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The question gives me a graph with a function and a vector and asks me to determine the sign of the directional derivative at various points. This part is easy. But then, it asks me to justify my answer using the dot product. How do I do that beyond simply saying "the dot product of the gradient and the vector is zero"?

I know this probably doesn't seem worthy of the holy Stack Exchange, but there are no Stack questions like it, the teacher decided not to release solutions, and he doesn't have office hours until the day before the exam so I'm totally stuck.

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  • $\begingroup$ You should type out all the relevant information from question $4$ in your post. You are much much more likely to get helpful answers if you do. $\endgroup$ Commented Feb 21, 2017 at 1:55
  • $\begingroup$ Use this definition/ property of the dot product: $$\vec v\cdot \vec w = \operatorname{sproj}_{\vec w}(\vec v)\|\vec w\| = \begin{cases}\|\operatorname{proj}_{\vec w}\vec v\|\|\vec w\|, & \text{the angle between $\vec v$ and $\vec w$ is $\le \frac {\pi}2$} \\ -\|\operatorname{proj}_{\vec w}\vec v\|\|\vec w\|, & \text{the angle between $\vec v$ and $\vec w$ is $\gt \frac {\pi}2$}\end{cases}$$ where $\operatorname{sproj}$ is the scalar projection operator. I.e., it basically comes down to the sign of the scalar projection. $\endgroup$
    – user137731
    Commented Feb 21, 2017 at 1:59

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Note that the gradient, $\nabla f$, points in the direction of maximum change of $f$ and is, therefore, normal to level curves.

The directional derivative, $\frac{df}{du}=\hat u\cdot \nabla f=|\nabla f|\cos(\theta)$, where $\hat u$ is a unit vector along $\vec u$, and $\theta$ is the angle between $\nabla f$ and $\hat u$.

As one example, at the point $Q$, the angle between the $\nabla f$ and $\hat u$ appears to be less than $\pi/2$ and hence the directional derivative should be positive.

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