Find the maximum rate of change of f(x,y) = y^2=x at the point (2,4) and the direction in which it occurs.?

Question

Asks for the answer in a unit vector.

How do i go about this question ? I'm struggling. Thankyou!

Sorry! In my tired stupor i typed it wrong, it is meant to say f(x,y) = y^2/x i thank the last answer i found it quite helpful.

Thanks to the last answer i worked out the gradient vector to be 16i + 4j and the unit vector to be 4/sqrt(17) i + 1/sqrt(17) j. I think this is correct if not please let me know :)

Answer 1

Well, to do this... whatever the actual equation is: you need find the gradient vector at that point, and then the corresponding unit vector.

To find the gradient, take the partial with respect to x, evaluated at the point and put that on i. Next, do the same with respect to y and put that on j. To find the unit vector corresponding to this gradient, divide each component by the magnitude of the vector, which would be the square root of the sum of the squares of the two components.

If you meant y^2+x, you would get <1,8> for the gradient vector, and <1/65, 8/65> for the unit vector.

Answer 2

Revised:
f = y²/x
The gradient would be
V = = <-y²/x² , 2y/x> @(2,4) = <-4,4>

||V|| = (4²+4²)^1/2 = 4√2

Unit vector is then 1/(4√2)<-4,4> = <-1/√2 , 1/√2>

The unit vector is the direction and ||V|| is the max rate of change. Hope this helps

Answer 3

Do you mean f(x,y)=y^2+x?

Answer 4

sorry i'm not a nerd like u guys