This is what is good about the internet for education. A good succinct lecture really gets the information across. If you want to learn we are in a golden age. I am very excited for it all.
Another nice graphic application of Grad is that it provides the normal direction of the surface T(x,y)=constant. This idea can be very useful for the geometrical understanding of our math problems. For any surface I can write as f(x,y)=constant, I can obtain the normal of this surface at any surface point just obtaining the Grad vector (whatever the complexity of this surface is). This directly applies to plasticity theory with respect to the yield surface and the normality rule.
Sir, you are the best I have ever known you have done this difficult subject as easy as possible for me and for my friends so you deserve a great thank 🎉🎉❤❤
Correct me if i'm wrong, but i think the radius vector is pointed from the center of the Earth to the mass. This way the minus sign in the equation of force make sense.
You are absolutely correct. The minus sign gains its significance when you define a inertial system (physics term) with unit-vectors with origin at the center of the earth. One equivalent statement comes from linear algebra and the algebra of vectors. The physical “inertial system” is the mathematical equivalence to a set of basis vectors that roughly speaking spans a vector space, with a origin that is constant. So if we would say that the earth rotates around the moon, i.e. we define an inertial system at the center of the moon, then the force would change sign depending on how we choose the basis vectors, or respectively in what way we define the inertial system. (This is crucial in my opinion to understand Steve’s video about Coordinates and Autoencoders (…)) I recommend Princetons book Jeremy Kasdin and Dereks “Engineering Dynamics - A comprehensive Introduction” for a more low level, yet richer, vector formalism used in dynamics. But a high level introduction as this series is worth a lot. Then perhaps you will see that the probability for needing to ask for correction about a fact about vectors and physics might decrease!
At 12:05, shouldn't vector r point outwards from the centre of the Earth? So the gravitational force has the direction towards the centre given the minus sign in the equation.
Dear Professor Brunton, I heard mathematicians treat the output of the gradient operator as a *row* vector (for reasons supposedly to do with differential geometry), while engineers and statisticians typically treat it as a *column* vector. I'm curious if there is a preference for either convention in the optimization / machine learning literature, as well as your own preference.
Row vectors and column vectors DO NOT exist. A vector is a vector, an abstract object, independent on the reference frame or the coordinates you use. Then, you can sort the coordinates (reminding which base vectors are associated with) as you wish: if you're free to choose your convention, choose the one you're more comfortable with, if you're using some libraries check out the documentation to find the convention they're using
Its the small details added that fill in the missing knowledge gaps for beginners. Why are the vector symbol over the grad removed? I needed to know that!!! Thanks teacher.
Gradient alone is not a vector, it is a function. When we apply specific values to a gradient function we obtain the gradient vector in a specific point in a coordinate system. The result represents a vector, then we use vector symbol over that.
I have a question about calculating a gradient of gravitational potetnital: V = -(GmM)/r I can decompose r into a form where I keep each axis separate using pythagorean theorem: r = (x^2 + y^2 + z^2)^(1/2) now V equals: V = -(GmM)*(x^2 + y^2 + z^2)^(-1/2) Now I can calculate a derivative in each direction (x, y, z): dV/dx = -(GmM)*(-1/2)*(x^2 + y^2 + z^2)^(-3/2)*2x = x(GmM)*(x^2 + y^2 + z^2)^(-3/2) = x(GmM)/(r^3) y and z can be calculated correspondingly. Now, what's the next step? How do I go from this vector: ∇V=[dV/dx, dV/dy, dV/dz] to: F = -r(mMG)/(r^3)
Well you basically already did it. You calculate dV/dx , dV/dy, dV/dz which will give you: x(GmM)/(r^3) y(GmM)/(r^3) z(GmM)/(r^3) respectively. you then substitude this into your vector ∇V=[dV/dx, dV/dy, dV/dz]. You will get (and here i factored out the (GmM)/(r^3) from each component): (GmM)/(r^3) * [x , y , z]. But [x , y , z] is just your position Vector. In other Terms: its r. So you get your desired Equation: ∇V = (mMG)/(r^3) * r (where the last r in the equation is a vector). Since F = - ∇V we have to add a minus sign and then we are done: F = -(mMG)/(r^3) * r as shown in the video.
You can randomly take a vector, with known unit length and started from a known initial point, on a surface and calculate its maximum increasing/decreasing speed using the time as unit length increment in both x and y direction. You will use it when you are calculating the directional derivative but eventually you will not see "t" cuz what you get will be a number determined by the magnitude of the delta, unit vector, and the angle between these two vectors.
