Find a vector such that directional derivative is zero

  • #1

Homework Statement



Let [tex]f (x, y) = e^x^2 + 3e^y[/tex] . At the point (0, 1) find:
(a) a vector u such that the directional derivative [tex]D_u f[/tex] is maximum and write down this maximum value,
(b) a vector v such that [tex]D_v f = 0[/tex]

Homework Equations



grad f / directional derivative formula


The Attempt at a Solution



I can find part (a), simply calculate the gradient vector, but part (b) I don't know what to do. The answer given is:

(b) Dv f is zero when v is tangent to the level curve passing through (0, 1), i.e. when v = i
(or some multiple of this).

I don't understand what this means.
 
Last edited:

Answers and Replies

  • #2
1,631
4
well, you probbably know that:

[tex]D_uf=grad{f}*u=|grad{f}||u|cos \theta=|grad{f}|[/tex]

you only need to fill in the details. theta is the angle between grad{f} and u.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
964

Homework Statement



Let [tex]f (x, y) = e^x^2 + 3e^y[/tex] . At the point (0, 1) find:
(a) a vector u such that the directional derivative [tex]D_u f[/tex] is maximum and write down this maximum value,
(b) a vector v such that [tex]D_v f[/tex]
You mean [tex]D_v f= 0[/tex]?

Homework Equations



grad f / directional derivative formula


The Attempt at a Solution



I can find part (a), simply calculate the gradient vector, but part (b) I don't know what to do. The answer given is:

(b) Dv f is zero when v is tangent to the level curve passing through (0, 1), i.e. when v = i
(or some multiple of this).

I don't understand what this means.
Along a level curve, the function is a constant. (That's what "level curve" means!). That's why the derivative in that direction is 0. You can also do this by using the formula for derivative in the direction of unit vector v: [itex]D_v f= \nabla f\cdot v[/itex]. The dot product of those two vectors will be 0 when they are perpendicular. Of course, a unit vector that makes angle [itex]\theta[/itex] with the x-axis is [itex]cos(\theta)\vec{i}+ sin(\theta)\vec{j}[/itex] so you can also say that the derivative in the direction making angle [itex]\theta[/itex] with the x-axis is [itex]cos(\theta) \partial f/\partial x+ sin(\theta)\partial f/\partial y[/itex].

It shouldn't be too hard to see that if you are standing on a mountain looking up the steepest direction, the steepest direction down is right behind you and the trail around the mountain, the level curve, is to your side- at right angles to the "steepest" way.

The gradient is always perpendicular to a level curve.
 
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  • #4
The explanation above is good.

So, if say the gradient vector was i + 2j, then a vector such that Dv f = 0 would be +- (2i - j) ?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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964
Yes, that would work.
 
  • #6
Ok, got it, thanks.
 

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