Q. 41 - Calculus

Question

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

$f(x,y)=(x+1)y^{2}P=(-1,0)$

Step-by-Step Solution

Answer

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

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

Step1: Expression of solution

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

  $\begin{array}{r}\mathrm{\nabla }f\left(P\right)\times u=\mathrm{\nabla }f(-1,0)\times u=\left({\left.\frac{df}{dx}\right|}_{(-1,0)}i+{\left.\frac{df}{dy}\right|}_{(-1,0)}j\right)\times \left(\frac{i+j}{\sqrt{i^2+j^2}}\right)\end{array}$

$\begin{array}{r}=\left[{\left(y^2\right)}_{(-1,0)}i+(2xy+2y{)}_{(-1,0)}j\right]\times \left(\frac{i+j}{\sqrt{2}}\right)\\ \end{array}$

Step 2: Equation

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

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

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