In matrix notation, as seen in this video, each row will always imply a different basis direction. It is when writing a vector in component form that i-hat, j-hat, and k-hat are written explicitly to make it clear to the reader/writer that a linear combination of the scaled Cartesian (rectangular) basis is used. To reiterate, in matrix notation, you should already know not to add the individual components of the vector together as they are on separate rows and thus imply a different 'direction'. In component form, the only reason you know not to add the individual components of the vector together is because the basis vectors (i-hat, j-hat, k-hat) are written explicitly.
Apologies if that's confusing. It felt weird to simply say that i-hat, j-hat, and k-hat are already implied in matrix form as each row represents its own basis vector, scaled by some factor.
Yeah it is but he writes it in a different way -GMmr(vector)/r^3 Since r(vector)= r r(hat) so r^3 cancel with r in the numerator and that will give you force=GMmr(hat)/r² r(hat) is present since force is a vector
He wrote the same thing. You are missing the directional unit vector r cap which is (r vector)/|r|. He just wrote r cap in terms of RHS and thus r² gets multiplied by another r making it r³ and instead of r cap we write r vector. Hope it's clear😊.
The (1/||v||)*v normalises the v vector (call this value v'). We normalise it because we do not care about its magnitude, we only care about it's direction, and as such the magnitude of v'=1. Now we have v'.(grad(f)). What we are doing with this dot product operation is projecting v' onto the gradient vector grad(f). This basically tells us how much of vector v' is pointed in the same direction as grad(f). If the vectors are at right angles to each other (orthogonal) then the answer will be 0 because no amount of v' points in the same direction as grad(f). If the answer is 1 then they point in the same direction, if it is -1 they point in the opposite direction.
@@godfreypigott Okay. A fair criticism, even if giving it in the form of rhetoric comes across as a little rude (expecially since it is directed at someone who is giving a well produced, advanced lecture course for completely free). He's definitely been through the idea of projection before in other videos and lecture material, and I'd argue that linear algebra (a topic he has covered in some detail) is a prerequisite for vector calculus. If a viewer doesn't have a good intuition for how the dot product works, this lecture course is probably one step too advanced for the viewer (as they've clearly skipped a fundamental foundation of the topic). If he went through everything in the absolute finest detail all the time, these videos would become lengthy and unwieldy, which would defeat the point. One of the reasons that these lectures are so good and so easy to absorb is that they don't overwhelm the viewer with superfluous information. (Editted to make Godfrey happy)
@@EdwardAlexanderSmall I just told you I understand this. So I don't know why you you want to belittle me with the second half of your second paragraph.
Best series I found for catching up on uni maths I forgot. Goes into enough detail whilst being short enough to fit in at work
This is what is good about the internet for education. A good succinct lecture really gets the information across. If you want to learn we are in a golden age. I am very excited for it all.
Another nice graphic application of Grad is that it provides the normal direction of the surface T(x,y)=constant. This idea can be very useful for the geometrical understanding of our math problems. For any surface I can write as f(x,y)=constant, I can obtain the normal of this surface at any surface point just obtaining the Grad vector (whatever the complexity of this surface is). This directly applies to plasticity theory with respect to the yield surface and the normality rule.
Thank you for this comment.
Sir, you are the best I have ever known you have done this difficult subject as easy as possible for me and for my friends so you deserve a great thank 🎉🎉❤❤
Sir, You are the Best, truly intuitive. Much appreciation!
I have an exam on this on monday. ≈50% of students fail the course, where I study. Thank you for the great explanations
Correct me if i'm wrong, but i think the radius vector is pointed from the center of the Earth to the mass. This way the minus sign in the equation of force make sense.
You are absolutely correct. The minus sign gains its significance when you define a inertial system (physics term) with unit-vectors with origin at the center of the earth. One equivalent statement comes from linear algebra and the algebra of vectors.
The physical “inertial system” is the mathematical equivalence to a set of basis vectors that roughly speaking spans a vector space, with a origin that is constant.
So if we would say that the earth rotates around the moon, i.e. we define an inertial system at the center of the moon, then the force would change sign depending on how we choose the basis vectors, or respectively in what way we define the inertial system.
(This is crucial in my opinion to understand Steve’s video about Coordinates and Autoencoders (…))
I recommend Princetons book Jeremy Kasdin and Dereks “Engineering Dynamics - A comprehensive Introduction” for a more low level, yet richer, vector formalism used in dynamics. But a high level introduction as this series is worth a lot.
