Understanding Differentiability in Vector Calculus: Homework Help and Solutions

[Ted123]

Homework Statement

[PLAIN]http://img261.imageshack.us/img261/1228/vectorcalc.png

Homework Equations

The Attempt at a Solution

I have the definition but what do I do with

$f({\bf a+h})-f(\bf{a})={\bf c\times h} + \|{\bf a+h} \| ^2 {\bf c} - \|{\bf a}\| ^2 {\bf c}$ ?

 

[HallsofIvy]

I recommend looking at $\vec{h}$ going to 0 along the coordinate axes separately.

If $\vec{h}= <h, 0, 0>$ then $\vec{c}\times\vec{h}$= <0, -c_yh, c_zh>[/itex], $||\vec{a}+ \vec{h}||^2\vec{c}= (a_x^2h^2+ 2a_xh+ h^2+ a_y^2+ a_z^2)\vec{c}$, and $||\vec{a}||^2\vec{c}= (a_x^2+ a_y^2+ a_z^2)\vec{c}$, so that $||\vec{a}+ \vec{h}||^2\vec{c}- ||\vec{h}||^2\vec{c}= (2a_xh+ h^2}\vec{c}$. Separate that into parts that are linear in h (D) and non-linear in h (E).

Do the same for $\vec{h}= < 0, h, 0>$ and $\vec{h}= < 0, 0, h>$. Show that $\displaytype\lim_{h\to 0}E/h= 0$ from all three directions.

 

[Ted123]

I recommend looking at $\vec{h}$ going to 0 along the coordinate axes separately.

If $\vec{h}= <h, 0, 0>$ then $\vec{c}\times\vec{h}$= <0, -c_yh, c_zh>[/itex], $||\vec{a}+ \vec{h}||^2\vec{c}= (a_x^2h^2+ 2a_xh+ h^2+ a_y^2+ a_z^2)\vec{c}$, and $||\vec{a}||^2\vec{c}= (a_x^2+ a_y^2+ a_z^2)\vec{c}$, so that $||\vec{a}+ \vec{h}||^2\vec{c}- ||\vec{h}||^2\vec{c}= (2a_xh+ h^2}\vec{c}$. Separate that into parts that are linear in h (D) and non-linear in h (E).

Do the same for $\vec{h}= < 0, h, 0>$ and $\vec{h}= < 0, 0, h>$. Show that $\displaytype\lim_{h\to 0}E/h= 0$ from all three directions.

What do you mean by the notation < ... > ?

 

[Bohrok]

What do you mean by the notation < ... > ?

They must be angle brackets for vectors.