I had a pretty slow week this week and didn’t get too much done. About 95% of my week was spent working on how to solve the area inside of polar coordinate functions. (Not sure if that’s how to properly say that…) I spent two days reviewing trig identities which are my absolute nemesis but, thankfully, I got a much better handle on them so hopefully they won’t seem so difficult going forward. I also briefly got started on the next section in the Prerequisite Review of Multivariable Calculus which was about vectors. I managed to get all the way through the article on Sunday but I think I’m going to leave it out of this week’s post and go through it with more detail next week. Compared to polar coordinates, the section I went through on vectors seemed relatively straightforward which is definitely a relief. 😮‍💨

I worked through two questions this week that had to do with the area inside polar coordinate functions. The first question I worked on came from a section titled Finding the area of a polar region or the area bounded by a single polar curve. Here’s a screen shot from the start of that video:

As you might imagine looking at the screen shot, the questions asks you to find the area within the function r(θ) = 1 – cos(θ). In order to solve the question, the first thing you have to remember is the formula/integral to find the area of polar coordinates which is area = _a∫^b (r(θ))²/2 dθ. In this case, the bounds being calculated are from θ = [0, 2π] which you can figure out by using a T-chart with1 – cos(θ) on one side and r(θ) on the other side. Starting with θ = 0π and moving through values up to θ = 2π, it becomes obvious that those are the bounds.

Step 2 to solve this question is to then input the bounds and r(θ) into the function _a∫^b (r(θ))²/2 dθ which looks like this:

• _a∫^b (r(θ))²/2 dθ 
  • = ½  ₀∫^2π (1 – cos(θ))² dθ
  • = ½ ₀∫^2π (1 – cos(θ)(1 – cos(θ)) dθ 
  • = ½ ₀∫^2π (1(θ)⁰ – 2cos(θ) + cos²(θ)) dθ

At this point, everything seemed straightforward to me except for finding the integral of ₀∫^2π cos²(θ) dθ. I tried using the inverse chain rule (a.k.a. u-sub which I’ve never really figured out) but couldn’t come up with a solution. I clicked forward into the video and realized Sal used the trig identity cos²(θ) = ½ (1 + cos(2θ)). I couldn’t remember how to derive that identity so at that point I had to go back and review all the trig identities… 🤬

My intuition/vague recollection told me that I needed to review a video where Sal draws out two triangles on top of each other. After looking through Khan Academy, I found an article that had a diagram that made the sine and cosine angle addition identities pretty simple to understand. Here’s a screen shot of that image and the notes I took working through it myself:

One key thing to remember when working through this proof is that the hypotenuse of the triangle in the middle of the square needs to be equal to 1. Then you start with ∠a and denote the opposite side and adjacent side as sin(a) and cos(a) respectively. You can then find the opposite and adjacent sides of ∠b using cos(a) as the hypotenuse. After going through the rest of the sides of the triangles and square, you get to the identities:

• Sin(a + b) = sin(a)cos(b) + sin(b)cos(a), and
• Cos(a + b) = cos(a)cos(b) – sin(a)sin(b)

I knew these identities were going to be key to figure out how Sal got to the identity cos²(θ) = ½ (1 + cos(2θ)) but I couldn’t figure out how he got there. I remembered that if you replace b with another a that would lead you to Cos(a + b) = cos(2a) = cos(a)cos(a) – sin(a)sin(a) = cos²(a) – sin²(a) but then couldn’t figure out what to do after that. I went for a walk with my dog and remembered that I needed to combine the angle addition identities with Pythag’s trig identity sin²(θ) + cos²(θ) = 1. That leads you to:

• Cos(a + b) = cos(2a) = cos²(a) – sin²(a)
  • …
    • sin²(θ) + cos²(θ) = 1
    • sin²(θ) = 1 – cos²(θ)
  • cos(2a) = cos²(a) – sin²(a)
  • cos(2a) = cos²(a) – (1 – cos²(θ))
  • cos(2a) = cos²(a) – 1 + cos²(θ))
  • cos(2a) = 2cos²(a) – 1
  • cos(2a) + 1 = 2cos²(a)
  • (cos(2a) + 1)/2 = cos²(a)
  • cos²(a) = ½ (cos(2a) + 1)

Here are some notes I took where I derived this identity plus a bunch of other ones:

One thing I made a note of is that it seems like most trig identities are derived from the two angle addition identities and Pythag’s trig identity. After I got a handle on the cos²(θ) = ½ (cos(2 θ) + 1) I was then finally able to go back and work through the initial question:

The second question I worked through this week came from the section Finding the area of the region bounded by two polar curves. I wasn’t quite able to figure it out on my own and had to watch through the video to solve it correctly but I was pretty close to getting it correct on my own. I’m not going to go into much detail, but when I first tried to solve this question I made it harder on myself by not simplifying the integrand. Watching how Sal solved it, I realized there were a few ways to cancel out some of the coefficients/scaler factors which I didn’t do initially. Like I said, I’m not going to go into any detail here, but here’s a screen shot from the video and the notes I took once I figured out how to solve it:

Lastly, as I mentioned at the beginning, I briefly worked through the next section of the prerequisite review which was called Vectors. In the hour that I worked through it, I learned something that blew my mind which is that Pythag’s Theorem can be applied to 3D shapes using the formula a² + b² + c² = d². 🤯 Even writing that out now, having said a² + b² = c² to myself about a million times over the past ~3 years, it seems crazy to me that you can just add in d². I worked through a proof that explains how adding d² works, but I’m going to wait until next week to post it so that I can write out better notes on how it works.

I’m happy that I got through the Parametric Equations and Polar Coordinates section of the Multivariable Calculus (100/4,800 M.P.), Prerequisite Review this week and am hopeful that, based on how quickly I got through the section on vectors, it won’t take me too long to get through the remaining sections. Not including Vectors, there are five more sections in the prerequisite review; Dot Products, Cross Products, ‘Matrices, intro‘, Visualizing Matrices, and Determinants. I’ve just decided that my goal this week is to get through the first two, Dot Products, and Cross Products. Hopefully they’re just as straightforward as Vectors!