Thank you for this comment.

Hey man!Are u indian?Just wondering if this helped you in your university maths?

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!

Glad you liked it!

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 :)

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.

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.

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!

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.

the video is mirrored.

I would assume that he mirrors the video itself

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.