What is the arclength of f(t) = (te^(2t)-e^t-3t,-2t^2)f(t)=(te2t−et−3t,−2t2) on t in [1,3]t∈[1,3]?

1 Answer
Yonas Yohannes
Mar 9, 2016

L = int_1^3 sqrt([(2t+1)e^(2t)-e^t-3]^2 +16t^2) dtL=∫31√[(2t+1)e2t−et−3]2+16t2dt
After setting the above Integral I used an online Integration Calculator for the numerical approximation. No closed integral exist, but the numerical approximation of L is: L = 1179.856725422904L=1179.856725422904

Explanation:

Given f(t) = (te^(2t)-e^t-3t,-2t^2)f(t)=(te2t−et−3t,−2t2) on t in [1,3]t∈[1,3]
Find the Arclength, L:
L= int_a^b sqrt((f'(t))^2 + (g'(t))^2) dt

f(t) = te^(2t)-e^t-3t
f'(t) = (2t+1)e^(2t)-e^t-3
(f'(t))^2 = [(2t+1)e^(2t)-e^t-3]^2

g(t) =-2t^2
g'(t) = -4t
(g'(t))^2 = 16t^2

L = int_1^3 sqrt([(2t+1)e^(2t)-e^t-3]^2 +16t^2) dt
No closed integral, but numerically you can Approximate, L:
L = 1179.856725422904

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