Then perhaps you will see that the probability for needing to ask for correction about a fact about vectors and physics might decrease!
Yes absolutely. Negative sign indicates F is opposite to the unit vector of r.
You're right!
At 12:05, shouldn't vector r point outwards from the centre of the Earth? So the gravitational force has the direction towards the centre given the minus sign in the equation.
Think you sir,,,,,,love from Bangladesh 🇧🇩🇧🇩
so glad to be seeing your lecture. thank you
Teachings efficient and therapeutic!
Do some more on ODE Sir
Always wonderful to learn from you, i tried the example of gravitational potential, quite a neat example. Thanks for the video :)
Hey man!Are u indian?Just wondering if this helped you in your university maths?
Thank you, Sir
I am eagerly waiting for the coming lectures.
Dear Professor Brunton, I heard mathematicians treat the output of the gradient operator as a *row* vector (for reasons supposedly to do with differential geometry), while engineers and statisticians typically treat it as a *column* vector. I'm curious if there is a preference for either convention in the optimization / machine learning literature, as well as your own preference.
Row vectors and column vectors DO NOT exist. A vector is a vector, an abstract object, independent on the reference frame or the coordinates you use. Then, you can sort the coordinates (reminding which base vectors are associated with) as you wish: if you're free to choose your convention, choose the one you're more comfortable with, if you're using some libraries check out the documentation to find the convention they're using
@@wp4297 i like to project my vectors onto a cylinder so i don't have to look at all of the columns at the same time
@@andrewferguson6901 I have not enough imagination for cylinders in high-dimensional spaces :)
Please make a video on double curl of a vector field on a cartesian point.
Its the small details added that fill in the missing knowledge gaps for beginners.
Why are the vector symbol over the grad removed? I needed to know that!!!
Thanks teacher.
Gradient alone is not a vector, it is a function. When we apply specific values to a gradient function we obtain the gradient vector in a specific point in a coordinate system. The result represents a vector, then we use vector symbol over that.
Great explanation!
Really great lecture. Thank you very much!
Keep it up. Awesome series
Coming to learn new knowledge！
Great lecture! one thing, shouldn't the field have only "one M"? i.e., it should be the gravitational force per unit of mass
Sir, please can u explain augmented Lagrangian (penalty function)
New epsode at MathFlix! 😀
love this series
Thank you sir its a great explaining
I have a question about calculating a gradient of gravitational potetnital:
V = -(GmM)/r
I can decompose r into a form where I keep each axis separate using pythagorean theorem:
r = (x^2 + y^2 + z^2)^(1/2)
now V equals:
V = -(GmM)*(x^2 + y^2 + z^2)^(-1/2)
Now I can calculate a derivative in each direction (x, y, z):
dV/dx = -(GmM)*(-1/2)*(x^2 + y^2 + z^2)^(-3/2)*2x = x(GmM)*(x^2 + y^2 + z^2)^(-3/2) = x(GmM)/(r^3)
y and z can be calculated correspondingly. Now, what's the next step? How do I go from this vector:
∇V=[dV/dx, dV/dy, dV/dz]
to:
F = -r(mMG)/(r^3)
Well you basically already did it. You calculate dV/dx , dV/dy, dV/dz which will give you:
x(GmM)/(r^3)
y(GmM)/(r^3)
z(GmM)/(r^3)
respectively.
you then substitude this into your vector ∇V=[dV/dx, dV/dy, dV/dz].
You will get (and here i factored out the (GmM)/(r^3) from each component):
(GmM)/(r^3) * [x , y , z]. But [x , y , z] is just your position Vector. In other Terms: its r.
So you get your desired Equation:
∇V = (mMG)/(r^3) * r (where the last r in the equation is a vector).
Since F = - ∇V we have to add a minus sign and then we are done:
F = -(mMG)/(r^3) * r as shown in the video.
@@philipbarthelma1903 Ahh I see! I got stuck on calculating partial derivatives (dV/dx,...) 🙄 the rest is clear. Thanks for the explanations!
Hi Steve, do you have a video for "Field"? like what is field and what is scalar or vector field? Thank you very much!
Thank you Steve. Enjoy your sabatical 💗
Great explanations for easy intuitive understanding of PDEs. I however was wondering whether we can add time in these equations, and how?
You can randomly take a vector, with known unit length and started from a known initial point, on a surface and calculate its maximum increasing/decreasing speed using the time as unit length increment in both x and y direction. You will use it when you are calculating the directional derivative but eventually you will not see "t" cuz what you get will be a number determined by the magnitude of the delta, unit vector, and the angle between these two vectors.
thanks for your great explaination
Glad you liked it!
