Question

For the function f(x,y)=4-x^2-2y^2 and P(1,1,1) let C' be the path of steepest descent on the surface beginning at P and let C be the projection of C' on the xy-plane. Find an equation of C in the xy-plane.

Answer 1

steepest change direction is the gradient vector sometimes symbolized as \(\nabla\), which is: \((f_x,f_y)\)
In this case :\(\nabla = (-2x,-4y)\)
Sincé you are ascked for descend you will have to take it with negative sign, so it will be:\(\nabla = (2x,4y)\)

Answer 2

thanks, but I understand finding the gradient but not the actual equation of c

Answer 3

you could use: \(\large \frac{dy}{dx}=-\frac{f_x}{f_y}\)
so: dy/dx=-2x/4y=-x/2y
Nw solve this differential equation to get the curve equation

Answer 4

2ydy=-xdx
y^2=-(1/2)x^2+C
now find C so that the curve passes through P

Answer 5

1=-(1/2)+C
so C=3/2

Answer 6

and curve equation is y^2=-(1/2)x^2+3/2