Find an equation for all the points $(x,y)$ for which the function $f(x,y)=x^2+y^2−3x−y$ has a maximum rate of change in a direction parallel to the vector $u = (1, 1)$

And here is my attempt to solve it:

  • we take the partial derivatives: $$ \dfrac {\partial}{\partial x} = 2x-3$$ $$ \dfrac {\partial}{\partial y} = 2y-1 $$

  • now we find the unit vector $\dfrac {\left|\vec{v} \right|} {\left|v \right|}$ $= \dfrac {1} {\sqrt 2}$

  • now use the formula that says: $$D_uf(a,b) = f_u′ (a,b) = u·\nabla f(a,b) = u·(f_x(a,b),f_y(a,b))$$

  • Hence the equation is $$\dfrac {2x+2y-4}{\sqrt 2} = \sqrt 2 \{x+y-2\}$$ which is wrong according to my professor. Any help will be highly appreciated.

  • 1
    $\begingroup$ To typeset the character $\partial$ use the command \partial. $\endgroup$ – DMcMor Apr 15 '17 at 21:22

The direction of maximum rate of change parallel to the gradient vector, which in this case is $\vec\nabla f= (2x-3,2y-1).$ So we need this to be proportional to $(1,1).$ This just means that $2x-3 = 2y-1$ which gives $y = x-1.$ So the solutions are all the points on this line.

  • $\begingroup$ i sent a PM to my teacher and he said the following : Your equation/answer can be rewritten as 0x + 0y = 0 which is trivially true for ALL points (x,y) in R^2, so, no - this is not the answer. The problem requires that grad f is parallel to (1,1), which can be written as grad f = k(1,1). Eliminate k and you should get the linear eq. y = x - 1. how do i eliminate k $\endgroup$ – Reddevil Apr 15 '17 at 21:29
  • $\begingroup$ @Reddevil Write out the two equations and it ought to become obvious. $\endgroup$ – amd Apr 15 '17 at 21:45
  • $\begingroup$ @Reddevil It's an equation for each component. $\nabla f = (k,k)$ means $2x-3 = k$ and $2y-1 = k.$ So eliminate $k$ from those two equations and you get the answer. This is what I meant when I said the vector being proportional to $(1,1)$ means its components are equal. $\endgroup$ – spaceisdarkgreen Apr 15 '17 at 21:54

See directional derivative, this question is fairly routine.

$$\nabla_{\mathbf v} f(x,y) = \nabla f(x,y) \cdot \mathbf v$$ gives the rate of change in direction $\mathbf v$ and is itself a map from $\Bbb R^2 \to \Bbb R$. The maximum is a critical point of the total derivative of the directional derivative above, or boundary point.

$$ \nabla_{\mathbf v} f(x,y) = \frac{\partial f(x,y)}{\partial x} v_x + \frac{\partial f(x,y) }{\partial y} v_y = (2x-3)v_x+(2y-1)v_y=2x+2y-4$$

Seems there is a min at negative infinity. Intuitively it seems $f(x,y)$ is a infinite parabola or extruded parabola, and has no maximum rate of change in direction $(1,1)$


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