Find a scalar field that satisfies the following conditions.

Asked Sep 20 '15 at 15:03

Active Sep 20 '15 at 15:03

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QUESTION

Find a scalar field $f$ that satisfies the following conditions:

Partial derivatives equal to $0$.

$f'(\vec x,\vec v)=3$, for $\vec v=\left(\frac {\sqrt2} 2,\frac {\sqrt2} 2\right)$.

ATTEMPT

$\frac {\partial f} {\partial x}=0\implies f=\alpha(y)$, analogous for $y$, so I get $f=\beta(x)$, substracting and rearranging, $\alpha(y)=\beta(x)$ so they must be constants, but this obviously doesn't satisfy the second condition. How can I solve this?

edited Jun 12 '20 at 10:38

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asked Sep 20 '15 at 15:03

YoTengoUnLCD

13,384

By $f'(\vec x,\vec v)=3$ you are talking about the directional derivative, right? – porridgemathematics Sep 20 '15 at 15:15
Correct! ${}{}{}$ – YoTengoUnLCD Sep 20 '15 at 15:15
You are correct. If the partials are identically zero everywhere, then $f$ is constant. And therefore, it's directional derivative is also $0$ for any direction. – Mark Viola Sep 20 '15 at 15:20
Right, it doesn't seem to make sense that the directional derivative for a constant function is positive, so perhaps the question is incorrect? – porridgemathematics Sep 20 '15 at 15:21
I thought that maybe from my conclusions I could only say that if such an $f$ exists, then it would have to be discontinuous, but I'm not sure im right. @Dr.MV doesn't that hold iff $f$ is differentiable? – YoTengoUnLCD Sep 20 '15 at 15:22
Well, the directional derivative is still the dot product of the gradient of the scalar field, and the direction vector. So it is in this case $\frac{1}{\sqrt2}$$\frac {\partial f} {\partial x}$ + $\frac{1}{\sqrt2}$$\frac {\partial f} {\partial y}$ – porridgemathematics Sep 20 '15 at 15:35
This should not ever be equal to $3$ – porridgemathematics Sep 20 '15 at 15:36
Hmm my def of the $f'(x,v)=\lim_{h \to 0} \frac {f(x+hv)-f(x)} h$, which equals $\nabla f \cdot v$ only if $f$ is differentiable – YoTengoUnLCD Sep 20 '15 at 15:38

Find a scalar field that satisfies the following conditions.

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