Math Is Fun Forum

KarimAzer

Guest

Gradient Vector...help&check please

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

(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

Beneath it, on the second line, I gain

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

That's the 'check' part of the title, the help part, is how do I go about finding the length of the gradient at this point(when X is 15 and Y is 50?)

Thanks

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Gradient Vector...help&check please

Hi;

Looks like KarimAzer means this:

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: Gradient Vector...help&check please

hi KarimAzer

In Latex, enclose your powers in these brackets {}.  Not needed for a single character.

What you have worked out is correct for the two components so the answer (for x = 15, y = 50) is

Interesting that the two components have the same digits.

The gradient vector is a vector so should be 2 x 1

http://mathworld.wolfram.com/Gradient.html

And what do you mean by its length?  I suppose you could compute the magnitude of the vector by

For x = 15 and y = 50 I made this 2.036909

Bob

Last edited by Bob (2011-10-14 00:46:33)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KarimAzer

Guest

Re: Gradient Vector...help&check please

Thank you both! I apologize for my awful subscript, I'll make sure I nail Latex next time until it's correct.

Thank you Bob; I'm glad I've got it correct so far because I keep missing out little things here and there.

Regarding the last part of the question; it asks what is the length of the gradient at this point; i'm unsure as to whether it means the magnitude or the directional derivitive. Bit of a dodgy question it seems

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: Gradient Vector...help&check please

hi

Let's take a risk and go for the magnitude.  That seems best to me.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

KarimAzer

Guest

Re: Gradient Vector...help&check please

hi

Let's take a risk and go for the magnitude.  That seems best to me.

Bob

Ok Bob, thank you.

And thank you for checking it through, I really wasn't sure if I'd differentiated it correctly!

KarimAzer

Guest

Re: Gradient Vector...help&check please

Also, isn't the square root of  2725 52.2015? Is that the magnitude?

KarimAzer

Guest

Re: Gradient Vector...help&check please

Ahhh scrap that ^^. I understand, I find each scalar component first using each entry....

Thank you!