Q. 42 - Calculus

Question

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

$f(x,y,z)=x^2+y^2-z^3 \text{ at }P=(0,0,0)$

Step-by-Step Solution

Answer

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

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

Step 1: Expression of solution

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

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

                                                       

Step 2: Equations

$\begin{array}{r}=\left[(2x{)}_{(0,0,0)}i+(2y{)}_{(0,0,0)}j+{\left(3z^2\right)}_{(0,0,0)}j\right]\times \left(\frac{i+j+k}{\sqrt{3}}\right)\\ \end{array}$

$=[0i+0j+0k]\times \left(\frac{i+j+k}{\sqrt{3}}\right)$