What is the Directional Derivative at a Given Point?

[jegues]

Homework Statement

See first figure.

Homework Equations

N/A

The Attempt at a Solution

See second figure. I defined direction of the line by which the two planes intersect as,

$$\vec{d}$$

and found that the point they are asking about is when,

$$t=1$$

and I'm stuck here. This is my first attempt at a question like this so I applogize in advance if I'm missing something obvious.

How do I figure out what direction is the direction of increasing x?

$$D_{u^{+}}f(x,y,z) = f_{x}(x,y,z)a + f_{y}(x,y,z)b + f_{z}(x,y,z)c$$

So,

$$D_{u^{+}}f(x,y,z) = ax^{2} + b2y -c2z$$

At the point (-1,1,3),

$$D_{u^{+}}f(-1,1,3) = a + 2b -6c$$

This is as far as I got.

Any suggestions?

 

[Office_Shredder]

To take a simpler example: If you have the function f(x,y)=x²+y², and I told you to take the directional derivative along the line y=x...

The obvious vector to take is the vector (1,1) (scaled to be a unit vector). But that's not the only direction you could pick, (-1,-1) also is a vector pointing along the line y=x. If I addeed: take the direction of increasing x, then you would want to pick the vector that points along the direction in which x increases. This means that the vector pointing in the direction of the line has a positive x coordinate, so the right direction would be (1,1)

 

[jegues]

To take a simpler example: If you have the function f(x,y)=x²+y², and I told you to take the directional derivative along the line y=x...

The obvious vector to take is the vector (1,1) (scaled to be a unit vector). But that's not the only direction you could pick, (-1,-1) also is a vector pointing along the line y=x. If I addeed: take the direction of increasing x, then you would want to pick the vector that points along the direction in which x increases. This means that the vector pointing in the direction of the line has a positive x coordinate, so the right direction would be (1,1)

Okay so the line I'm given is as follows,

$$\vec{l} = <0, -1, 2> + <-1,2,1>t$$

and for x to increase along the line l,

$$t \leq -1$$.

so if I simply compute t = -1, I'll get a vector with the direction,

$$\vec{u} = <1,-3,1>$$

Scaling this to be a unit vector,

$$\hat{u} = \frac{1}{\sqrt{11}}<1,-3,1>$$

Would this be a vector of magnitude 1 pointing in the direction of increasing x?

 

[jegues]

Bump, still looking for some help on this one.

 

[jegues]

Bumpity bump!

 

[Raskolnikov]

Okay so the line I'm given is as follows,

$$\vec{l} = <0, -1, 2> + <-1,2,1>t$$

and for x to increase along the line l,

$$t \leq -1$$.

so if I simply compute t = -1, I'll get a vector with the direction,

$$\vec{u} = <1,-3,1>$$

That's not right. Your $$\vec{u}$$ is the vector from the origin to the point (1, -3, 1) on line l where t = -1. You want the direction of your line itself for which x is increasing. So you should just have $$\vec{u} = <1, -2, -1>.$$ Make that into a unit vector and finish up.

 

[jegues]

That's not right. Your $$\vec{u}$$ is the vector from the origin to the point (1, -3, 1) on line l where t = -1. You want the direction of your line itself for which x is increasing. So you should just have $$\vec{u} = <1, -2, -1>.$$ Make that into a unit vector and finish up.

Okay so scaling it as a unit vector,

$$\hat{u} = \frac{1}{\sqrt{6}} < 1, -2, -1>$$

Now how can I use this to solve for a, b and c? That's what I want to do next right?

 

[Raskolnikov]

No...that's what you just found. $$\vec{u} = <a, b, c>$$

 

[jegues]

No...that's what you just found. $$\vec{u} = <a, b, c>$$

Oh okay! I think I get it now.

Thanks again.

 

[jegues]

Sorry for ressurecting this thread, but I've got another quick question.

When someone asks me for the directional derivative of something at a given point, is my final answer a scalar?

In this case would it simply be the scalar provided by,

$$\nabla f(-1,1,3)\cdot \hat{u}$$

?

Thanks again!

 

[Raskolnikov]

Yes, it's the scalar product of the gradient and a unit vector.
$$D_{u}f= \nabla f\cdot\frac{\bold{u}}{\left|\bold{u}\right|}$$