91影视

The given is the unit vector u. The objective is to find why -u is a unit vector and what is the relationship between $D_{\mathbf{u}}f(a,b)$and $D_{-\mathbf{u}}f(a,b)$. The method used to solve is modulation and direction derivatives of the function

Let $u=伪i+尾j$be a unit vector in $R^2$.

To prove that $-u=-伪i-尾j$is also a unit vector

Since $\mathit{u}\mathbf{=}\mathit{伪}\mathit{i}\mathbf{+}\mathit{尾}\mathit{j}$is a unit vector, so

$\left|u\right|=|伪i+尾j|$

$=\sqrt{{伪}^2+{尾}^2}$

$=1$

Also

$|-u|=|-伪i-尾j|$

$=\sqrt{(-{伪}^2)+(-{尾}^2)}$

$=\sqrt{{伪}^2+{尾}^2}$

$=1$[Since $\sqrt{{伪}^2+{尾}^2}=1$]

Thus, -u is a unit vector

(b)

Let $f(x,y)$be a two variable function and $u=(伪,尾)$be a unit vector for $D_{u}f(a,b)$.

The objective is to establish the relationship between$D_{u}f(a,b)$and $D_{-u}f(a,b)$.

The direction derivatives of the function $f(x,y)$at $(x_0,y_0)$in the direction of $u=<伪,尾>$is given by

$D_{u}f(x_0,y_0)=\underset{h鈫�0}{\lim }\frac{f(x_0+伪h,y_0+尾h)-f(x_0,y_0)}{h}$

Thus,

$D_{u}f(a,b)=\underset{h鈫�0}{\lim }\frac{f(a+伪h,b+尾h)-f(a,b)}{h}......\left(1\right)$

for $-u=<-伪,-尾>$

$D_{-u}f(a,b)=\underset{h鈫�0}{\lim }\frac{f(a-伪畏,b-尾畏)-f(a,b)}{畏}$$=\underset{畏鈫�0}{\lim }\frac{f(a+伪(-畏),b-尾(-畏\left)\right)-f(a,b)}{畏}$

Let $h=-畏$then

localid="1650433862247" $$\begin{gathered} D_{-a}f(a,b)=\underset{h鈫�0}{\lim }\frac{f(a+伪h,b+尾h)-f(a,b)}{-h} \\ =\underset{h鈫�0}{\lim }\frac{f(a+伪h,b+尾h)-f(a,b)}{h} \\ =-D_{a}f(a,b)[From(1\left)\right] \end{gathered}$$