Question

What is the maximum rate of change of `f(x,y) = y^2/x` at the point `2,4`?

Answer 1

I think you're asking about the directional derivative here, and the maximum rate of change which is the gradient , leading to the normal vector `vec n`.

So for scalar `f(x,y) = y^2/x`, we can say that:

`nabla vec f = langle - y^2/x^2, (2y)/x rangle = vec n`

And:

`vec n_{(2,4)} = nabla f_{(2,4)} = langle -4, 4 rangle`

So we can conclude that:

`abs( vec n_{(2,4)}) = abs( langle -4, 4 rangle ) = 2 sqrt2`