Axis Through x in the direction u .... in R^k .... ....

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The discussion centers on the interpretation of a statement from Shmuel Kantorovitz's book "Several Real Variables," specifically regarding the concept of an "axis" through a point \(x\) in the direction of a unit vector \(u\) in \(\mathbb{R}^k\). The parametric representation of this line is given by \(\gamma(t) = x + tu\). Participants clarify that this terminology is appropriate as it relates to the directional derivative of a function \(f\) restricted to this line, where \(t\) serves as an independent variable axis in a new coordinate system. The directional derivative is expressed as the partial derivative \(\partial f/\partial t\).

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I am reading the book "Several Real Variables" by Shmuel Kantorovitz ... ...

I am currently focused on Chapter 2: Derivation ... ...

I need help with an interpretation of a statement by Kantorovitz near to the start of Chapter 2 ...

The start to Chapter 2 in Kantorovitz reads as folows:
View attachment 7800
https://www.physicsforums.com/attachments/7801
At the end of the above quoted text we read the following:

" ... ... In general, for any unit vector $$u \in \mathbb{R}^k$$ and $$x \in \mathbb{R}^k$$, the axis through $$x$$ in the direction $$u$$ is the directed line with the parametric representation

$$\gamma \ : \ t \in \mathbb{R} \rightarrow \gamma (t) := x + tu \in \mathbb{R}^k$$ ... ... "
My question is as follows:

Why does Kantorovitz refer to the above line as "the axis through $$x$$ in the direction $$u$$" ... surely it is just a line as in my diagram below showing the line through $$x$$ in the direction $$u$$ in $$\mathbb{R}^3$$ ... it is not an axis but simply a line ..https://www.physicsforums.com/attachments/7802The required equation of the line, I think, arises as follows:Consider an arbitrary point, $$P$$, on the line given by $$\gamma (t)$$ where $$t \in \mathbb{R}$$.$$u \in \mathbb{R}^k$$ is a vector parallel to the direction of the line ...... we have that $$\gamma (t) = OP$$$$\Longrightarrow \gamma (t) = OP_0 + tu$$$$\Longrightarrow \gamma (t) = x + tu $$
Is that a correct interpretation of the line/axis through x in the direction u ... ?

Peter
 
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Hi Peter,

As I understand, you are wondering why the line through $P_0$ is called an axis. The idea is that we will consider the restriction of $f$ on that line; in this case, $f$ is a function of $t$ only, and you could imagine plotting a graph of $f$ in a coordinate system where $t$ would be the axis of the independent variable and you would plot the value of $f$ on another axis (outside $\mathbb{R}^k$). The directional derivative of $f$ is simply the derivative of $f(t)$ on that graph.

You could also imagine setting up a new coordinate system centered at $P_0$ with one axis $t$ in the direction $\mathbf{u}$. The directional derivative of $f$ is simply the partial derivative $\partial f/\partial t$.
 
castor28 said:
Hi Peter,

As I understand, you are wondering why the line through $P_0$ is called an axis. The idea is that we will consider the restriction of $f$ on that line; in this case, $f$ is a function of $t$ only, and you could imagine plotting a graph of $f$ in a coordinate system where $t$ would be the axis of the independent variable and you would plot the value of $f$ on another axis (outside $\mathbb{R}^k$). The directional derivative of $f$ is simply the derivative of $f(t)$ on that graph.

You could also imagine setting up a new coordinate system centered at $P_0$ with one axis $t$ in the direction $\mathbf{u}$. The directional derivative of $f$ is simply the partial derivative $\partial f/\partial t$.

Thanks castor28 ... well ... that cleared up that matter ... ! ... appreciate the assistance ...

Thanks again ...

Peter
 

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