This week I reviewed two concepts, vectors (i.e. position on a Cartesian plane) and polar coordinates. Both concepts made WAY more sense to me this time around which I’m really happy about, but in a certain way I feel like I didn’t make much progress this week since I didn’t learn anything new. I also feel like I should have been able to review more topics in the span of 5 days. I’m certain I studied more than 5 hours this week but the quality of focus I had when studying wasn’t great. Like I said, I’m happy that vectors and polar coordinates clicked more into place for me this week (although I can tell that still have a ways to go before I intuitively understand both concepts) but, going forward, I want to put more time, energy, and focus into studying so that I can make quicker progress.

My week started off reviewing a video tilted Parametric Curve Arc Length. The video goes through how to derive and calculate the formula that determines the length of an arc and, more specifically, the length of a curve that denotes an object’s position. Here’s a screen shot of the video with the notes I took from it below:

The big insight I had working through this question was that, when it comes to position with respect to time, the x-axis and y-axis are individually denoting the object’s y-position with respect to time (i.e. y(t) = t) and the object’s x-position with respect to time (i.e. x(t) = t). I can now somewhat visualize the position graph being broken down into two sub-graphs each denoting the respective x(t) and y(t) positions. Understanding that each axis is its own function helps me to understand how/why the line can have two ‘outputs’ (i.e. values on the range/along the y-axis) for one input (i.e. values on the domain/along the x-axis). You can see on the screen shot that KA doesn’t mark the axes as y(t) and x(t) which I think would be useful for people to understand this concept. Maybe it’s not technically correct tow rite it that way, but thinking about it like this definitely helps me. As you can see from the rest of my notes, in order to find the length of the curve you need to know/find the derivatives of x(t) and y(t). You can then use Pythag’s Theorem and integrals to determine the length of the curve.

I began reviewing parametric equations on Wednesday and got to a section titled Solving Motion Problems Using Parametric and Vector-Valued Functions. (A title that just rolls off the tongue./s) The questions I reviewed in this section had to do with the same type of thing as above and had me calculate the object’s velocity in either the x-direction or y-direction given the function of the path of the object and its specific (x, y) position (which I think of now as (x(t), y(t))). Here are three questions from that section:

Question 1

What helped me most on this question was thinking about the velocity as either x’(t) or y’(t) as opposed to dx/dt and dy/dt. I find Lagrange notation much easier to work with than Leibniz notation so swapping the two when necessary made things much easier. In this question I also made the connection that there were three given values, x(t), y(t) and y’(t), and, after taking the derivative of both sides of the equation with respect to time, I would have to use simple arithmetic to find x’(t).

Question 2

After working through the question above, this question also became very simple to think through. It’s simply asking me to use the same technique as above to find the missing derivative and then plug both derivatives into Pythag’s Theorem which is the value of the magnitude of the particle’s velocity. I definitely find these questions simple to think through now which wasn’t the case when I first learned them so I’m pretty pumped about that. 💪🏼

Question 3

I’m still having a harder time understanding this question as well as the first two from above. I understand the math behind it in that if you take the integral of v’(t) from t = [a, b] you’ll get the object’s distance travelled (which is literally what I worked through at the beginning of the week), however whereas I can visualize the formula and how it works going from position to velocity, I have hard time visualizing going from velocity to position. In any case, that’s exactly what’s happening in this question and, even though it’s not 100% intuitive for me right now, I’m starting to wrap my head around why this works.

The last thing I reviewed this week was polar coordinates which I got to on Saturday. Here’s a screen shot and the notes I took from the a video I watched titled Polar Function Derivatives:

First off, I had the same type of insight here in that the function r = sin(2θ) can actually be thought of as r(θ). This makes it much easier for me to think through what’s going on. I’m not going to go into detail here but I spent all of Saturday working through the different values of sin(2θ) and got a better feel for why the radius of that function makes the four-leaf clover. This video also works through how to convert polar coordinates into cartesian coordinates and why it works. (This is why there’s a blue rectangle on the chart which Sal used to show that x = r(θ)cos(θ) and y = r(θ)sin(θ).) And lastly, the video also goes over how to find the derivative of the function at a given point which I now find pretty straightforward.

All in all, even though I didn’t make any tangible progress though Multivariable Calculus (100/4,800 M.P.), I’m still happy with how this week went. I’m really glad that I gained more clarity on both vectors and polar coordinates, especially considering how difficult they both were when I first started working on them. There are a few more videos from the unit Parametric Equations, Polar Coordinates, and Vector-Valued Functions (which is the unit I was reviewing this week which I forgot to mention) which I’m hoping to get through early this coming week. After that I’ll be starting on the next article from the first unit in Multivariable Calculus which is titled Vectors and Notation. Including that article, there are six articles in total I need to get through in the first unit of Multivariable Calculus so fingers crossed it won’t take me too long!