The gradient?

Calculus Level 3

The gradient of a function of two variables, f ( x , y ) \nabla f(x,y) , is always perpendicular to the graph of the function.

True False

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1 solution

Hamza A
Dec 25, 2018

An important distinction is to be made here. The gradient is perpendicular to the level curves rather than the graph of a function

Yeah....level curves........or as I prefer to use......perpendicular to the contour lines..........

Aaghaz Mahajan - 2 years, 5 months ago

The statement is not even false but meaningless, since the graph resides in R 3 \mathbb{R}^3 while the gradient resides in R 2 \mathbb{R}^2 .

Otto Bretscher - 2 years, 5 months ago

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Yeah as my professor puts it "The gradient lives in the domain"

Hamza A - 2 years, 5 months ago

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Exactly! So maybe you should change the answer to a third option, "the statement is meaningless."

Otto Bretscher - 2 years, 5 months ago

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@Otto Bretscher I think that this question may help in clearing the concepts of some people who are confused regarding gradients..........

Aaghaz Mahajan - 2 years, 5 months ago

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@Aaghaz Mahajan It does not really help to say that the statement is false. The main point is that the gradient and the graph live in different spaces.

As a Swiss, I like to explain this stuff in terms of landscapes ("mountains and valleys") and maps. The gradient lives on the map, and it is meaningless (not even false) to say that it is perpendicular to the landscape.

Otto Bretscher - 2 years, 5 months ago

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@Otto Bretscher Well,Sir....that is obvious....but what I was saying is that maybe some people are not clear with the definition of gradient itself........let alone where it "lives" in........But, that is my opinion.... :) Hummus can always change the options to whatever he seems is correct..............

Aaghaz Mahajan - 2 years, 5 months ago

@Otto Bretscher The statement itself is false regardless if it is deemed meaningless given the context that the gradient lives in the domain. The 2 dimensional gradient here by definition has a z component of 0.

Hamza A - 2 years, 5 months ago

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@Hamza A No! In maths we don't do that; R 2 \mathbb{R}^2 is not a subspace of R 3 \mathbb{R}^3 , although this identification is sometimes made in introductory physics courses, unfortunately. In maths, ( x , y ) (x,y) is not identified with ( x , y , 0 ) (x,y,0) ; as we keep saying: they live in different spaces.

Otto Bretscher - 2 years, 5 months ago

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@Otto Bretscher I see. Can you report it and say that in the report? Thank you.

Hamza A - 2 years, 5 months ago

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@Hamza A Let me do that; it will be interesting to see what the experts from Brilliant think. Another point: The term "three-dimensional function" is unusual at best; the common term is "function of two variables."

Otto Bretscher - 2 years, 5 months ago

1 pending report

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