mltr Cứu Mạng chutieu

take gradient of f (x,y) = (2x, 2y)
length of vector = sqrt ((2x)^2 + (2y)^2) @(0,1) = sqrt (0+4) = 2
direction grad f (0,1) = (0, 2)
similar for other problems

I found it out earlier today. Thanks anyway mltr and thanks for helping me out the other times too

you're welcome ... I'm glad you figured it out ... don't hesitate to ask if you have any questions, be sure to include what subjects you're studying. It helps to optimize the way to find the solutions.
June 15th 2011, 12:02 am
I. Given the function f(x,y)=4x^2y^2 .
A. What is the slope of the path at ( 0,1) if one is traveling in the positive y direction?
B. What is the slope at the same point if one is traveling in the positive x direction?
C. Form the vector <partial derivative of x, partial derivative of y> at this point
1. How long is this vector?
2. In which direction does it point?
II. Repeat the process with a new function f(x,y)=e^(x^2y) .
III. Repeat again with the function f(x,y)=2x .
IV. What conclusion can you draw about the relationship between the vector <partial derivative of x, partial derivative of y> and the surface f(x,y) to which it is related?
I do not know how to find C 2 for I II III and IV. I was wrong when I decided to take calc 3 in summer with only 6 weeks. too much hw to handle.