I didn’t get as far this week as I hoped I would but still had a fairly productive week, nonetheless. I made it through six videos and two exercises which covered two topics. The first was vector valued functions which represented speed/position and finding the derivative of those functions which represented their velocity (I still haven’t completely wrapped my head around this). The second topic I worked on was a concept known as the multivariable chain-rule. For the most part, I enjoyed working through all the videos and exercises this week, although working on understanding how/why the derivative of a speed/position vector valued function equates to that function’s velocity was hard to understand and pretty frustrating. All in all though, it was a pretty solid week. 😬

Here’s a screen shot from the fourth video, Vector Valued Function Derivative Example, in the section Differentiating Parametric Curves:

What’s happening in this screen shot is there are two functions, r ⃗₁(t) = tî + t²ĵ which is on the left and r ⃗₂(t) = 2tî + 4t²ĵ which is on the right. The graphs below each function represent position, a.k.a. the x-axis can be thought of as x(t) and the y axis as y(t). Although the graphs for each function look the same, i.e. the path taken by the object is the same, because r ⃗₂(t) has coefficients in front of each variable, the object is going faster along the same path, a.k.a. it takes less time for the object to got the same distance as in r ⃗₁(t).

I’m going to struggle to explain this part, but when you find the derivative of r ⃗₁(t) and r ⃗₂(t), you can use Pythag’s theorem to determine the object’s velocity which is the hypotenuse/vector at a given point along the path. (Pretty sure that made no sense…) If you compare the two velocity vectors from r ⃗₁(t) and r ⃗₂(t), you can see that the length/speed of the velocity vector in r ⃗₂(t) is twice as large as it is in r ⃗₁(t). One thing to note is that (and I could be wrong about this) the objects in question get to the same position in r ⃗₂(t) at t = ½ as in r ⃗₁(t) when t = 1.

Like I said, I’m a bit confused by this and don’t know if I’m explaining this properly or if I’m even correct. Regardless, here are the notes I took from this vid:

Even though I don’t really know why these questions work the way they do, I got pretty good at understanding how to solve them. Here are three example questions from the exercise from this section:

Question 1

Question 2

Question 3

(In this question, for some reason after I answered it I wasn’t able to scroll up to the top of the webpage so the function of p(t) got cut off. The function was p(t) = (3*sin(2t) , 3*cos(2t) , t²).)

On Friday I started the next section titled Multivariable Chain Rule. I only made it through two videos in this section but they both seemed pretty straightforward. Here are two screen shots from the first video and the notes I took from it:

Again, I wouldn’t say I 100% understand what’s going on with this, but in the example above the gist of what’s happening is that you have two functions, x(t) and y(t), which are themselves variables within the function f(x(t) , y(t)). To find the derivative of f(x(t) , y(t)), you find the derivative of x(t) which would be x’(t) and the partial derivative of f with respect to x, and then do the same thing for y'(t) and the partial derivative of f with respect to y. Because you’re finding the partial derivatives of f with respect to x and y respectively, what you’re doing is finding the gradient of f(x(t) , y(t)), but I find it easier to think through it like this:

• d/dt f(x(t) , y(t)) =
  • = f’(x(t)) * x’(t) + f’(y(t) * y’(t)

I’m pretty sure using Lagrange notion like that is wrong, but I find it WAY easier to think through it like that right now. (As a side note, when a function is embedded (not sure if that’s the right word) within another function, i.e. x(t) and y(t) in this example, they’re known as Intermediary Functions.)

Here’s a note I took that shows the formula for the multivariable chain rule:

Here are two questions I worked through in the first exercise from this section:

Question 4

You can see in my notes that as I was working through this question I was confused at one point along the way and didn’t know/couldn’t remember that the gradient of a multivariable function was considered the derivative of that function. I was about to Google it but thought I’d be able to find the answer quicker if I asked ChatGPT. I did and screenshotted the response which you can see below. I just thought it was interesting how ChatGPT helped me understand the answer in about three seconds which got me thinking how AI is going to change things going forward.

Question 5

I’m pretty happy with where I’m at right now but do wish me progress through MC was faster. I’m now 35% of the way through this unit, Derivatives of Multivariable Functions (740/2,100 M.P.), so I still have a ways to go to get through the whole thing. Given the rate of my progress, I’m starting to think it’s unlikely that I’ll get through Multivariable Calculus by the time September rolls around. If that’s the case, it won’t be the end of the world. It’s frustrating that it’s taking me so long, but I’ve made peace with the fact that learning calculus is hard. It’s annoying but just one of those is-what-it-is type of things. 😡 🤷🏻‍♂️