# Directional Derivative Example .... SHifrin, Ch.3, Section 1, Example 3 ....

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In summary, Peter needs help understanding an example in Chapter 3, Section 1 of Multivariable Mathematics. He was able to find help from other users on the forum, but would still appreciate some clarification on why it is critically important that $1+\dfrac t{\|a\|} > 0$ for \left\| a+t\dfrac{a}{\|a\|}\right\| to equal \|a\|+ t.
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I am reading the book: Multivariable Mathematics by Theodore Shifrin ... and am focused on Section 3.1 Partial Derivatives and Directional Derivatives ...

I need some help with Example 3 in Chapter 3, Section 1 ...

Example 3 in Chapter 3, Section 1 reads as follows:View attachment 8587In the above text we read the following:" ... ... $$\displaystyle \lim_{ t \to 0 } \frac{ \| a + t \frac{a}{ \| a \| } \| - \| a \| }{ t } = \lim_{ t \to 0 } \frac{ ( \| a \| - t ) - \| a \| }{ t }$$ ... ... "
Can someone please explain exactly how $$\displaystyle \lim_{ t \to 0 } \frac{ \| a + t \frac{a}{ \| a \| } \| - \| a \| }{ t } = \lim_{ t \to 0 } \frac{ ( \| a \| - t ) - \| a \| }{ t }$$ ... ... ... indeed, specifically how $$\displaystyle \| a + t \frac{a}{ \| a \| } \| = ( \| a \| - t )$$ ... ...
Help will be appreciated ...

Peter

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Hi Peter,

This is a nice question.

Peter said:
I am reading the book: Multivariable Mathematics by Theodore Shifrin ... and am focused on Section 3.1 Partial Derivatives and Directional Derivatives ...

I need some help with Example 3 in Chapter 3, Section 1 ...

Example 3 in Chapter 3, Section 1 reads as follows:In the above text we read the following:" ... ... $$\displaystyle \lim_{ t \to 0 } \frac{ \| a + t \frac{a}{ \| a \| } \| - \| a \| }{ t } = \lim_{ t \to 0 } \frac{ ( \| a \| - t ) - \| a \| }{ t }$$ ... ... "
Can someone please explain exactly how $$\displaystyle \lim_{ t \to 0 } \frac{ \| a + t \frac{a}{ \| a \| } \| - \| a \| }{ t } = \lim_{ t \to 0 } \frac{ ( \| a \| - t ) - \| a \| }{ t }$$ ... ... ... indeed, specifically how $$\displaystyle \| a + t \frac{a}{ \| a \| } \| = ( \| a \| - t )$$ ... ...
Help will be appreciated ...

Peter

The short answer is that $\left\| a+t\dfrac{a}{\|a\|}\right\|=\|a\| +t\dfrac{\|a\|}{\|a\|}=\|a\| +t$ because $a$ and $\dfrac{a}{\|a\|}$ point in the same direction and $a$ is a unit vector. There is no need to think about vectors in the plane or some higher dimensional space to justify this. Think about two positive numbers--say $x$ and $y$--on the real line added together. As vectors both positive numbers point to the right, so $|x+y|=|x|+|y|.$ In other words, two vectors sitting on a line are two vectors sitting on line, regardless of whether that line is in 1, 2, 20, or $n$ dimensions.

If that explanation doesn't cut the mustard, so to speak, then it can be rigorously shown by calculating the norm via the dot product: $$\left\|a+t\dfrac{a}{\|a\|}\right\|=\left[\left(a+t\dfrac{a}{\|a\|}\right)\cdot \left(a+t\dfrac{a}{\|a\|}\right)\right]^{1/2}.$$

Alternatively, $\left\| a+t\dfrac{a}{\|a\|}\right\|= \left\|\left(1+\dfrac t{\|a\|}\right)a\right\| = \left|1+\dfrac t{\|a\|}\right|\|a\| = \left(1+\dfrac t{\|a\|}\right)\|a\| = \|a\|+ t$, provided that $t>-\|a\|$ (so that $1+\dfrac t{\|a\|} > 0$).

Opalg said:
Alternatively, $\left\| a+t\dfrac{a}{\|a\|}\right\|= \left\|\left(1+\dfrac t{\|a\|}\right)a\right\| = \left|1+\dfrac t{\|a\|}\right|\|a\| = \left(1+\dfrac t{\|a\|}\right)\|a\| = \|a\|+ t$, provided that $t>-\|a\|$ (so that $1+\dfrac t{\|a\|} > 0$).

GJA, Opalg ... thanks for the help ...

But ... just a clarification ...

Why is it critical/necessary that $1+\dfrac t{\|a\|} > 0$ in order for $$\displaystyle \left\| a+t\dfrac{a}{\|a\|}\right\| = \|a\|+ t$$ ... ... ?EDIT: OK .. sorry... it's obvious... no problem ...

Peter

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## 1. What is a directional derivative?

A directional derivative is a measure of the rate of change of a function in a specific direction. It tells us how the function changes as we move along a particular direction in its domain.

## 2. How is the directional derivative calculated?

The directional derivative is calculated by taking the dot product of the gradient vector of the function and the unit vector in the desired direction.

## 3. What is the significance of the directional derivative?

The directional derivative is useful in optimization problems, as it tells us the direction in which the function is changing the fastest. It can also help us understand the behavior of a function in a specific direction.

## 4. Can you provide an example of a directional derivative?

Suppose we have a function f(x,y) = x^2 + y^2. The directional derivative of this function at the point (1,2) in the direction of (1,1) can be calculated as follows:
D(1,1) = ∇f(1,2) · u = (2x, 2y)(1,1) · (1/√2, 1/√2) = (2, 2) · (1/√2, 1/√2) = 2/√2 + 2/√2 = 2√2

## 5. How is the directional derivative related to the partial derivatives?

The directional derivative can be thought of as a generalization of the partial derivatives. If we take the directional derivative in the direction of the x-axis, we get the partial derivative with respect to x. Similarly, if we take the directional derivative in the direction of the y-axis, we get the partial derivative with respect to y.

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