Question

Consider the function Upper F (x comma y )equalse Superscript negative x squared divided by 6 minus y squared divided by 6 and the point Upper P (negative 3 comma 3 ). a. Find the unit vectors that give the direction of steepest ascent and steepest descent at P. b. Find a vector that points in a direction of no change in the function at P.

Answer 1

Answer:

a.

$\vec{u}=\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}]$ (ascent)

$\vec{u}=-\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=-\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}]$  (descent)

b.

$\vec{v}=\frac{1}{\sqrt{2}}[\hat{i}+\hat{j}]$

Step-by-step explanation:

a. The function is given by:

$F(x,y)=e^{-(x^2/6+y^2/6)}$

the point is P(-3,3)

a. The unit vector that gives the direction of the steepest ascent is necessary to compute the gradient of F(x,y):

$\bigtriangledown F(x,y)=e^{-(x^2/6+y^2/6)}(-\frac{x}{3})\hat{i}+e^{-(x^2/6+y^2/6)}(-\frac{y}{3})\hat{j}\\\\\bigtriangledown F(x,y)=-\frac{1}{3}e^{-(x^2/6+y^2/6)}[x\hat{i}+y\hat{j}]$

The, it is necessary to evaluate in the point P, and to compute the norm of the vector in order to get the unit vector:

$\bigtriangledown F(-3,3)=-\frac{1}{3}e^{-(\frac{9}{6}+\frac{9}{6})}[-3\hat{i}+3\hat{j}]\\\\\bigtriangledown F(-3,3)=e^{-3}[\hat{i}-\hat{j}]\\\\|\bigtriangledown F(-3,3)|=\sqrt{(e^{-3})^2+(e^{-3})^2}=\sqrt{2}e^{-3}\\\\\vec{u}=\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}]$   (ascent)

for the steepest descend you have

$\vec{u}=-\frac{\bigtriangledown F(-3,3)}{|\bigtriangledown F(-3,3)|}=-\frac{1}{\sqrt{2}}[\hat{i}-\hat{j}]$

b.

the vector with the direction of no change is a vector perpendicular to grad(F):

$\bigtriangledown F(-3,3)\cdot \vec{v}=0\\\\e^{-3}v_1-e^{-3}v_2=0\\\\v_1=v_2$

furthermore, v is an unit vector:

$\sqrt{v_1^2+v_2^2}=1\\\\v_1 ^2+v_1^2=1\\\\2v_1^2=1\\\\v_1=\frac{1}{\sqrt{2}}=v_2$

then, the vector is:

$\vec{v}=\frac{1}{\sqrt{2}}[\hat{i}+\hat{j}]$