Q. 39 - 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 $u$.

$f(x,y)=xy+2x-y\text{ at }P=(1,-2)$

Step-by-Step Solution

Answer

The given function's directional derivative is $\nabla f\left(P\right).u=0$.

Step 1: Unit vector.

We've $f(x,y) =xy+2y-y$ placed a unit vector $P(1,-2)$ at the supplied location $u=(\alpha ,\beta )=(1,1)$.

Step 2: Derivation of Function.

With directional unit vector $u$, the directional derivative of a function at point $p$ is given by,

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

$=\left[(y+2{)}_{(1,-2{)}^2}i+(x-1{)}_{(1,-2)}j\right]\times \left(\frac{i+j}{\sqrt{2}}\right)$

$=\left[\right(-2+2)i+(1-1\left)j\right]\times \left(\frac{i+j}{\sqrt{2}}\right)$

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

$\nabla f\left(P\right).u=0$