Consider the function f (x comma y )equals3 x squared minus 2 y squared minus 3 and the point (negative 2 comma 1 ). 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.

we calculate the gradient [tex]\nabla f(x,y) = \frac{\partial f(x,y)}{\partial x} \hat{\textbf{x}}+ \frac{\partial f(x,y)}{\partial y} \hat{\textbf{y}} + \frac{\partial f(x,y)}{\partial z} \hat{\textbf{z}}[/tex]

for each term we consider all variables different to the one we are derivating as constants. For each term we have

[tex]\frac{\partial f(x,y)}{\partial x} = 3 \frac{\partial }{\partial x}(x^2) = 6x\\ \frac{\partial f(x,y)}{\partial y} = -2 \frac{\partial }{\partial x}(y^2) = -4y\\\frac{\partial f(x,y)}{\partial z} = 0\\[/tex]

Therefore:

[tex]\nabla f(x,y) = 6x \hat{\textbf{x}} -4y \hat{\textbf{y}}[/tex]

gives the direction of maxium increase.

a) with x = -2, y= 1

[tex]\nabla f(-2,1) = <-12,-4>[/tex] which magnitude is [tex]\sqrt{(-12)^2+(-4)^2 } = \sqrt{160} = 4 \sqrt{10}[/tex]

so the unitary vector in the direction of the steepest ascent is

[tex]u_{1} = \frac{\nabla f(-2,1)}{|\nabla f(-2,1)|} = \frac{1}{4 \sqrt(10)}*<-12,-4>[/tex]

and the unitary vector in the direction of steepest descent is

[tex]u_{2} = \frac{- \nabla f(-2,1)}{|\nabla f(-2,1)|} = \frac{1}{4 \sqrt(10)}*<12,4>[/tex]

finally, the vector in no change direction is basically doing one of the following possibilities with [tex]u_{1}[/tex]:

if we have a vector <a,b> the perpendicular vector (direction of no change) will be either <-a,b> or <a,-b>

so i will select <-a,b>

[tex]u_{no change} = \frac{1}{4 \sqrt(10)}*<12,-4>[/tex]