I didn’t reach my goal this week of getting through the section Divergence but did have a very productive week, nonetheless. It was disappointing that it took me until Friday to get through the four articles from the section Differentiating Vector-Valued Functions (Articles), but they gave me a much better intuitive understanding of vectors, their derivatives and curvature, in general, so I still think it was time well spent. I then managed to get through four of the six videos from Divergence, and both of the two exercises. I could have finished it off and watched the last two videos in the section but wanted to wait until next week so that I can understand divergence more and give a better explanation of it then. All in all, I’m happy with my effort this week and feel like I have a better idea of what’s going on with multivariable calculus than I did at the start of the week.
Below is the first note I took this week which came from the first article titled Derivatives of Vector-Valued Functions. It shows that the derivative of a position vector is its velocity vector. In retrospect, this seems pretty obvious to me but for whatever reason this concept really sunk in for me after reading the article. Here are a few screen shots from the article and the note I took below them:
The next article was titled Curvature and helped me get closer to fully understanding how/why curvature works. I still don’t intuitively understand it and can’t visualize how the formula works in my head, but I think I now understand the gist of what’s happening with curvature. Here’s a screen shot from the article that breaks it down:
To put it into my own words as simply as possible, curvature is:
- A tiny change in the unit vector over a tiny change in time, divided by a tiny change in the arc-length over the same tiny change in time. – Will Malmo
In the same article Grant walked through how to find the curvature of a given function and got to a point where he needed to find the derivative of the unit tangent vector. Below is a screen shot I took where he worked through how to do this. The only reason I made a note of it was because I thought the notation he used to find the derivative of the vector (where it looks like d/dt([vector])) was interesting and something I hadn’t seen before:
The third article was titled Multivariable Chain Rule, Simple Version but I definitely did not find it simple. The first note I wrote down was on something that I knew already but clicked for me a bit more this week:
I think it’ll be useful for me going forward to always picture a f(x(t), y(t)) type of function as an xyz-plane where the coordinates of x and y (which themselves are two unique functions of x(t) and y(t)) output a z-value/coordinate. I don’t know if that made sense, but I can now visualize a xyz-surface in my head and it actually makes a bit of sense now. (But still a somewhat confusing. ) This started to make more sense to me partially after looking at the screen shot below. However, in the image below the last number line is denoted with f whereas I think of that number line as z now:
Here are the notes I took where I had a mini-insight (I think) into why the multivariable chain rule works the way it does:
Again, I don’t completely understand what’s going on (or if I’m actually correct about this for that matter), but I can visualize taking the derivative of x(t) and why that indicates the change in x on f(x(t), y(t)), and then visualize a cross section of f(x(t), y(t)) and visualize why, to find the change in z, you’d only be calculating the partial derivative of x (a.k.a. holding y constant).
(I feel like none of that made sense and that I could be wrong about everything I just wrote.)
The final article was titled Partial Derivatives of Parametric Surfaces and I think I may have figured out how they work. Here are two pages from my notes:
In the article Grant shows the following video where he transforms a 2D input function to its 3D counterpart. Here’s the vid:
He goes on to find the derivatives of t and s, the two input vectors, at a given point. Here are the images of those vectors where the derivative vector is shown in yellow:
Like I said in my note, I think combining those two vectors could be interpreted as some type of parallelogram which would indicate the slope of the 3D object at that point. I don’t know if that would indicate velocity in some way though.
Is started the next section, Divergence, on Friday. I’m not going to go into detail about it but my general understanding of what I’ve learned so far is that if more of whatever’s in a vector field is flowing away from a certain point, it is DIVERGING at that point and if more is flowing into a certain point, it is CONVERGING at that point. Here’s a screen shot from one of the questions I worked through in the first exercise:
Between the two exercises I finished, I somehow managed to get all of the eight questions correct on my first attempt but didn’t really know what I was doing so it was pretty lucky. Regardless, here are three questions I worked on from the second exercise:
Question 1
Question 2
Question 3
Like I said at the start, I’m going to wait until next week to try to explain divergence after I’ve watched the final two videos in that section. I’m glad that the general idea of what divergence is seems pretty simple to understand, but I think the math could end up being a bit tricky.
I’m now 54% of the way through Derivatives of Multivariable Functions (1,140/2,100 M.P.). I’m optimistic that I’ll be able to get through the rest of it relatively quickly – hopefully midway through August. At this point I’m only 24% of the way through Multivariable Calculus (1,140/4,800 M.P.) so there’s no way I’ll finish this course by the start of September (a.k.a. the four-year mark of working on KA!). I’m fine with that though. My new goal is to get through it before the start of the New Year. Then there’ll only be one more math course left on linear algebra before I’ll be DONE! My plan is to get into the physics courses after that!