0
$\begingroup$

For the following function: $f(x,y)=\ln(x-y)$, at $(2,1)$ what is the unit vector for the minimum directional derivative of $f$ ?

Firstly to find the gradient of $f$ at $(2,1)$: $$ \nabla f(x,y) = \begin{bmatrix} \frac{\partial}{\partial x}(\ln(x-y))\\ \frac{\partial}{\partial y}(\ln(x-y))\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x-y}\\ \frac{-1}{x-y}\\ \end{bmatrix} $$

$$ \nabla f(2,1) = \begin{bmatrix} 1\\ -1\\ \end{bmatrix} = \langle 1,-1\rangle $$

The minimum directional derivative is: $$-|\nabla f(2,1)| = -|\langle 1,-1\rangle|=-\sqrt 2$$

To find the unit vector: $$ u = \frac{\nabla f(2,1)}{-|\nabla f(2,1)|} = \frac{\langle 1,-1\rangle}{-\sqrt2} = \frac{\langle -1,1\rangle}{\sqrt2} $$

$\therefore$ The minimum directional derivative of $f(x,y)$ is $-\sqrt 2$ and occurs in the direction of unit vector $\frac{\langle -1,1\rangle}{\sqrt2}$

Is this the correct procedure to solve the problem? Thanks in advance.

$\endgroup$
1
$\begingroup$

Your proof is correct but is, in my opinion, missing a few words here and there explaining why your procedure is correct.

As an example : it seems to me that you're implicitly using the property that the directional derivative $\partial_v f$ at a point $a$ is the dot product $\nabla f (a) \cdot v$. This is not always the case ! But it is true when the function $f$ is differentiable at $a.$ So your proof should at least contain the words " Since $f$ is differentiable at ..."

Similarly you could be elaborate a bit more on certain parts:

The minimum directional derivative is: $$−|\nabla f(2,1)|=−|⟨1,−1⟩|=−\sqrt 2$$

Why is this true ? You don't need the write a huge paragraph of course ;) but a few extra words here and there would make your proof far more readable.

$\endgroup$

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.