This week felt like a blur. I made it through the first four sections of the unit Applications of Derivatives, all of which were interrelated but also had their differences. The first and third sections were titled Meaning of the Derivative in Context and Non-motion Applications of Derivatives respectively only had a handful of videos between them. They weren’t difficult and not worth going over in detail here. The second and fourth sections were titled Straight-line Motion and Introduction to Related Rates respectively and were much more interesting but also fairly difficult for me to understand. I was able to figure out the gist of what was being taught in both of those sections but for a large part of the week I was pretty confused which was frustrating.

The second section of the unit that I worked through this week, Straight-line Motion, had 7 videos and 2 exercises. All of the videos and questions in the unit were concerned with a particle’s position on an x-axis that was moving to the ‘left’ or the ‘right’ and the speed, velocity, and acceleration of the particle as it moved. I’d be given a function that described the particles position at a specific time, t, with the condition that time could not be negative, a.k.a. t ≥ 0. The KA videos I watched described the difference between speed, velocity, and acceleration but I had to Google the 3 terms to get a better understanding of their differences:

• Speed
  • How fast an object is moving in any direction.
• Velocity
  • How fast an object is moving to OR away from a specific point.
  • This means velocity can be negative. For example, if a particle’s position is moving further to the ‘left’ or ‘down’ an XY-coordinate plane, it will have a negative velocity relative to the origin of the coordinate plane.
• Acceleration
  • Describes if the velocity is ‘speeding up’ or ‘slowing down’ relative to a specific point.
  • For example, if a particle on an x-axis had a negative velocity (a.k.a. it’s moving to the left of the origin) and it’s speeding up in that direction, the acceleration is INCREASING in the NEGATIVE direction. If, however, the particle is moving in the negative direction but it’s speeding up in the positive direction (i.e. it’s accelerating back towards the origin) than the acceleration is INCREASING in the POSITIVE direction:

 Here’s an example of the type of question I was given:

Plugging the function into Desmos results in a graph that looks like this:

The hard part to understand about this type of question is that the y-axis on the graph is used to describe the particles position on a separate, hypothetical x-axis, and the x-axis on the graph above is used to describe the time at which the particle is being measured at (i.e. at x = 2, 2 seconds have passed).

(Note: the graph shouldn’t go into the negative side of the x-axis but I don’t know how to set Desmos to do that.) For example:

• When t = 0:
  • f(0) = 0³ – 4(0)² + 3(0) – 2
    • = -2
    • (This means that at time equals 0, the particle is at -2 on the hypothetical x-axis.)
• When t = 2
  • f(2) = 2³ – 4(2)² + 3(2) – 2
    • = 8 – 16 + 6 – 2
    • = -4
    • (When time equals 2, the particle is located at -4 on the hypothetical x-axis.)

To find the velocity of the particle, V(t), you take the derivative of the function f(t):

• V(t) = d/dt[f(t)] = d/dt[t³ – 4t² + 3t – 2]
  • V(t) = f’(t) = 3t² – 8t + 3

I always have to remind myself that the value of V(t), as shown on the graph, (i.e. the value of the y-coordinate along the function) equals the slope of the corresponding x-coordinate along f(t). This means, for example, that at t = 0, the slope of f(0) equals 3 which means at exactly t = 0 the particles instantaneous velocity is 3 meters/second (the question doesn’t give distance or time so I just made that up) in the positive direction. At t = 1, the slope of f(1) = -2, i.e. at the 1 second mark the particle’s velocity is moving -2 meters/second.

To find the acceleration, you take the derivative of V(t):

• A(t) = d/dt[V(t)] = d/dt[3t² – 8t + 3]
  • A(t) = V’(t) = 6t – 8

As you can see, the acceleration of the particle is increasing at a constant rate. The acceleration begins at -8 meters/second/second (this is hard for me to understand but acceleration is measured in m/s²) when t = 0 but as each second passes the acceleration increases by 6 beginning at -8.

Once you understand that you need to differentiate f(t) to find the velocity and differentiate V(t) to find the acceleration, it’s not difficult to solve the four questions given so I’m not going to bother doing that here.

The fourth section of the unit was called Related Rates which I initially had a very hard time understanding but started to get the hang of by the end of the week. The general idea of what was taught is that, if you have two functions that are related to each other in an equation, if you can find the derivative of one of the quantities you can then solve for the derivative of the other quantity. The example I was given is the equation for the area of a circle, A = πr², and then thinking of the radius and area expanding both as functions of time, t, leaving you with A(t) = π[r(t)]². An example question I was given was, “A pebble is dropped into lake which produces a circular ripple. At t­­₀­ the ripple has a radius of 8cm and is expanding at 3cm/sec. What is the rate at which the area is expanding at t₀?”:

• A = πr²
  • “Area equals pi times radius-squared.”
• A(t) = π[r(t)]²
  • “Area, with respect to t (time) equals pi times radius-, with respect to t, squared.”
• d/dt[A(t)] = d/dt[π[r(t)]²]
  • d/dt[A(t)] = π * d/dt[[r(t)]²]
  • A’(t) = π * 2r(t) * dr/dt
    • (Note: This is the chain rule where [r(t)]² = a(b(x)) and a(x) = x² and b(x) = r(t) which leaves you a’(b(x)) * b’(x) = 2r(t) * dr/dt.)
  • A’(t) = π * 2r(t) * r’(t)
    • (Note: Now you can input 8cm into r(t) and 3cm/sec on r’(t).)
  • A’(t) = π * 2(8cm) * 3cm/sec
    • = 48π cm²/sec
    • (Therefore at t₀ the area is expanding at 48π cm²/sec.)

For the most part I can understand these questions and am able to do the algebra without too much difficulty. One thing that throws me off, however, is that the questions often use the variables x and y as functions of t which I forget to think of as x(t) and y(t). Here’s an example of a question that I got wrong where this was the case:

I’m sure I’ll get it eventually but I get thrown off when I have to differentiate -2x which I forget is a composite function where x is a function of t which leaves you with -2 * dx/dt and not simply -2. I’m sure with a bit more practice I’ll get the hang of it.

It’s looking pretty likely that I’ll be able to get through this unit, Applications of Derivatives (480/1500 M.P.) before the end of the month. If I push hard, I think I can get pretty close to finishing all the videos and exercises this coming week. I looked further ahead and saw that there are 4 more units in the course so if I manage to get through a unit each month I will be finished by the end of November. Some of the units are huge though so I’m thinking it’s likely that won’t happen but you never know!