This week I passed the Differential Equations unit test on Tuesday which I was very happy about (although there were a number of questions where I wasn’t completely confident in my answer). After that, I started on the following unit, Applications of Integrals, but only made it through 4% of it which was disappointing. Of the few videos and one exercise I worked through in the second unit, my impression is that the unit will bring all of what I’ve learned in calculus together and I’ll apply the math I’ve learned in practical types of applications/situations (hence the name of the unit, Applications of Integrals). Assuming that’s the case, I think this unit is going to be very helpful for me to start connecting all the calculus dots and help me to figure out how it all works and why. 

There were only 11 questions on the unit test which is the main reason why I was able to get through it all in one day. Looking back at my notes, the only question I really struggled with was question 10:

I ended up doing this question wrong but sort of managed to get to the correct answer. It’s probably hard to decipher my notes but on the left side of the first page, I used the separation-of-variables technique to conclude that y = –2/(x² + C). (I’m actually not sure if that’s true as I’m not sure if C would go into the denominator with x² or if it would stay as its own term.) Since the question asked if y = 2/(9 + x²) was a solution to dy/dx = xy², it made sense in my mind to simply input C = 9 or C = –9 to see if I could get them to look the same. I realized that I they wouldn’t look the same as one expression was negative and the other was positive so I concluded that the given equation, y = 2/(9 + x²), wasn’t a solution. I was correct but that wasn’t the proper way to solve this question. Here’s what I was supposed to do:

• Step 1 – Find the derivative y in the given solution y = 2/(9 + x²):
  • d/dx[y] = d/dx[2/(9 + x²))
    • y’ = 2 * d/dx[1/(9 + x²)]
      • = 2 * d/dx[(9 + x²)^–1]
      • = 2 * d/dx[a(b(x))]
        • a(x) = x^-1
        • a’(x) = –x^–2
        • b(x) = (9 + x²)
        • b’(x) = 2x
      • = 2 * a’(b(x)) * b’(x)
      • = 2 * –(9 + x²)^–2 * 2x
      • =2 * 2x * /(–(9 + x²)²
      • y’ = –4x/(9 + x²)²
• Step 2 – In the given differential equation y’ = xy², replace y’ and y with –4x/(9 + x²)² and 2/(9 + x²) respectively and simplify each expression to see if they equal each other:
  • y’ = xy²
    • y’ = –4x/(9 + x²)²
    • y = 2/(9 + x²)
  • –4x/(9 + x²)² = x * (2/(9 + x²))²
  • –4x/(9 + x²)² = x * 2²/(9 + x²)²
  • –4x/(9 + x²)² = x * 4/(9 + x²)²
  • –4x/(9 + x²)² = 4x/(9 + x²)²
  • –4 * x/(9 + x²)² = 4 * x/(9 + x²)²
  • –4 ≠ 4
• Conclusion
  • Inputting y’ = –4x/(9 + x²)2 and y = 2/(9 + x²) results in –4 ≠ 4 meaning that y = 2/(9 + x²) is not a solution to the differential equation y’ = xy².

As I mentioned in my intro, there were a number of other questions that I wasn’t completely confident on how to answer but had a good enough understanding of what was going on to come up with the correct answer. I didn’t have to look anything up to answer any of the questions which was also a good sign. 👍🏻

On Wednesday I began the unit Applications of Integrals. The first section was titled Average Value of a Function and it taught me a formula that you can use to find out the average value of a function between a certain interval using integrals. The formula is:

• f_ave(b – a) = 1/(b – a) *  _a∫^b f(x) dx

I’m still not 100% sure how/why this formula works but applying it in the exercise I worked on seemed pretty straight forward. Here are two questions from that exercise:

Question 1

Typically I would type out the math for this question but, since my notes are pretty legible (for once), I’m not going to bother. I should note that graphing the function f(x) = 14 – 6x² on Desmos and looking at the interval between x = [-1, 3] is what really helped me to understand that the area between f(x) and the x-axis on that interval would cancel out and therefore equal 0. Seeing it graphed is what really helped me put it all together. 

Question 2

The next section of the unit was titled Straight-Line Motion and went over the relationship between an object’s position, velocity, and acceleration. I had learned about these concepts before but, having a stronger grasp on calculus now, it all made a lot more sense this time around. The first thing that was covered were a few definitions of a handful of key terms and and explanation of how they relate to each other, and calculus and integrals:

What’s not made clear from my notes is that velocity is used to measure displacement and speed is used to measure distance. To find the displacement of an object, you find the antiderivative of velocity and to find the distance an object travels you find the antiderivative of the absolute-value of velocity. Here are how position, velocity, and acceleration relate to each other through derivatives and integrals:

This section had 5 videos and 1 article which I got through but also has 2 exercises which I wasn’t able to start. The math in this section doesn’t seem too difficult but the concepts of what’s going on and visualizing it all happening it still over my head. This does seem like a very practical use of calculus, however, so I’m hoping that getting through this section will really help me to get a stronger grasp on calculus. 

My goal this week is to first get through Straight-Line Motion and then the following section, Non-Motion Applications of Integrals. If I do, I’ll be ~25% of the way through Applications of Integrals (80/1,900 M.P.) which would be a pretty decent start after just 2 weeks. I definitely won’t get through Calculus 1 before May 5^th but, nonetheless, I’m making progress and will get through it eventually. My new goal is to get through Calculus 1 AND Calculus 2 before September (i.e. the start of the new school year, i.e. the 3 year mark of me working on KA). I think this is another ambitious goal but potentially achievable. I’m already 46% of the way through Calc 2, plus there’s a section in Calc 2 that’s also titled Applications of Integrals so I’m assuming once I finish this unit I’ll be a lot closer to being done Calc 2, as well. After >2.5 years, I’m finally starting to see the light at the end of the calculus tunnel and I am SO. JACKED. 💪🏼😤