0

Im on the mountain $$ z= e^{-2x^2-y^2} $$ at the point $$ (1/2,1/\sqrt{2},e^{-1}) $$ which direction should i go,so that i will remain at the same hieght line ????

thanks

hint the vector should be at XY plane

Styxer
  • 65

1 Answers1

0

Hint: you want the vector $\bf v$ such that $\frac{\partial z}{\partial {\bf v}}(p) = 0 $, where $p$ is the given point.

Ivo Terek
  • 77,665
  • thats what i thought, but i dont understand why its true? please explain, – Styxer Oct 13 '14 at 08:54
  • @uberStyx0. You want to keep $z$ constant and move, don't you ? – Claude Leibovici Oct 13 '14 at 08:58
  • @ Claude_Leibovici i also know that but why it should be equals to zero, theres something basic that im missing,if its equals to zero, doesnt it means that the theres no movement at all? – Styxer Oct 13 '14 at 09:07
  • The direcional derivative is the derivative of the function $$t \mapsto f(p + t{\bf v})$$ so, this derivative being zero means that, in the $\bf v$ direction specifically, the derivative is zero, hence there is no variation in this direction. – Ivo Terek Oct 13 '14 at 09:15
  • Great!!!!, thanks but i have i have got something else, i said that u=a+b(vector) and i got $$ -2e^{-1}a+e^{-1}*(-\sqrt(2))b $$ but how do i find what is a and b? – Styxer Oct 13 '14 at 09:26