Gradient and parallel points of a function

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SUMMARY

The discussion focuses on finding points where the gradient of the function f(x,y) = x² + y³ + 2xy is parallel to the vector u = (1,1). The gradient is calculated as ∇f = (2x + 2y, 3y + 2x). The user initially set the gradient equal to u, leading to the equations 1 = 2x + 2y and 1 = 3y + 2x. The correct approach is to set ∇f = c * u for some scalar c, which allows for multiple solutions rather than a single point.

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Homework Statement


Find the set of points where the gradient of f is parallel to u =(1,1) for
f(x,y)=x2 + y3 + 2xy

Homework Equations


the gradient of f(x,y)=((partial derivative of f wrt x), (partial derivative of f wrt y))
u=grad f

The Attempt at a Solution


fx = 2x+2y
fy = 3y+2x

1=2x+2y
1=3y+2x

x=1/2
y=0

I set the partials equal to the respective values of u and solved and got one set of points. I have a feeling this is wrong. I got x= 1/2 and y= 0

I looked through the notes and cannot find anything to help me with this so I came here in hope of getting pointed in the right direction.
 
Last edited:
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You don't need grad(f) = u, they're just parallel, so grad(f) = c*u, for any scalar c.
 

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