Find the line integral of f(x,y) = ye^(x^2) along the curve r(t) = -4t i - 3t j from t = -2 to t = 0

Question

Vocaloid · Answer

Please do not comment unless you are actually familiar with the subject and are making a genuine attempt to help.

sillybilly123 · Answer

$\int ~ ds ~ ~ ~ \{y e ^ {x^2} \}$

By Def:

$ ds = \sqrt{dx^2 + dy ^2 } ~ \cdot ~ dt$

$x= - 4 t, ~ dx = - 4 ~ dt$

$y= - 3 t, ~ dy = - 3 ~ dt$

$ ds = \sqrt{16  + 9  } ~ dt = 5 ~ dt$

$\implies 5 \int ~   dt ~ ~  (- 3t) e ^ {16 t^2} $

$= - \dfrac{15}{32}\left[ e ^ {16 t^2} \right]_{-2}^{0}$

$=  \dfrac{15}{32}  e ^ {64} $

if that's rubbish, do holler :)

Vocaloid · Answer

thank you C:

sillybilly123 · Answer

Ouch.  By Def 🤨🥴

$ds = \sqrt{\dot x^2 + \dot  y ^2 } ~ \cdot ~ dt ~ ~ \because  ds^2 = dx^2 + dy^2$