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

a.

`\vec{u}=(\bigtriangledown F(-3,3))/(|\bigtriangledown F(-3,3)|)=(1)/(√(2))[\hat{i}-\hat{j}]` (ascent)

`\vec{u}=-(\bigtriangledown F(-3,3))/(|\bigtriangledown F(-3,3)|)=-(1)/(√(2))[\hat{i}-\hat{j}]` (descent)

b.

`\vec{v}=(1)/(√(2))[\hat{i}+\hat{j}]`

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))(-(x)/(3))\hat{i}+e^(-(x^2/6+y^2/6))(-(y)/(3))\hat{j}\\\\\bigtriangledown F(x,y)=-(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)=-(1)/(3)e^{-((9)/(6)+(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}=√(2)e^(-3)\\\\\vec{u}=(\bigtriangledown F(-3,3))/(|\bigtriangledown F(-3,3)|)=(1)/(√(2))[\hat{i}-\hat{j}]` (ascent)

for the steepest descend you have

`\vec{u}=-(\bigtriangledown F(-3,3))/(|\bigtriangledown F(-3,3)|)=-(1)/(√(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:

`√(v_1^2+v_2^2)=1\\\\v_1 ^2+v_1^2=1\\\\2v_1^2=1\\\\v_1=(1)/(√(2))=v_2`

then, the vector is:

`\vec{v}=(1)/(√(2))[\hat{i}+\hat{j}]`