Calculus III Q: find directions were f(x,y) = 16x^2 - 4y has zero change at (1,1,12)

[miley]

Hi could anyone help me with the following problem?

Find the directions were the function f(x,y) = 16x^2 - 4y has zero change at the point (1,1,12). Express the directions with unit vectors.

My thought was to take the partial derivatives and set them equal to zero:
fx = 32x = 0
fy = 4 = 0

But that obviously doesn't work! Does anyone have any ideas or could anyone help me work through this problem? Thank you!

(Sorry to the moderator, I didn't realize someone had to approve the question first, so I sent it in twice.)

 

[miley]

Calculus III question

Hi, I am having trouble on this problem. Could anyone help me out?

Find the directions where the function f(x,y) = 16x^2 - 4y has zero change at the point (1,1,12). Express the directions with unit vectors.

My thinking was to take the partial derivatives and set them to equal zero, like:
fx = 32x = 0
fy = 4 = 0

But that obviously doesn't work because 4 is not 0. So I am confused about what to do. Could anyone help me out or guide me through the problem please?

 

[HallsofIvy]

What, exactly, do you mean by "zero change"? There cannot be any change at a single point. Do you mean "zero rate of change"? The rate of change of f(x,y) in the direction of unit vector v is given by the dot product \(\displaystyle \nabla f\cdot v\). Here, \(\displaystyle \nabla f= 32x\vec{i}- 4\vec{j}\). At x= 1, = 1, that is \(\displaystyle 32\vec{i}- 4\vec{j}\). Taking unit vector v to be <p, q> we have \(\displaystyle \nabla f(1, 1)\cdot v= 32p- 4= 0\). That is, p= 8. Since that already is larger than 1, there cannot be a unit vector of the form < 8, u>.

 

[miley]

What, exactly, do you mean by "zero change"? There cannot be any change at a single point. Do you mean "zero rate of change"? The rate of change of f(x,y) in the direction of unit vector v is given by the dot product \(\displaystyle \nabla f\cdot v\). Here, \(\displaystyle \nabla f= 32x\vec{i}- 4\vec{j}\). At x= 1, = 1, that is \(\displaystyle 32\vec{i}- 4\vec{j}\). Taking unit vector v to be <p, q> we have \(\displaystyle \nabla f(1, 1)\cdot v= 32p- 4= 0\). That is, p= 8. Since that already is larger than 1, there cannot be a unit vector of the form < 8, u>.

Thanks for answering. I assume the teacher meant to ask in what direction is the rate of change zero from that point. Could you please explain where you got that the rate of change of f(x,y) in the direction of unit vector v is given by the dot product of the gradient of f and v?