Vector field of gradient vector and contour plot

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The discussion centers on the relationship between the gradient vector and contour plots for the scalar field defined by the equation xy/3. Participants note that the gradient vector should be perpendicular to the contour graph, although this is not clearly evident in the provided plot. Some suggest that missing level curves may affect the visualization, and decreasing the sampling interval could help. Others assert that the plot does indeed show the gradient as orthogonal to the level curves. The conversation emphasizes the importance of accurately representing scalar fields and their gradients in contour plots.
Leo Liu
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Given the equation ##\frac{xy} 3##. It is a fact that the gradient vector function is always perpendicular to the contour graph of the origional function. However it is not so evident in the plot above. Any thought will be appreciated.
 
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Leo Liu said:
However it is not so evident in the plot above.
Why do you say that? It looks true to me in the plot.
 
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FactChecker said:
Why do you say that? It looks true to me in the plot.
Some of the level curves are missing. Maybe I should decrease the sampling interval for the level curves?
 
The level curves are all hyperbolas. Try imagining a couple more of them and see if it's crosses through the gradients perpendicularly.
 
Leo Liu said:
Given the equation ##\frac{xy} 3##. It is a fact that the gradient vector function is always perpendicular to the contour graph of the origional function. However it is not so evident in the plot above. Any thought will be appreciated.
##xy/3## is a scalar field, it is not an equation. An equation requires an equal sign.

Apart from that, it is quite apparent from the plot that the gradient is orthogonal to all of the level curves you have provided. I do not understand what more you want from such a plot.
 
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Thread 'Does this series converge uniformly?'

First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...
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