Q. 69 - Calculus

Question

Let $(a, b)$ be a point in the domain of the function of two variables, $f(x,y)$, and $\mathbf{u}$ be a unit vector for which $D_{u}f(a,b)$ exists. Prove that$D_{-\mathrm{u}}f(a,b)=-D_{\mathrm{u}}f(a,b).$

Step-by-Step Solution

Answer

Hence proved is $=-D_{\mathrm{u}}f(a,b)$

Step1: Variable function.

Let $(x,y)$ be a two-variable function and $\mathbf{u}=⟨\alpha ,\beta ⟩$,  be a unit vector for  $D_{-u}f(a,b)$The objective is to prove that

$D_{-\mathbf{u}}f(a,b)=-D_{\mathbf{u}}f(a,b)$

The direction derivative of a function $f(x,y)$ at $\left(x_0,y_0\right)$in the direction of $\mathbf{u}=⟨\alpha ,\beta ⟩$ is given by

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

Thus,

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

Step2: Find the Equations.

For $-\mathbf{u}=⟨-\alpha ,-\beta ⟩$

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

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

Let $h=-\eta$then

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

$=-D_{\mathrm{u}}f(a,b)$ from$\left(1\right)$