Question

Let f be a function of two variables that has continuous partial derivatives and consider the points

Answer 1

Answer:

The directional derivative of f at A in the direction of $\vec{u}$ AD is 7.

Step-by-step explanation:

Step 1:

Directional of a function f in direction of the unit vector $\vec{u}=(a,b)$ is denoted by $D\vec{u}f(x,y)$,

$D\vec{u}f(x,y)=f_{x}\left ( x ,y\right ).a+f_{y}(x,y).b$.

Now the given points are

$A(8,9),B(10,9),C(8,10) and D(11,13)$,

Step 2:

The vectors are given as

AB = (10-8, 9-9),the direction is

$\vec{u}_{AB} = \frac{AB}{\left \| AB \right \|}=(1,0)$

AC=(8-8,10-9), the direction is

$\vec{u}_{AC} = \frac{AC}{\left \| AC \right \|}=(0,1)$

AC=(11-8,13-9), the direction is

$\vec{u}_{AD} = \frac{AD}{\left \| AD \right \|}=\left (\frac{3}{5},\frac{4}{5} \right )$

Step 3:

The given directional derivative of f at A $\vec{u}_{AB}$ is 9,

$D\vec{u}_{AB}f=f_{x} \cdot 1 + f_{y}\cdot 0\\f_{x} =9$

The given directional derivative of f at A $\vec{u}_{AC}$ is 2,

$D\vec{u}_{AB}f=f_{x} \cdot 0 + f_{y}\cdot 1\\f_{y} =2$

The given directional derivative of f at A $\vec{u}_{AD}$ is

$D\vec{u}_{AD}f=f_{x} \cdot \frac{3}{5} + f_{y}\cdot \frac{4}{5}$

$D\vec{u}_{AD}f=9 \cdot \frac{3}{5} + 2\cdot \frac{4}{5}$

$D\vec{u}_{AD}f= \frac{27+8}{5} =7$

The directional derivative of f at A in the direction of  $\vec{u}_{AD}$ is  7.