Question

Suppose a tub has the shape of an elliptical paraboloid given by z = ax2+by2 (where a, b are some positive constants). If a marble were released at the point (1, 1, a + b) on the inside surface of the tub, in what direction would it begin to roll? Your answer should be a unit vector in the requested direction.

Answer 1

Answer:

It would roll in this direction.

$\nu = (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})$

Step-by-step explanation:

It would roll to the direction of maximum decrease, which is the -1 times the direction of maximum increase, which is given by the gradient of the function.  

Since  

$z = ax^2 + by^2$

For this case, the gradient of your function would be

$\nabla z = (2ax , 2by)$

And  -1  times the gradient of your function would be

$-\nabla z = (-2ax , -2by)$

Then, at

 $(1,1,a+b),\\x = 1 \\y = 1$

So it would go towards

$v = (-2a,-2b)$

The magnitud of that vector is

$|v| = 2\sqrt{a^2+b^2}$

and to conclude it would roll in this direction.

$\nu = (-a/\sqrt{a^2+b^2},-b/\sqrt{a^2+b^2})$