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.

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

6 0

3 years ago

Carlos finished 1/3 of his art project on Monday. Tyler finished 1/2 of his art project on Monday. Who finished more of hood art

vladimir2022 [97]

Tyler finished more of the art project.

Carlos finished 1/3
In terms of percentage 1/3 *100% = 33.3%

Tyler finished 1/2
In terms of percentage 1/2 *100% = 50%

50% > 33.3%
50% is greater than 33.3%

4 0

3 years ago

Read 2 more answers