0
$\begingroup$

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.

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

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $(1,1)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.