Gradient and directional derivative

Question

A hill can be modeled with the equation H=100−x^4−3y^2 , where H denotes the elevation. A ball is placed on a hill at x0, y0. Find the initial direction in which the ball would roll if it would be released from rest. Express this as a cartesian vectors.

 

What is have so far:

 

f(x,y)=100-x⁴-3y²

grad f(x,y)=<-4x³, -6y>

so -4x³ in ax direction and -6y in ay direction. 

 

How do i know which way the ball moves and how to express this properly in cartesian vectors?

Answer 1

Denise,

              The ball, when placed on the surface H(x,y) has potential energy mgH where m is the mass of the ball. 

Newtons 2nd law is m*a=-grad(mgH(x,y)) where a is the acceleration of the ball. The mass cancels and the acceleration(vector) is equal to the- gradient of H which is 4gx0³i+6gy0j where i and j are unit vectors in the x and y directions and is the direction of motion of the ball at x0,y0

 

Hope this helps

 

Jim