Gradient Vector Check?

nabran · **Joined:** Thu, 13 Oct 2011 17:01:28 UTC **Posts:** 16

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

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

3/2Y^1/4X^-1/4

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

outermeasure · **Posted:** Fri, 14 Oct 2011 13:02:00 UTC

nabran wrote:

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

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

3/2Y^1/4X^-1/4

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

You mean $F(X,Y)=2X^{3/4}Y^{1/4}$ or $F(X,Y)=2X^{\frac{3}{4}}Y^{\frac{1}{4}}$ not F(X,Y)=2X^3/4Y^1/4

--- note the braces. And similarly with your other lines.

So you should get
$\displaystyle \nabla F(X,Y)=\begin{pmatrix}\frac{3}{2}X^{-\frac{1}{4}}Y^{\frac{1}{4}} \\ \frac{1}{2}X^{\frac{3}{4}}Y^{-\frac{3}{4}}\end{pmatrix}$ .

nabran · **Joined:** Thu, 13 Oct 2011 17:01:28 UTC **Posts:** 16

Yaaaay! Thank you; I won't make the mistake again now with my subscript, I can see how you did it! Sorry.

I'm glad I did something correct, so that means my next two parts of this question are correct.Also, I have two questions remaining on this question.

What does the first line in the vector actually mean? If I increase X by 1; the marginal product decreases by ^1/4 each time? I don't really get what this equation shows.

Also, how do you calculate the length of the gradient at the point (X=15, Y=50) in the second entry (Sorry, i got mixed up before, 0.203 refers to the second entry in the vector, not the first). There's not a topic about this in the book, is that simply 50^2-15^2? I'm thinking it can't be since it's a vector and it has more planes.

(I'm sure you'll be glad to know, once I understand this last part of the Q and complete the last part of the other Q which I'm trying now, I can start learning about 'Positive Definiteness' and so will be gone for a long time! - Hehe!)

nabran · **Joined:** Thu, 13 Oct 2011 17:01:28 UTC **Posts:** 16

Done it ^^ thank you!

qwe · **Joined:** Wed, 26 Oct 2011 23:05:31 UTC **Posts:** 1

lol...u must be a student of Uni of Birm trying to do the problem sheet, rite?:P gotcha! lol

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