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

[Math Amateur]

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

 

[GJA]

Hi Peter,

This is a nice question.

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: $$\left\|a+t\dfrac{a}{\|a\|}\right\|=\left[\left(a+t\dfrac{a}{\|a\|}\right)\cdot \left(a+t\dfrac{a}{\|a\|}\right)\right]^{1/2}.$$

 

[Opalg]

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 $1+\dfrac t{\|a\|} > 0$).

 

[Math Amateur]

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