How do I find directional derivatives using partial derivatives?

[Arden1528]

I am completely stuck on these. I am supposed to find the directional derv. of 1+2x(y)^(1-2) with point (3,4) and vector (4,-3)

I understand the formula to find this. You have to find the partial derv. of x and y then plug in (3,4) to get a value. Then take the partial derv. of x and multiply it by a, and partial derv. of y multiplied by b.

fx(3,4)a+fy(3,4)b

fx=partial derv of x
fy = partial derv. of y

the answer in the back of the book is 23/10, I get 23/3. So it must be in my partials? any help is very much apperciated.

 

[Hurkyl]

Don't forget that your direction vector is supposed to be a unit vector.

(4, -3) isn't quite a unit vector, so you have to scale it so it is a unit vector before you use it in the problem.


You might want to check the rest of your work again too.

 

[Arden1528]

What can I do to the vector (4,-3) to make it a unit vector? Multiply it till the legnth is 1?

 

[StephenPrivitera]

Originally posted by Arden1528
What can I do to the vector (4,-3) to make it a unit vector? Multiply it till the legnth is 1?

Basically.
You want sqrt(x²+y²)=1 where x and y are the i and j components of the vector.
So
x²+y²=1
Presently you have
x²+y²=25
If you divide each side by 25 then
sqrt((x/5)²+(y/5)²)=1
So if the vector is (4/5,-3/5) then you have a unit vector.

 

[Arden1528]

Thank you so much. If I could buy you a beer and cigar I would. That is all I needed, know it all makes sense. I can not tell you how much I am greatfull. I have that feeling of solving that ever so long math problem, it's great.