Partial derivatives of vector fields, component by component

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Here we step through each partial derivative of each component in a vector field, and understand what each means geometrically.

[@itsmagic2]

Small correction: The partial derivative of Q in respect to x (dQ / dx) at the selected point (2,0) is actually equal to "-4" and not "-2" as shown .

[@Alb-Patriot]

i was just going to write that

[@AmanYadav-bv8lo]

Correct but we have to compromise that, is it not?

[@vgtboy]

3:40

[@datascientistdr.dhannani2944]

Grant Sanderson is the best maths teacher! Hard working n enthusiastic❤

[@omerzaman2373]

I just have one thing to say, I LOVE YOU.

[@Quancept]

thank you. fantastic tutorial!

[@poiuwnwang7109]

@3:38 Should dQ/dx=-4 instead of -2?

[@PhucHoang-ng4vh]

hmm , let choose an arbitrary (x,y) in the first quadrant, now look at x-component (i call it f_1 which is xy),
we have partial derivative d(f_1)/dx = y > 0 -> means a little nudge in x direction cause f1 increase, but i came to the graph and see that the vectors kind of shorter in x direction, they are kind of 'pointing downward'
can anybody explain it to me ? 🤨

[@mounirmohammedamine6108]

Is there any way to have the curvature formula based on partial derivatives with two parameters?

[@krishnadasnair8765]

At 6:47 isn't the y component increasing as we move towards positive y? If yes then how is partial of Q wrt y zero?

[@toufeeqsiddique7520]

Not increasing,, but kind of remains same ... that's why it came out to be 0

[@desrucca]

If you plot it using line instead of vector fields, it will looks like a curve where point (2 , 0) is the critical point (i.e maximum value) **horizontally** of that curve
Since vertical's critical point is df/dx = 0. Then, horizontal's critical point basically means df/dy = 0

[@desrucca]

It is *decreasing* in both positive-negative y direction.
Positive y : decreasing by anti-clockwise rotation
Negative y : decreasing by clockwise rotation

[@karimrafat3800]

I am two years late, but I think I understand it. look at what the change in y component with respect to y came to be for every point not only at a specific point. it is not a constant, but a function of y.
partial of Q wrt y is equal to 2y. but for specific points where y = 0 it gives zero (which means along the x-axis.
the function equal to 2y means that as you move towards positive for larger and larger values of y the y component of the vector v increases faster (in positive direction) whereas for negative values of y it decreases and at some point, you should notice that it reverses its direction.

[@zhenniqi136]

好！！！！！Awesome！！！！！！！

[@technoshrink]

Why do these have so few views?

[@allyourcode]

Yeah, KA should be getting more views.
I think it's because Khan Academy doesn't try to spread its videos on TH-cam. E.g. they NEVER ask people to like and subscribe. They are more focused on getting views from their own website. This seems like a mistake on their part; imho, they should be trying to direct people to their website from their TH-cam videos, but none of the video descriptions link back to the lesson(s) that the video is used in :/

[@18x9]

freat video

[@karthikKarthik-by6ws]

this is the comment after 2 years