Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

First I will write what is written in the book and then I will ask the question:

Suppose $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ is a real-valued function. Let $v$ and $x \in \mathbb{R}^3$ be fixet vectors and consider the function from $\mathbb{R}$ to $\mathbb{R}$ defined by $t \mapsto f(x+tv)$. The set of points of the form $x+tv, t\in \mathbb{R}$ is the line $L$ through the point $x$ parallel to the vector $v$.

The function $t \mapsto f(x+tv)$ represents the function $f$ restricted to the line $L$. For example, is a bird flies along this line with velocity $v$ so that $x+tv$ is its position at time $t$, and if $f$ represents the temperature as a function of position, then $f(x+tv)$ is the temperature at time $t$. We may ask: How fast are the values of $f$ changing along the line $L$ at the point $x$ ? Because the rate of change of a function is given by a derivative, we could say that the answer to this question is the value of the derivative of this function of $t$ at $t=0$ (when $t=0$, $x+tv$ reduce to $x$). This would be the derivative of $f$ at the point $x$ in the direction of $L$, that is, of $v$.

What is written in bold is my issue. I don't understand the question very well and I can't translate the answer from the text correctly. Why is the answer to that question the value of the derivative of that function of $t$ at $t=0$. Why $t=0$ and not $t=1$?

Thanks :)

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

I don't know if this will make things more clear for you; hopefully it will help a bit.

The derivative of a function $f: \mathbb{R} \to \mathbb{R}$ at a point $a$ is (as you probably already know): $$ f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{ x- a} = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}. $$

So we look at the difference quotient while we approach the point in questions.

Now for a function $f: \mathbb{R}^3 \to \mathbb{R}$ we also have derivatives. But now there is a direction on the derivative. Consider the point $\bf{a} \in \mathbb{R}^3$ and the direction from that point given by a direction vector $\vec{v}$. Again we want to consider a difference quotient while we vary the $\bf{x}$ along the line with (unit) direction vector $\vec{v}$. That is we are considering the line $L$ given by teh vector equation: $\bf{a} + t\vec{v}$. So looking at the values of $f$ along that line we get the directional derivative: $$ \lim_{t \to 0}\frac{f(\bf{a} + t\vec{v}) - f(\bf{a})}{t}. $$

We need $t$ to approach $0$ because we want the values of $f$ near the point $\bf{a}$.

share|improve this answer
add comment

It is not at $t=0$ nor $t=1$, $t$ is the variable, and you take the limit $t\to 0$ to define the derivative.

When you write the derivative, $t$ doesn't appear anymore. It's written $\partial_{L}f(x)$. It's just like a real function: $f(x)$, when you define the derivative, you take the limit as $\varepsilon$ goes to $0$:

$$ f'(x) = \lim_{\varepsilon \to 0} \frac{f(x+\varepsilon)-f(x)}{\varepsilon}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.