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  1. unknownunknown
    • one year ago
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    In the second link example the gradient is <3,12> instead of <1,4>. So why in the first link is it treated as <1,2,2> instead of <2,4,4>? (and yes, this is in reference to the gradient not the unit vector).

  2. phi
    • one year ago
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    I think they made a mistake. the *direction* of the gradient can be any size (unit length is nice), but if you want to calculate how much the function changes in the tangent plane with delta x or delta y, you need the correct magnitude of the gradient, (and correct direction). That is, they should use \( 6 \hat{u} \)or equivalently < -2, -4,-4>

  3. IrishBoy123
    • one year ago
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    i wouldn't personally get too hung up on "directions" because the magnitude of the direction vector is not important when you are only talking about direction itself, which is the focus of the initial discussion, but i agree totally that you always need to be aware of the distinction. which leads to the second part of the first answer which looks very dubious, ie the rate of change question. we can check the answer if we have \(T(x,y,z) = x^2 + 2y^2+ 2z^2 \) then we can easily say that \(T(x+ dx,y+dy,z+dz) = (x+dx)^2 + 2(y+dy)^2+ 2(z+dz)^2 \) so: \(dT(x,y,z) = T(x+ dx,y+dy,z+dz) - T(x,y,z)\) with: \(ds = \sqrt{dx^2 + dy^2 + dz^2}\) \(dT(x,y,z) = x^2+2xdx + dx^2 + 2y^2+4ydy+2dy^2+ 2z^2 + 4zdz +2dz^2 - x^2 - y^2 - z^2\) \(= 2xdx + 4ydy+ 4zdz\), as you would expect and of course you get the same result by writing out the partials: \(\delta T = T_x \delta x + T_y \delta y + T_z \delta z\) as there are no constraints on dx, dy dz, ie they are just infinitesimal changes in the scalar field, we can set them all to \(\delta\) so long as we are consistent with the algebra. [to be completely clear about this, we don't need to tie dx, dy,dz into the original \(x^2 + 2y^2+ 2z^2\), we could simply set them to some small number such as 0.0000001 and run this in a spreadsheet and that should work as a numerical solution.] thus, using \(\delta\), we get:\(ds = \sqrt{\delta^2 + \delta^2 + \delta^2} = \sqrt{3} \delta\) and \(dT(x,y,z) = (2x + 4y + 4z) \delta\) such that \(\frac{dT}{ds}_{(1,1,1)} = \frac{1}{\sqrt{3}}(2(1) + 4(1) + 4(1)) = \frac{10}{\sqrt{3}} \ne 3\)

  4. unknownunknown
    • one year ago
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    Thanks for the help guys.

  5. unknownunknown
    • one year ago
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    I got -6 as well.

  6. IrishBoy123
    • one year ago
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    that second paper you posted is pretty useless. this guy is usually awesome. i haven't watched this one but now i will. https://www.youtube.com/watch?v=XZ1QwS1IKgw

  7. unknownunknown
    • one year ago
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    Yeah, all this is located from http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/ Unfortunately though so far in my studies, errors are incredibly rampant in most of the pdfs, making learning difficult

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