Q. 40 - Calculus

Question

In exercise $39$-$42$, show that the directional derivative of the given function at the specified point $P$ is zero for every unit vector $u.$

$\begin{array}{r}\begin{array}{r}f(x,y)=3x^2-4xy+2y^{2}P=(0,0)\\ \\ \end{array}\end{array}$

Step-by-Step Solution

Answer

Directional derivation of function at point $P$ with directional unit vector is given by,

                                 $\mathrm{\nabla }f\left(P\right)\times u=0$

Step1: Expression of solution

We have given is $f(x,y)=3x^2-4xy-2y^2$ at specified point $P$$= (0,0)$ with unit vector $u=(1,1)$

$\begin{array}{r}\mathrm{\nabla }f\left(P\right)\times u=\mathrm{\nabla }f(0,0)\times u=\left({\left.\frac{df}{dx}\right|}_{(0,0)}i+{\left.\frac{df}{dy}\right|}_{(0,0)}j\right)\times \left(\frac{i+j}{\sqrt{i^2+j^2}}\right)\\ \end{array}$

Step 2: Calculation

$\begin{array}{r}\mathrm{\nabla }f\left(P\right)\times u=\left[(6x-4y{)}_{(0,0)}i+(-4x-4y{)}_{(0,0)}j\right]\times \left(\frac{i+j}{\sqrt{2}}\right)\\ \end{array}$

$\begin{array}{r}=\left[\right(0-0)i+(0-0\left)j\right]\times \left(\frac{i+j}{\sqrt{2}}\right)\end{array}$

$=[0i+0j]\times \left(\frac{i+j}{\sqrt{2}}\right)$

$\mathrm{\nabla }f\left(P\right)\times u=0$