Directional Derivatives and Sperical Coordinates

[minor girder]

Hey! Im having a bit of a problem with a question from my Calculus Homework.

If i have a function, F(x, y)= Whatever function of x ,y, and a direction to go in v, i know to find the directional derivative, $D_{vF}(x,y)=(F_x,F_y)\cdot (v)$

My university lately has been doing questions about optimizing these directional derivatives, ie finding the direction such that the directional derivative is minimised/maximised.

Its fairly simple to see that using $v=(cos(\theta),sin(\theta))$ in 3 dimensions (x,y are inputs, z is the output) is the way to go.

My question is how should i go about this in 4 dimensions? I have 3 dimensions of freedom, so i am thinking of using (Cos,sin,phi)?, for 0<Phi<2Pi how do we think this would go?

Ive attatched the question itself, but i dont want solutions to that question, i am moreso wondering if using (Cos,Sin,Phi) would be the way to go.

[viscid mesa]

Makon W

[wanton aspen]

It's simply asking for the negative gradient in (0,1,1)

[minor girder]

It’s asking which direction that is in. How are you meant to find that?

Because i can do that in R3, by using the vector cos(theta), sin(theta). The directional derivative is then a R2 function of Theta, which makes it very easy to find the min/max, just solve for f’(theta)=0

But generalising this up to a function in R4 is a little harder

[tender flare]

What is the 4th dimension here?

[minor girder]

B=f(xyz) XYZ is the input, B is the output. 4 dimensions.

[tender flare]

ah mb, I read the z^2 as y^2.

[tender flare]

same procedure still applies. You have a function f(x,y,z) and you know that the direction of steepest ascent at some point p in R^3 is equal to the gradient of f. So steepest descent will be the negative of that direction

I just reread the post, it seems like you are just struggling with computing the gradient?

I would stay in cartesian here since the partials look quite straight forward

for example: for partial y treat the non y variable functions as constants so you can let c = e^{-x^2}, d = 4/z^2

partial f/ partial y = partial_y ( c*e^{-y^2} + d ) = -2yc e^{-y^2}

[wanton aspen]

The components are simply the partial derivatives with respect to each variable, respectively.

[viscid mesa]

bacc (unhelpful)

[wanton aspen]

Since you have 3 independent variables so w = f(x,y,z) you will have 3 partials with respect to x,y and z