Can I use the gradient to model the gradient of sand density on Arrakis?
Thank you so much
Great video! Thanx! 😊
Thank you!
thank you sir
Is the concept of grad similar to maxima and minima?
UC Santa Cruz - Physics 1987
Thank you
Thanks.. Some books for it?
many thanks!!
Hi. So F is minus the directional derivative of V ? Excellent video.
Great!
YES.
How are you so good at writting backwards?
the video is mirrored.
May I ask where is the I,j,k in nebla (delta ) why you don't put them in the column vector like [d/dx i, d/dy j, d/dz k]
In matrix notation, as seen in this video, each row will always imply a different basis direction. It is when writing a vector in component form that i-hat, j-hat, and k-hat are written explicitly to make it clear to the reader/writer that a linear combination of the scaled Cartesian (rectangular) basis is used.
To reiterate, in matrix notation, you should already know not to add the individual components of the vector together as they are on separate rows and thus imply a different 'direction'. In component form, the only reason you know not to add the individual components of the vector together is because the basis vectors (i-hat, j-hat, k-hat) are written explicitly.
Apologies if that's confusing. It felt weird to simply say that i-hat, j-hat, and k-hat are already implied in matrix form as each row represents its own basis vector, scaled by some factor.
I didn’t understand how the gravitational force is equal to mMG/r^3 , Isn’t suppoed to be 1/r^2
Me too
Yeah it is but he writes it in a different way -GMmr(vector)/r^3
Since r(vector)= r r(hat) so r^3 cancel with r in the numerator and that will give you
force=GMmr(hat)/r² r(hat) is present since force is a vector
@@victorlc.vabeiryusa7458ohh I see, thank you so much ❤
Some hints to write V = -mMG/r into cartesian coord (x,y,z) ? I have no clue where to start thx
r = x^2 + y^2 +z^2
@@ItahangLimbudon't you mean r^2=x^2+y^2+z^2?
@@MarceloKatayama sorry forget r^2
Is he writingevery thing mirrored???
I would assume that he mirrors the video itself
F = -GMm/r2 (NOT r cube)
He wrote the same thing. You are missing the directional unit vector r cap which is (r vector)/|r|. He just wrote r cap in terms of RHS and thus r² gets multiplied by another r making it r³ and instead of r cap we write r vector. Hope it's clear😊.
Some ants actually do fly Steve 🙂 6:18
Some ants do fly.
Now when I need to write gravity I can turn it into del mass instead.
(Some) ants certainly can fly, to my eternal horror.
But *WHY* is the directional derivative given by that dot product?
The (1/||v||)*v normalises the v vector (call this value v'). We normalise it because we do not care about its magnitude, we only care about it's direction, and as such the magnitude of v'=1.
Now we have v'.(grad(f)). What we are doing with this dot product operation is projecting v' onto the gradient vector grad(f). This basically tells us how much of vector v' is pointed in the same direction as grad(f).
If the vectors are at right angles to each other (orthogonal) then the answer will be 0 because no amount of v' points in the same direction as grad(f). If the answer is 1 then they point in the same direction, if it is -1 they point in the opposite direction.
It's a projection of the gradient onto the unit vector so it's the dot product.
@@EdwardAlexanderSmall I actually know how it works. My point was that he should be explaining this instead of just giving the result.
@@godfreypigott Okay. A fair criticism, even if giving it in the form of rhetoric comes across as a little rude (expecially since it is directed at someone who is giving a well produced, advanced lecture course for completely free).
He's definitely been through the idea of projection before in other videos and lecture material, and I'd argue that linear algebra (a topic he has covered in some detail) is a prerequisite for vector calculus. If a viewer doesn't have a good intuition for how the dot product works, this lecture course is probably one step too advanced for the viewer (as they've clearly skipped a fundamental foundation of the topic).
If he went through everything in the absolute finest detail all the time, these videos would become lengthy and unwieldy, which would defeat the point. One of the reasons that these lectures are so good and so easy to absorb is that they don't overwhelm the viewer with superfluous information.
(Editted to make Godfrey happy)
@@EdwardAlexanderSmall I just told you I understand this. So I don't know why you you want to belittle me with the second half of your second paragraph.
If there are two sources of heat and the bee is unlucky, it could just stay at a non optimal point
♥♥♥♥♥♥
Small correction at 6:20. Some ants do fly.
Worst Earth ever.