Question

How to find the directional derivative of ? : $f\left( x,y \right)={{e}^{-xy}},P\left( 1,-1 \right),v=-i+\sqrt{3}$

Answer 1

Hint: Let the function $f\left( x,y \right)$ be the height of a mountain range at each point $x=x,y$. If you stand at some point $x=a$, the slope of the ground in front of you will depend on the direction you are facing. It might slope steeply up in one direction, be relatively flat in another direction, and slope steeply down in yet another direction. The partial derivative of $f$ will give the slope $\dfrac{\partial f}{\partial x}$ in the positive $x$ direction and the slope $\dfrac{\partial f}{\partial y}$ in the positive $y$ direction. We can generalize the partial derivatives to calculate the slope in any direction. The result is called the directional derivative.

Complete step by step solution:
Theorem : If $f$ is a differentiable function of $x,y$ , then $f$ has a direction derivative in the direction of any unit vector $\overrightarrow{u}= < a,b > $ and ${{D}_{u}}f\left( x,y \right)={{f}_{x}}\left( x,y \right)a+{{f}_{y}}\left( x,y \right)b$
We are finding out the slope of the function in two different directions using partial differentiation. These directions are given by unit vectors which are of the form $\overrightarrow{u}= < a,b > $ . When we are not given the unit vector but are given an angle $\alpha $, then we obtain the unit vector in the following way :
$\overrightarrow{u}= < \cos \alpha ,\sin \alpha > $ .
Now we are given the angle in the form of a complex number.
We have to find it’s angle.
When a complex number is of the form $a+ib$ , we all know it’s principal argument which is represented by $\theta ={{\tan }^{-1}}\dfrac{b}{a}$ .
The principal argument of the given complex number would be the following :
 $\begin{align}
  & \Rightarrow \tan \theta =\dfrac{-1}{\sqrt{3}} \\
 & \Rightarrow \theta ={{\tan }^{-1}}\dfrac{-1}{\sqrt{3}} \\
 & \Rightarrow \theta ={{120}^{\circ }} \\
\end{align}$
We got our angle.
So the unit vector would be the following :
$\begin{align}
  & \Rightarrow \overrightarrow{u}= < \cos \alpha ,\sin \alpha > \\
 & \Rightarrow \overrightarrow{u}= < \cos \left( {{120}^{\circ }} \right),\sin \left( {{120}^{\circ }} \right) > \\
 & \Rightarrow \overrightarrow{u}= < \dfrac{-\sqrt{3}}{2},\dfrac{1}{2} > \\
\end{align}$
These are the prerequisites for doing directional derivatives which we found out.
Let us use the formula of directional derivative and find it out.
\[\begin{align}
  & \Rightarrow {{D}_{u}}f\left( x,y \right)={{f}_{x}}\left( x,y \right)a+{{f}_{y}}\left( x,y \right)b \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( \dfrac{\partial \left( {{e}^{-xy}} \right)}{\partial y} \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( \dfrac{\partial \left( {{e}^{-xy}} \right)}{\partial y} \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( -{{e}^{-xy}}y \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( -{{e}^{-xy}}x \right)\left( \dfrac{1}{2} \right) \\
\end{align}\]
Let us now substitute the value of $x,y$with the given point $P\left( 1,-1 \right)$.
Upon doing so, we get the following :
\[\begin{align}
  & \Rightarrow {{D}_{u}}f\left( x,y \right)={{f}_{x}}\left( x,y \right)a+{{f}_{y}}\left( x,y \right)b \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( \dfrac{\partial \left( {{e}^{-xy}} \right)}{\partial y} \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( \dfrac{\partial \left( {{e}^{-xy}} \right)}{\partial y} \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( -{{e}^{-xy}}y \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( -{{e}^{-xy}}x \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( -{{e}^{-\left( 1 \right)\left( -1 \right)}}\left( -1 \right) \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( -{{e}^{-\left( 1 \right)\left( -1 \right)}}\left( 1 \right) \right)\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\left( \dfrac{-\sqrt{3}}{2}e \right)+\left( -\dfrac{1}{2}e \right) \\
 & \Rightarrow {{D}_{u}}\left( {{e}^{-xy}} \right)=\dfrac{-e}{2}\left( \sqrt{3}+1 \right) \\
\end{align}\]
$\therefore $ The directional derivative of $f\left( x,y \right)={{e}^{-xy}},P\left( 1,-1 \right),v=-i+\sqrt{3}$ is \[{{D}_{u}}\left( {{e}^{-xy}} \right)=\dfrac{-e}{2}\left( \sqrt{3}+1 \right)\].

Note: Directional derivative is a complex concept . We should fully understand the concept and read more examples to understand where and how it is used. Once we grasp the applicative knowledge, we can easily do the sums. It is important to know the formulae and all about the unit vectors used in it. We should have good knowledge on partial differentiation before learning this concept. We should be aware of all of partial differentiation concepts and it’s different varieties of problems.