Well, I didn’t get through the 7 videos and 5 exercises that I was hoping to this week. It took me 3 days and 26 pages of notes to get through 2 exercises (😳) which set me back, but the good news is the final two sections of the unit weren’t too difficult to understand (although I haven’t got all the way through the second section and I think it may end up getting a bit tricky). I was happy with the effort I put in this wee, however, and managed to get through 6 videos and 3 exercises so I still think there’s a chance I can finish the unit by next Saturday!
The two exercises I worked through from Tuesday to Thursday had to do with related rates and their corresponding derivatives. At the end of last week, I mentioned that I was struggling to 1) properly apply the process necessary to solve these questions which required me to use multiple formulas for 2D and 3D shapes (i.e. I’d have to find the derivative of the rate of change for the circumference of a circle, the area of a circle, and the perimeter of a circle) and 2) wrapping my head around the concept that each value would be a function of time (i.e. C’(t), A’(t), and P’(t)). After working through these questions for 3 days, however, I definitely started to get the hang of it and the questions felt much less overwhelming. Here are 2 examples from my notes:
(Note: I didn’t mention in my note what the process is to solve this question. The question asks to find the rate at which the area is changing. The formula for the area of a triangle is A = (b * h)/2 so in order to find the rate at which the area is changing I need to know the rate that the base and height are changing, a.k.a. the derivatives of b(t0) and h(t0). The question gives the rate that the base is changing so I first need to find the derivative of h(t0) and then I can find the rate of change of the area.)
- Step 1 – Draw Triangle
- As mentioned in the description of the question and as you can see from my drawing, the ladder (the hypotenuse) is 5 meters long (forgot to add m on the diagram for meters), the height of the ladder on the wall is 3 meters high, and the distance the base of the ladder is from the wall is 4 meters long. The question states that the base is sliding away from the wall at 6m/s which you can see labelled on my diagram.
- (Note: the velocity that the top of the ladder is sliding down the wall, -8m/s, is calculated in Step 2 but I added it to my diagram.)
- As mentioned in the description of the question and as you can see from my drawing, the ladder (the hypotenuse) is 5 meters long (forgot to add m on the diagram for meters), the height of the ladder on the wall is 3 meters high, and the distance the base of the ladder is from the wall is 4 meters long. The question states that the base is sliding away from the wall at 6m/s which you can see labelled on my diagram.
- Step 2 – Find the derivative of h
- Ladder2 = base2 + height2
- d/dt[L2] = d/dt[b(t0)2] + d/dt[h(t0)2]
- (Note: since the ladders’ height doesn’t change with respect to time, it’s a constant and therefore it’s derivative is 0.)
- 0 = 2b * b’(t0) + 2h * h’(t0)
- 0 = 2(4) * (6) + 2(3) * h’(t0)
- 0 = 8 * 6 + 6 * h’(t0)
- 0 = 48 + 6 * h’(t0)
- -48 = 6 * h’(t0)
- -48/6 = h’(t0)
- h’(t0) = -8m/s
- d/dt[L2] = d/dt[b(t0)2] + d/dt[h(t0)2]
- Ladder2 = base2 + height2
- Step 3 – Find the derivative of A
- Area = (base * height)/2
- d/dt[A(t0)] = d/dt[(b(t0) * h(t0))/2]
- A’(t0) = ½ d/dt[b(t0) * h(t0)]
- = ½(b’(t0))(h(t0)) + (b(t0))(h’(t0))
- = ½ ((6)(3) + (4)(-8))
- = ½ (18 + -32)
- = ½ (-14)
- A’(t0) = -7m2/s
- d/dt[A(t0)] = d/dt[(b(t0) * h(t0))/2]
- Area = (base * height)/2
(Note: In order to figure out how much the volume is increasing at t0, you have to find the radius of the sphere to plug it into the equation V = 4/3πr3, and in order to find the radius of the sphere you need to use the formula for the surface area of the sphere, S = 4πr2, because the question gives you the surface area and its’ rate of change.)
- Step 1 – Find the radius at t0
- S = 4πr2
- 36π = 4πr2
- 36π/4π = r2
- 9 = r2
- √9 = √r2
- r = 3
- S = 4πr2
- Step 2 – Find the volume at t0
- V = 4/3πr3
- = 4/3π(3)3
- = 4/3π(27)
- V = 36π
- V = 4/3πr3
- Step 3 – Find the derivative of the radius at t0
- S’(t0) = d/dt[4πr(t0)2]
- 14π = 4π * d/dt[r(t0)2]
- 14π/4π = 2r * r’(t0)
- 14π/4π = 2(3) * r’(t0)
- 14π/4π = 6 * r’(t0)
- 14π/4π/6 = r’(t0)
- r’(t0) = 7/12
- S’(t0) = d/dt[4πr(t0)2]
- Step 4 – Find the derivative of the volume at t0
- d/dt[V(t0)] = d/dt[4/3πr(t0)3]
- V’(t0) = 4/3π d/dt[r(t0)3]
- = 4/3π 3r(t0)2 * r’(t0)
- = 4/3π(3)2 * 7/12
- = 4/3π(9) * 7/12
- = (4 * π * 9 * 7)/(3 * 12)
- V’(t0) = 21π
- d/dt[V(t0)] = d/dt[4/3πr(t0)3]
After finally getting through the questions on related rates, the next thing I learned about is a concept known as Local Linearity. The gist of this concept is that if you “zoom in” far enough on a curved function you can use the formula y = mx + b to get an approximate value for 2 points on the x-axis that are close to each other. Sal used the functions f(x) = x2and f(x) = x1000 to show that if you zoom in far enough on these functions they appear to be linear:
There are exceptions to this “rule” however which I’m pretty sure are any point discontinuities. What this means is, no matter how far you zoom into a point discontinuity, it will never form a straight line at that point. Sal used the example of y = abs(x – 1) at the point (1, 0) to demonstrate this:
I’m not exactly sure why this concept is important to know, but I’m assuming it will come in handy later on.
The last thing I learned about this week is what’s known as L’Hôpital’s Rule which helps you figure out the limits of values that are thought of as indeterminate form, i.e. when a limit equals 0/0 or ∞/∞ or –∞/∞. I only made it through 2 videos in this section and I don’t have a very good grasp on how the formulas work or why they work, but here’s a page from my notes that shows the conditions required for L’Hôpital’s Rule to be applicable and an example of how it can be used:
I’m not going to bother explaining either page of my notes now because, to be completely honest, I don’t really know how the rule works. I have 5 more videos and 3 exercises left to get through on L’Hôpital’s Rule so I’m sure I’ll be able to give a better explanation of it by the end of next week.
It’s possible I could get through Applications of Derivatives (960/1500 M.P.) by the end of next Saturday but it will most likely be a photo finish, so to speak. I will likely need to pass the unit test on my first attempt in order to finish in time. Although this unit has been a bit of a slog, I feel like I have strong enough grasp on everything I’ve been taught that I could potentially pass the unit test on my first try. ALSO! Hard to believe I’m beginning Week 100 tomorrow!! Triple digits seems pretty crazy. And only 4 more weeks until I hit the 2 year mark! I think I’m officially a nerd now. 🤓