It took me 3 days, 2 attempts, and ~40 pages of notes but I finally passed the Applications of Derivatives unit test. I scored 12/15 on my first attempt but luckily didn’t have to redo any exercises in order to keep my overall score at 80% (as I’ve said before, each exercise needs to be at 80% before doing the unit test otherwise you won’t finish the unit with a perfect score). I was happy that I got all 15 questions correct on my second attempt, however, but still felt slightly unsure of some of my answers as I went along.

In general, there were three types of questions that came up in the test; questions involving L’Hôpital’s Rule, related rates involving shapes, and related differentiable functions. The following are an example of each type of question:

Q. 6) Find the limit as x approaches ½ of (x * cos(π*x))/(e^x – √e):

(Note: inputting x = ½ into the limit results in 0/0, as shown as the top of the photo, which is why you need to try using L’H.R.)

• Lim x->½  (x * cos(π*x))/(e^x – √e) = Lim x->½ (a(x) * b(c(x))/d(x)
  • a(x) = x
  • a’(x) = 1
  • b(x) = cos(x)
  • b’(x) = -sin(x)
  • c(x) = π*x
  • c’(x) = π
  • d(x) = e^x – √e
  • d’(x) = e^x
• Lim x->½ (a(x) * b(c(x))/d(x) = Lim x->½ (a’(x) * b(c(x))) + (a(x) * b’(c(x) * c’(x))/d’(x)
  • = ((1) * cos(π*x)) + (½ * -sin(π*x) * π)/e^x
    • @ x = ½
  • = cos(π/2) + (½ * -sin(π/2) * π)/e^1/2
  • = 0 + (½ * -π * (1))/√e
  • = (-π/2)/√e
  • = – π/2√e

Q. 7) The surface area of a sphere is increasing at a rate of 14πm². At a certain instant, t₀, it’s surface area is equal to 36πm². At what rate is the sphere’s volume increasing at t₀?

(Note: Given that we know the sphere’s surface area and the rate at which the surface area is increasing, you need to use the formula for a sphere’s surface area in relation to it’s radius, S = 4πr², and then, after figuring out the rate of change of the radius, use the formula for the volume of a sphere, 4πr³/3, to figure out the rate of change of the volume.)

• Givens and Unknowns:
  • S(t₀) = 36πm²
  • S’(t₀) = 14πm²
  • r(t₀) = ?
  • r’(t₀) = ?
  • V(t₀) = ?
  • V’(t₀) = ?
• Step 1 – Find the radius:
  • S = 4πr²
    • 36πm² = 4πr²
    • 36πm²/4π = r²
    • 9m² = r²
    • √(9m²) = √r²
    • 3m = r(t₀)
• Step 2 – Find the rate of change of the radius:
  • d/dt [S(t₀)] = d/dt [4π(r(t₀)²)]
    • S’(t₀) = 4π *2r * r’(t₀)
    • 14πm² = 4π *2(3m) * r’(t₀)
    • 14πm² = 24mπ * r’(t₀)
    • 14πm²/24mπ = r’(t₀)
    • 7/12m = r’(t₀)
• Step 3 – Find the rate of change of the volume of the sphere:
  • d/dt [V(t₀)] = d/dt [4π(r(t₀)³)/3
    • V’(t₀) = 4π/3 * 3(r(t₀))² * r’(t₀)
      • = 4π/3 * 3(3m)² * 7/12m
      • = 4π/3 * 3(9m²) * 7/12m
      • = 4π/3 * 27m² * 7/12m
      • = (4π * 27m² * 7m)/(3 * 12)
      • = 756πm³/36
      • = 36πm³

Q. 9) The differentiable functions x and y are related by the following equation:

• y = √x
• dx/dt = 12 = x’(t)

What is dy/dt (a.k.a. y’(t)) when x = 9?

(Note: When answering these questions, you need to remember that both x and y are functions of t and therefore you’ll need to use derivative rules on them where appropriate. In this case, it means using the power rule on √x, a.k.a. x^1/2.)

• Givens and unknowns:
  • x(t) = 9
  • y(t) = ?
  • x’(t) = 12
  • y’(t) = ?
• Step 1 – differentiate both sides of the equation:
  • d/dt [y] = d/dt [x^1/2]
  • y’(t) = 1/2 * x^-1/2 * x’(t)
    • = 1/(2*√x) * x’(t)
    • = 1/(2*√9) * 12
    • = 12/(2*3)
    • = 12/6
    • = 2

(Note: In some of these questions you actually need to solve for y(t) in order to find the derivative of the unknown but in this question it was unnecessary.)

Overall, as frustrating as it was at times, I’m glad I had to do the unit test 3 times. Doing so definitely helped me wrap my head around a few things like related rates, unit conversion, etc. I also got much better at algebra involving logarithms and using derivative rules. That being said, I’m also very relieved to be moving on.

I began the next unit, Applications of Derivatives, on Saturday and only managed to get through a handful of videos in the first section. The only thing I learned about is what’s known as the Mean Value Theorem (MVT) which, in my mind, is a theorem that states, “if there’s a function that’s continuous and differentiable between a certain interval, than there’ll be at least one point along that interval where the slope of the function is equal to the average slope of the entire interval.”:

I went through one video that talked about the differences between Intermediate Value Theorem (IVT), Extreme Value Theorem (EVT), and Mean Value Theorem. I had learned about IVT a few weeks ago but I don’t really remember it and, as far as I can tell from searching my notes, I don’t think I’ve ever been taught EVT so right now I don’t really understand any of these theorems. That being said, here are their definitions:

I just realized that I’m know 53% of the way through Calculus 1 which is pretty exciting! Looking through Application of Derivatives (0/1800 M.P.), I’m 99% sure I won’t get through it before the end of August so my hope is to get through it before the end of September. 🤷🏻‍♂️ I have just two more weeks left before I hit the TWO YEAR (!!!) mark of working on KA. It’s pretty crazy to think of how far I’ve come, the direction I’m going, and thinking about where I’ll be in another 2 years. It would be great if I could get 60% of the way through Calculus 1 by the end of Week 104. To do so, I believe I would need to get through 16 exercises in this unit which I think is possible. Fingers crossed! 🤞🏼