What is the arc length of #r(t)=(te^(t^2),t^2e^t,1/t)# on #tin [1,ln2]#?

Question

Emilia Aguilar · Accepted Answer

Arc Length #~~ −2.42533 \ \ # (5dp)

The arc length is negative due to the lower bound #1# being greater than the upper bound of #ln2# We have a parametric vector function, given by: # bb ul r(t) = << te^(t^2), t^2e^t, 1/t >> # In order to calculate the arc-length we will require the vector derivative, which we can compute using the product rule: # bb ul r'(t) = << (t)(2te^(t^2)) + (1)(e^(t^2)) , (t^2)(e^t) + (2t)(e^t) , -1/t^2 >> # # \ \ \ \ \ \ \ \ = << 2t^2e^(t^2) + e^(t^2) , t^2e^t + 2te^t , -1/t^2 >> # Then we compute the magnitude of the derivative vector: # | bb ul r'(t) | = sqrt( (2t^2e^(t^2) + e^(t^2))^2 + (t^2e^t + 2te^t)^2 + (-1/t^2)^2) ) # # " " = sqrt( e^(2 t) t^4 + 1/t^4 + 4 e^(2 t) t^3 + 4 e^(2 t) t^2 + 4 e^(2 t^2) t^2 + e^(2 t^2) + 4 e^(2 t^2) t^4 ) # Then we can compute the arc-length using: # L = int_(1)^(ln2) \ | bb ul r'(t) | \ dt # # \ \ = int_(1)^(ln2) \ sqrt( e^(2 t) t^4 + 1/t^4 + 4 e^(2 t) t^3 + 4 e^(2 t) t^2 + 4 e^(2 t^2) t^2 + e^(2 t^2) + 4 e^(2 t^2) t^4 ) \ dt # It is unlikely we can compute this integral using analytical technique, so instead using Numerical Methods we obtain an approximation: # L ~~ −2.42533 \ \ # (5dp) The arc length is negative due to the lower bound #1# being greater than the upper bound of #ln2#

Aurora Alvarado · Answer

undefined The arc length of $ r(t) = (te^{t^2}, t^2e^t, \frac{1}{t}) $ on the interval $[1, \ln 2]$ can be found using the formula for arc length: $ L = \int_{a}^{b} \sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2} dt $, where $ (x(t), y(t), z(t)) $ are the parametric equations defining the curve.

In this case, differentiate each component of $ r(t) $ with respect to $ t $, then plug them into the arc length formula and integrate over the given interval $[1, \ln 2]$ to find the arc length.