Difficult matrix proof...

[nabran]

Gradient Vector Check+Help?

Hello! I'm finding this question challenging.... if someone could let me know if I'm on the right path it'd be brilliant

So; I've trying to find the gradient vector at a general point (X,Y) in

F(X,Y)=2X^(3/4)Y^(1/4)

(that's meant to be X to the power of three quarters, and Y to the power of a quarter)

I just want to check that my ordering and layout are correct for the answer, on the first line of my 2 by 1 vector I have

(3/2)Y^(1/4)X^(-1/4)
Beneath it, on the second line, I gain
(1/2)X^(3/4)Y^(-3/4)

I have to input X=15 and Y=50, I got an answer of 0.203. I just want to know if what I've done is correct, because I'm not too sure about my vector gradient, thanks

I think I've done that well, just a bit unsure.... if I have done it correctly, could someone please show me how they'd go about finding the length of the gradient at this point (i.e, when X=15 and Y=50)

 

[nabran]

Thank you, I'll keep trying on the next part of the question, I'd really appreciate feedback

 

[nabran]

New problem set above!!

 

[akoaysigod]

A gradient is a vector, so when you evaluate it at (15,50), you should get another vector. And what do you mean length? From where to where? It's been a while but you should have a formula in your book for the norm, length, magnitude, or whatever they call it, which will look something like

||V||=sqrt((v1)^2+(v2)^2+(v3)^2+...+(vn)^2) or something similar.

 

[renegade05]

HELLO nabran,

First I am not sure why you titled this thread "matrix proof". What are we proving?

Anyway, your partial derivatives are correct.

Now:
\(\displaystyle \large \vec{\nabla}f(a,b)=<f_x(a,b),f_y(a,b)>\)
\(\displaystyle \large \vec{\nabla}f(15,50)=<f_x(15,50),f_y(15,50)>\)
\(\displaystyle \large \vec{\nabla}f(15,50)=<2.0268,0.0268>\) Note: This is to 4 decimal places, not exact.

So this is your gradient vector at the point (15,50,f(15,50))

The length of the gradient represents the slope at the point (a,b,f(a,b))

Applied to your problem we have:

\(\displaystyle \large ||\vec{\nabla}f(15,50)||=\sqrt{2.0268^2+0.0268^2} = 2.0270\)

The interpretation of this is: at the point (15,50,f(15,50)) every unit moved in the horizontal direction you will move 2.0270 in the vertical direction if you follow the gradient vector.