Thread: [SOLVED] finding the gradient vector

Well this is the question-

df/dx( 2,-3) = 7 and df/dy (2,-3)=11

i) Find the equation of tangent I get z-f(2,-3)=7(x-2)+11(y+3)
ii) Find the directional derivative Duf(2,-3) in the direction of u=5i-2j
I get 7(5/root29) -11(2/root29) then just simpfly

now my real question is this

Find a non-zero vector v such that Dvf(2,-3)=0

What is the proper working out for this i can just make up any vector provided when I take the gradient vectors dot product i get zero..

can someone show me the proper way ?

For any vector a i + b j

a perpendicular vector is then simply -b i + a j

Of course there are others

so if i write any vector but provided their dot products equate to 0 its still true ??

I like that -bi +aj method ??

soo what i did still makes it right ?

Yes any perpindicular vector will do-- but I find when I have an automatic

I use it.

In 2 space all perpindicular vectors will be a scalar multiple of -b i + a j

Thanks I will use that way it seems more efficient