Walking down the hill...

[Guest]

Hello everyone i tried for a long time today to solve this problem, but I dont really know how to start.


Problem:
z= 4 - x^2 - 2*y^2 is the equation of a hill in a coordiate system where the positive z-axis points straight up. A person starts in (1,1,1) and walks down the hill, always walking in the direction where the hill is the steepest.
Find this way down the hill. (Think of the projection of the way on the x-y-plane as a curve y=f(x))


My toughts:
I see that the grad z is the vector wich points in the direction of the steepest way of the hill in every single point but I dont know how to connect them and make a curve.
I furthermore think that this should not be very difficult since the equation is not very complex.

Anyway, any help would be greatly apprecieated.
And by the way, excuse my bad english

 

[arthur ohlsten]

z=4-x^2-2y^2 let us rewrite the equation
x^2+2y^2=4-z

for z=0
x^2+2y^2=4 divide by 4
x^2/4 + y^2/2=1 a ellipse plot this

as z increases the ellipse's axis get smaller with a limit of z=4

WE HAVE A SYMETRIC HILL WITH CUTS PARALLEL TO THE XY PLANE GIVING ELLIPSES

two points on the line of decent is the top 0,0,4 and 1,1,1
the line has direction of -i -j + 3k
its equation is [x-1]/-1 =[ y-1]/-1 = [z-1]/3

[1-x]=1-y = [z-1]/3 answer


Arthur

please check for errors