After 5 months of working on Integrals, I FINALLY made it through the unit test. I picked up this week with four questions left to get through on the test and managed to get all four them correct on Tuesday, although I struggled with the last question. Since I got the very first question of the test wrong last week, my score ended up as 97% which I’m ok with even though I usually redo unit tests until I score 100%. After finishing the test, I then started the following unit, Differential Equations, on Wednesday and got through 10 videos and 4 exercises. So far, the math in this unit hasn’t been too difficult but the concepts seem fairly abstract and I have a feeling it’s going to get pretty hard. One good thing is that this unit’s only 1,100 M.P. whereas Integrals was 3,200 M.P. so I’m hoping that it won’t take me longer than a few weeks to get through it.

The final question on the unit test asked me to figure out which of four integrals written in standard form was equivalent to a given integral written in limit notation. (I have no clue if I’m saying that properly.) Here was the question:

It took me a bit of time looking at the limit just to wrap my head around what it was saying. This type of limit notation reminded me of Riemann Sums which helped me think of the limit as the value of the height of the area, (4 + 5i/n)^1/2, times the width of the area, 5/n. Looking at the given answers, it was clear that the interval being calculated had a width of 5 since the difference of all the upper and lower bounds was 5, but initially I couldn’t figure out if it was from [0, 5] or [4, 9]. I vaguely remembered that when I worked on this type of notation with Riemann sums that the start of the interval, ex. a in [a, b], would be in the ‘height’ section of the notation. Remembering this, I assumed that the integral being calculated was from [4, 9]  since there was a 4 in the limit notation so the answer was therefore either the second or third answer. I wasn’t 100% sure which one was correct but my gut told me that the correct answer wouldn’t already contain a 4 so I went with the third choice and I turned out to be correct.

Although I was disappointed that I got the first question wrong on the test and that I didn’t fully understand 100% of the questions, I still feel good about moving forward to the next unit. I think it’s likely that my understanding of the questions/concepts that I didn’t fully understand going through the test will be concretized as I move forward through the following units. Plus, for the most part I feel like I have a decent understanding of integrals and how they work. 

(Not to mention that it’s been 5 MONTHS of working through this unit……..) 

I started the unit Differential Equations on Wednesday. As I mentioned, although the math hasn’t been too difficult yet, the general concepts of what I’ve learned so far have been a bit hard to think through. My understanding at this point is that differential equations have to do with combining two functions/derivatives and then finding a third function/derivative that serves as a solution to the combination of first two functions/derivatives. (That is 100% the wrong way to say it but, like I said, I’m still trying to figure it out. It’s something like function A * derivative B = function (c).)

Below is an example question from the first exercise I worked on. I somehow missed screen capturing the actual question at the top, but the question was something like, “A hot drink’s temperature, T, is cooling down to room temperature, T_a, at a constant rate. Which of the four answers represents the rate that it’s cooling down?”

(That’s not exactly what the question was, but I can’t remember how it was phrased. 😔)

As you can see, these types of questions have to do with figuring out how functions/derivatives relate to each other. In the next exercise of the following section, Verifying Solutions for Differential Equations, I was given a derivative that was equal to a function that contained its own antiderivative and was then asked if a given antiderivative was a solution to the equation. (That made no sense…) Here are 3 example questions:

Question 1:

• Step 1 – Find the derivative of the given potential solution:
  • y = –2x/3 + 1/9
    • d/dx[y] = d/dx[–2x/3 + 1/9]
    • dy/dx = –2/3x⁰
    • y’ = –2/3
• Step 2 – Substitute y’ = –2/3 from the given potential solution back into the differential equation, y’ = 2x + 3y – 1, and the potential solution, y = –2x/3 + 1/9, in place of y in the differential equation:
  • y’ = 2x + 3y – 1
    • –2/3 = 2x + 3(–2x/3 + 1/9) – 1 
      • –2/3 = 2x + (–6x/3 + 3/9) – 1
      • –2/3 = 2x + (–2x) + 1/3 – 1
      • –2/3 = 2x + (–2x) + 1/3 – 3/3
      • –2/3 = –2/3
  • Conclusion – Since the end result are two equivalent expressions, the given solution, y = –2x/3 + 1/9, is in fact a solution to the differential equation.

Question 2:

Question 3:

Working through these questions helped me to get stronger grasp on the notations y’, dy/dx, and f’(x) and understand that they’re all literally the same thing. I’ve always found Lagrange notation, f’(x), easier to understand then Leibniz notation, dy/dx, so this exercise was good practice for me to relate the 3 types of notations to each other and understand that they’re all the same thing.

The last thing I briefly learned about this week was something called slope fields. Again, I really don’t have a solid understanding of what this concept it all about but the math hasn’t been too difficult to follow so far. Here’s a screen shot from the first video I watched about slope fields:

In the screen shot, Sal started by writing the derivative dy/dy = –x/y in the top left corner. He then drew a Cartesian plane and started to write out coordinates for different points from the equation and indicated their related derivatives (i.e. each point’s slope) using a little dash on the graph. As Sal continued drawing little dashes on the graph, it started becoming clear that this equation, dy/dy = –x/y, formed concentric circles around the origin.

As far as I can tell, the purpose of the slope field is to show which way differential equations slope. Apparently there are multiple solutions in most, if not all, differential equations. I think that drawing out a slope field helps you get an idea of the direction of one of the multiple potential solutions based on which way all the little slopes are going.

I’m 22% of the way through Differential Equations (240/1,100 M.P.) so I’m hoping it won’t take me much longer than a few weeks to get through this unit. The following unit, Applications of Derivatives, is the final unit of the course but is 1,900 M.P. so I highly doubt I’ll be able to get through both units plus the Course Challenge by the end of April, which was my goal. On the bright side, I’m actually 36% of the way through the following course, Calculus 2, which means it shouldn’t take me nearly as long to get through that course as it has to get through Calc 1. My progress over the past few months has felt super slow and I need to keep reminding myself that, even though it feels like it’s taking forever, I’m getting closer and closer to finally understanding calculus.