Q. 4 - Calculus

Question

 Let $\mathbf{u}$ be a unit vector in ${\mathrm{ℝ}}^2$.

(a) Explain why $-\mathbf{u}$ is a unit vector.

(b) If (a, b) is a point in the domain of the function of two variables, f(x, y), at which $D_{\mathbf{u}}f(a,b)$ exists, what is the relationship between $D_{\mathbf{u}}f(a,b)$and $D_{-\mathbf{u}}f(a,b)$ ?

Step-by-Step Solution

Answer

(a) $-u$ is a unit vector because $|-u|=1$

(b) $$\begin{gathered} D_{-a}f(a,b)=-D_{u}f(a,b) \\ {D}_{u}f(a,b)=-D_{-a}f(a,b) \end{gathered}$$

Step 1 : Introduction

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

Step 2 :

Let $u=\alpha i+\beta j$ be a unit vector in $R^2$.

To prove that $-u=-\alpha i-\beta j$ is also a unit vector

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

$\left|u\right|=|\alpha i+\beta j|$

$=\sqrt{{\alpha }^2+{\beta }^2}$

$=1$

Also

$|-u|=|-\alpha i-\beta j|$

$=\sqrt{(-{\alpha }^2)+(-{\beta }^2)}$

$=\sqrt{{\alpha }^2+{\beta }^2}$

$=1$ [Since $\sqrt{{\alpha }^2+{\beta }^2}=1$]

Thus, -u is a unit vector

Step 3 :

(b) 

Let $f(x,y)$ be  a two variable function and $u=(\alpha ,\beta )$ 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 $u=<\alpha ,\beta >$ is given by

$D_{u}f(x_0,y_0)=\underset{h\to 0}{\lim }\frac{f(x_0+\alpha h,y_0+\beta h)-f(x_0,y_0)}{h}$

Step 4 :

Thus,

$D_{u}f(a,b)=\underset{h\to 0}{\lim }\frac{f(a+\alpha h,b+\beta h)-f(a,b)}{h} ......\left(1\right)$

for $-u=<-\alpha ,-\beta >$

$D_{-u}f(a,b)=\underset{h\to 0}{\lim }\frac{f(a-\alpha \eta ,b-\beta \eta )-f(a,b)}{\eta }$ $=\underset{\eta \to 0}{\lim }\frac{f(a+\alpha (-\eta ),b-\beta (-\eta \left)\right)-f(a,b)}{\eta }$

Let $h=-\eta$ then

$$\begin{gathered} D_{-a}f(a,b)=\underset{h\to 0}{\lim }\frac{f(a+\alpha h,b+\beta h)-f(a,b)}{-h} \\ =\underset{h\to 0}{\lim }\frac{f(a+\alpha h,b+\beta h)-f(a,b)}{h} \\ =-D_{a}f(a,b) [From (1\left)\right] \end{gathered}$$