Question

In what direction is the derivative of f( x,y)= xy+y^2 at P(3, 2) equal to zero?

Answer 1

Where you given a direction vector?

Answer 2

doesnt the normal at a given point on a surface point towards the highest point?

Answer 3

not the normal, but the gradient ....

fx = 1, fy = 1+2y

<fx,fy> = <1,5>

Answer 4

the dot product of gradf and the direction vector comes to mind, and they want the to be zero, so the direction vector must be the perp ....

<1,5> perps to <5,-1> or <-5,1>

Answer 5

I'm trying to remember how to do this...
"In what direction is the derivative of.... ....equal to zero"

So we want to set up a directional derivative, and set it equal to zero, and solve for the vector.
\[\huge Dir_u f(x,y)\qquad = \qquad \nabla f \cdot \;\hat u\]
\[\huge Dir_u f(x,y)\qquad = \qquad <f_x,f_y>\cdot \;\hat u=0\]This reads "The directional derivative of f, in the direction of u, equals ..."

Hmm let's see. Taking the partials gives us,\[\large Dir_u f(x,y) \qquad = \qquad <y+y^2,\quad x+2y>\cdot \;\hat u\]

Evaluated at the point (3,2),
\[\large Dir_u f(3,2) \qquad = \qquad <2+2^2,\quad 3+2\cdot2>\cdot \;\hat u\]And then setting this equal to zero,\[\large <6,\quad 7>\cdot \;\hat u \qquad = 0\]

And this is only true when u is perpendicular to our vector <6,7>.
I think @amistre64 had the right idea.
I just think there was a little miscalculation in the partials.

Answer 6

Woops, I messed up my partials also... The second part of \(f_x\) was a constant, which should have become 0. So \(f_x\) should give us \(y\), not \(y+y^2\).

Answer 7

yeah, i messed up on the fx fy parts :) the rest of my recollection seems to have been sound tho ;) thnks