Find a vector such that directional derivative is zero

[username12345]

Homework Statement

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

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.

 

[sutupidmath]

well, you probbably know that:

$$D_uf=grad{f}*u=|grad{f}||u|cos \theta=|grad{f}|$$

you only need to fill in the details. theta is the angle between grad{f} and u.

 

[HallsofIvy]

Homework Statement

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

You mean $$D_v f= 0$$?

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: $D_v f= \nabla f\cdot v$. The dot product of those two vectors will be 0 when they are perpendicular. Of course, a unit vector that makes angle $\theta$ with the x-axis is $cos(\theta)\vec{i}+ sin(\theta)\vec{j}$ so you can also say that the derivative in the direction making angle $\theta$ with the x-axis is $cos(\theta) \partial f/\partial x+ sin(\theta)\partial f/\partial y$.

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.

 

[username12345]

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) ?

 

[HallsofIvy]

Yes, that would work.

 

[username12345]

Ok, got it, thanks.