Gradient Vector is Orthogonal to the Level Curve

[BennyT]

Homework Statement

Let f(x,y)=arctan(x/y) and u={(√2)/2,(√2)/2}
d.) Verify that ∇fp is orthogonal to the level curve through P for P=(x,y)≠(0,0) where y=mx for m≠0 are level curves for f.

Homework Equations

The Attempt at a Solution

∇f={(y)/(x^2+y^2),(-y)/(x^2+y^2)}
m=1/tan(k) where k=f(x,y) and tan(k)≠0
I'm stuck and very confused. The homework is doomed and turned in, but I still really want to understand how to think about problems like this one. Any help is appreciated.

 

[Orodruin]

What is the tangent vector of the level curve? What is the requirement to be orthogonal to the level curve?

 

[BennyT]

What is the tangent vector of the level curve? What is the requirement to be orthogonal to the level curve?

See that's the thing. I know this is a simple question, but I'm having a total blank. Thank you for your help and I understand if you can't show me more.

 

[HallsofIvy]

Uhh, do know what a level curve is?

 

[BennyT]

Uhh, do know what a level curve is?

I think I do. Its the function f (x,y) equal to some constant, say k. Maybe this is right, maybe not. But when k=f (x,y), k=arctan (x/y) or y=x/(tan (k)), right? So I believe it is the function set so that the set of x and y values must equal the value k. I don't know, thank you for your help.

 

[Orodruin]

Yes, but you really do not need to know what it is to solve this. You have already been told that the line y = mx is a level curve. This is a straight line. What is the tangent vector of a straight line? What is a tangent vector?

 

[HallsofIvy]

I think I do. Its the function f (x,y) equal to some constant, say k. Maybe this is right, maybe not. But when k=f (x,y), k=arctan (x/y) or y=x/(tan (k)), right? So I believe it is the function set so that the set of x and y values must equal the value k. I don't know, thank you for your help.

You also should know that the directional derivative for f(x,y), in direction of unit vector $\vec{u}$, is $\vec{u}\cdot \nabla f$. In particular, that directional derivative is 0 if and only if $\vec{u}$ is perpendicular to $\nabla f$. And, of course, since the function is constant on a level curve, the derivative along it (in its direction) is 0.