For the first time in a while, I actually achieved the goal I set for myself this week! 🥳 I rewatched and worked through the questions in the final three videos from Curvature and then made it through all the videos and exercises in the following section, Partial Derivatives of Vector-Valued Functions. The latter section was pretty straightforward for most part. I was able to quickly understand how to work through the questions from that section but couldn’t completely understand why some of the concepts worked. For the most part, I also think I wrapped my head around curvature which was definitely a relief. All in all, it was one of the better weeks I’ve had in a long time. ☺️
Here are two pages from my notes where I wrote down an insight I had at the beginning of the week:
Just now I asked ChatGPT what the difference between a tangent vector and unit tangent vector is. I’m still not 100% sure, but it seems like what I wrote down in my notes above is essentially correct. Here’s another formula/way of thinking about curvature that I wrote down:
Below is the question that Grant worked through in the 7th and 8th videos from the Curvature series. It has to do with finding the curvature of a helix that has the vector function s ⃗(t) = [cos(t), sin(t), (t/5)]. As you can see, the steps used to solve this question are written down on the last page of my notes and correspond with the circled numbers throughout the first three pages of my notes:
I’m able to understand how to work through this question but, like I said earlier, still don’t really understand why it works. I don’t understand why curvature is the derivative of the unit tangent vector of the derivative of arc-length… I feel like I’m pretty close to figuring it out though.
I tried using the same technique as in the question above to solve the following question, which was from the nineth and final video from Curvature, but it didn’t work. I realized I needed to use the formula K = (x’y’’ – x’’y’)/((x’)2 – (y’)2)3/2 to solve it which turned out to be much easier:
As you can see, I took the solution and simplified it a bit further. (At least, in my mind my final solution seems simpler. 🤔) I wasn’t sure if my solution was actually correct so I ran it through Symbolab and was pretty pleased with myself that my algebra and trig was correct. 🤓
On Thursday I started the following section, Partial Derivatives of Vector-Valued Functions, which turned out to be a pretty enjoyable section to work through. There were only five videos and two exercises in it so it was relatively small. What this section covers is summed up pretty clearly with the description written under the first video. It states, “When a function has a multidimensional input, and a multidimensional output, you can take its partial derivative by computing the partial derivative of each component in the output.”
Here are some screen shots from a question in the first exercise I worked through:
I ended up getting through this exercise on my first try but found it somewhat tricky to think through a few of the questions. There was one particular question that was really hard for me to visualize where the function’s formula was f(x, y) = –sin(x) + cos(y). The way that I had to think through that one (and all the questions for that matter) was by individually setting x and y in my mind to 0 and then thinking through what the shape of the output would look like across the y- and x- axis, respectively.
On Friday I watched two videos titled Partial Derivative of a Parametric Surface. Once again, I still haven’t fully wrapped my head around everything from these videos, but they gave me a stronger intuition on how derivatives work when dealing with a function that has 2D inputs and 3D outputs. Here are a bunch of screen shots from those videos and explanations under each screen shot about what’s going on:
In the top left corner you see the function for v ⃗(t, s) and that it outputs a 3D vector. The (x, y, z) axes in this screen shot is actually showing a 2D representation of t and s with both variables ranging from 0 to 3, which is expressed in the top right corner. The bottom right corner indicates a point (t, s) = (1, 1) and you can see that point marked out in red on the graph. You can also see in the middle of the top of the screen shot that there’s a vector [0, 1, 0]. This is the output of v ⃗(t, s) when inputting (1, 1) . The bottom left corner shows the vector for the partial derivative of t, but that part is irrelevant at this point. (It’s a part of the the final screen shot here.)
It’s hard to tell but this screen shot shows the 3D transformation of v ⃗(t, s) = [t2 – s2, st, ts2 – st2]. The red dot now has the coordinates of [0, 1, 0].
Going back to the 2D input space, the red line indicates the variable t when s = 1.
When going back to the 3D surface, you can see how the red line now stretches and twists across the 3D surface. This indicates the values of [x, y, z] when s is held constant at 1 and t varies from 0 to 3.
Finally, when inputting the (t, s) coordinates (1, 1) into the vector for the partial derivative of t, you can see that it outputs [2, 1, –1]. I could be wrong about this part, but I’m pretty sure this means that, considering that inputting (1, 1) into v ⃗(t, s) outputs [0, 1, 0], using those same coordinates in the vector in the partial derivative of t tells you that the slope of the tangent line to the surface of the 3D output of v ⃗(t, s) = [2, 1, –1] along what WERE the gridlines for the t-axis in the (t, s) input.
……. 🫤
I don’t know if any of that made sense. Like I said, I still only have a vague idea of what’s going on. I think I’m pretty close to having it figured out but am not quite there yet.
The second exercise I worked through gave me questions where I had to find the partial derivatives of certain functions. They were all pretty easy and I got through them all on my first attempt. Here are a few of those questions:
Question 1
Question 2
Question 3
Finally, in the last video of the series Grant explained that you can take four derivatives of vector fields which are (I think) functions with two inputs and two outputs but the outputs are parametric themselves. (I’m pretty sure I said that wrong. I mean to say that the output uses BOTH variables in each row of the output vector. I don’t know if I’m right about this though.) Here are two screen shots from that video:
I understand how to solve the partial derivative for each component of the function but I don’t really know what the point is so I can’t explain it here.
This is one of the longer posts I’ve written in a while so I’m going to wrap this up quickly. I’m pumped with where I’m at right now and the progress I’ve been making through Derivatives of Multivariable Functions (980/2,100 M.P.). The next section is made up of four articles and the section following that has six videos and two exercises. I’m hoping to get through both sections by the end of next week. If I’m able to have as productive a week next week as I did this week, I definitely think that’s within reach!