Question

Find the sum direction and magnitude of the greatest rate of change of the function z+ln(x+y^2) at (1,1)

Answer 1

it should not say sum... it should just say find the direction and magnitude
my bad

Answer 2

I know you need to take the directional derivative and turn (1,1) into a unit vector and I did that, but i'm stuck at this point. I have \[DuF(x,y) = (1+2y)\div(\sqrt{x}(x+y ^{2}))\]

Answer 3

I’m not very knowledgeable in vector calc, but this comes from
http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx#Gradient_Defn

There’s a theorem in vector calculus that says that
the gradient of a function (grad(f(x)) points in the direction of maximal change,
and that the magnitude of this change is ||grad(f(x))|| at that point x.

If f(x, y, z) = z + ln(x + y^2), then grad(f) = <1/(x + y^2), 2y/(x + y^2), 1>.

Plugging in our point (1, 1), we see that the direction of maximal change is
<1/2, 1, 1>,
and the magnitude of that change is ||<1/2, 1, 1>|| = sqrt(9/4) = 3/2.

Again, if you want, you can check out that link above for more explanation and
the proof for the theorem.

Answer 4

thank you thank you thank you