $, we use the formula for arc length:

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

Substitute $ x = t - 2t^3 $ and $ y = 4t + 1 $, then find $ dx/dt $ and $ dy/dt $.

$$ dx/dt = 1 - 6t^2 $$
$$ dy/dt = 4 $$

Now, we can plug these derivatives into the arc length formula and integrate over the interval $[-2, 4]$.

$$ L = \int_{-2}^{4} \sqrt{(1 - 6t^2)^2 + (4)^2} \, dt $$

$$ = \int_{-2}^{4} \sqrt{1 - 12t^2 + 36t^4 + 16} \, dt $$

$$ = \int_{-2}^{4} \sqrt{36t^4 - 12t^2 + 17} \, dt $$

This integral may not have a simple closed-form solution, but it can be evaluated numerically using appropriate computational methods or software.