Today I subbed in an AP Physics class. It's in my new district. As usual, I won't do "A Day in the Life" for high school classes that aren't math, though admittedly Physics class is closely allied with math.
Two of the classes are AP Physics 1, and the other is AP Physics C. All classes are doing review questions for the upcoming AP exam -- and now you might be wondering, how is the AP still not complete yet?
Apparently, there are three administrations of the AP this year, not just one. The first administration was at the traditional time -- the first two full weeks in May. The second administration is right now -- the next two weeks in May. And the final testing period is the first two weeks in June.
Each school can choose when to administer the exams. And so it appears that this school has chosen the second testing period -- the AP Physics exams will be given this upcoming Monday. In some ways, this is logical -- the week after the AP will be finals week, so we won't have that usual wasted time in May between the last AP and finals.
Many of the AP Physics kids are out taking other AP's this week, of course, and so there are plenty of absences in all classes. The regular teacher tells me to expect zero in-person students in fifth period Physics C (odd classes meet at this school on Wednesdays), with just a handful of online students -- and that's exactly what happens. He tells me that I can direct these students to a review problem set on the AP Classroom website -- and then they are allowed to log out of the Zoom meet immediately.
As for the song for today -- I default to my science standby, "Earth, Moon, and Sun." But only the Physics 1 classes get the song. That's because the microphone on Zoom suddenly stops during Physics C -- and as I just said earlier, all of the students in that period are online. Thus I couldn't sing to them before they logged out.
Today is Sunday, the third day of the week on the Eleven Calendar:
Resolution #3: We remember math like riding a bicycle.
I tell the Physics 1 students about the AP Physics C class that I took my senior. There was one thing that my AP Physics teacher drilled in our heads -- vectors operating at right angles are independent. This is a phrase that I definitely remember like riding a bike -- and it helps the students solve a "starter" Warm-Up problem that is given today:
Two blocks, one of mass M and one of mass 2M, slide on a frictionless horizontal surface with the same initial velocity. The blocks slide up two inclines and are launched onto a table of height H_0 above the floor. A) Which block lands on the table first? B) Which block travels further along the table?
This problem has a diagram which makes it easier to understand (what matters here is that the incline of the M block is 45 degrees while the incline of the 2M block is 30). Since I won't post the diagram, neither will I post a complete solution here. But what matters is that vectors operating at right angles are independent -- so we only need to consider the vertical and horizontal components separately. For example, Part A, landing on the table, only requires considering the vertical component (since gravity, which causes blocks to land on the table, operates vertically). The initial velocity of the blocks are v_0, but the vertical component of the 30-degree block is v_0 sin(30) = v_0/2.
In Physics C, I tell them another story from my own days as a student in this class (although I'm not sure whether they heard it due to the broken mic). Back then, we had only one administration of the AP, so we have plenty of time between the AP and finals. Our teacher filled that time with some simple electricity labs. Notice that Physics C includes both mechanics and electricity/magnetism, while Physics 1 is just mechanics. Also, Physics C leans more towards Calculus while 1 and 2 are Algebra. So while Physics 1 has a few sophomores, Physics C is for juniors and seniors only.
Today on her Daily Epsilon on Math 2021, Rebecca Rapoport writes:
Find x.
(Here are the givens from the diagram -- a triangle has sides x, 180, 181 and Area = 1710.)
Our streak of borderline Geometry-Trigonometry questions continues. Yesterday's problem was definitely Algebra//Pre-Calc trig since it involved sin(5pi/4) -- sines of angles greater than 180, or angles measured in radians, are not taught in Geometry.
Today's problem meanwhile looks as if it could be solved geometrically. Indeed, the triangle appears to be a right triangle with 181 as the hypotenuse. Then we could use the Pythagorean Theorem:
x^2 + 180^2 = 181^2
x^2 + 32400 = 32761
x^2 = 361
x = 19
And then we confirm it by using the area -- (1/2)(180)(19) = (90)(19) = 1710, which matches the area given in the problem. Therefore x = 19 -- and of course, today's date is the nineteenth.
But this isn't a proper solution method -- after all, I made a big deal last Thursday about not assuming that it's a right angle, and now suddenly I'm assuming it. Then again, I do go back and prove it by using the given area.
Some readers might point out that two sides and an area aren't sufficient to prove congruence -- so just because both Rapoport's triangle and the right triangle I found have two sides and an area in common, that doesn't mean that they're congruent. Indeed, in that same problem from Thursday, we had exactly that situation -- two triangles with two sides and an area in common, one acute, the other obtuse.
Then again, we run into this problem if we solve this properly -- that is, using the area formula from trig, A = (1/2)xy sin(theta), rather than assuming the triangle is right:
1710 = (1/2)(180)(181)sin(theta)
sin(theta) = 0.1049
And this has two solutions -- theta = 6 or theta = 174. At this point, we use some common sense -- it's wrong to assume that an angle is right just because it appears so (since it could be 89 or 91), but it's okay to assume that it's acute from its appearance (especially if we're hinging on 6 vs. 174 degrees). I admit that this comes up in many Geometry texts -- a problem where we must make a judgment that an angle is clearly acute (or obtuse).
Once we do so, we use Law of Cosines to find the missing side:
x^2 = 180^2 + 181^2 - 2(180)(181)cos(6)
x^2 = 361
x = 19
Notice that in this proper solution, we never need to prove that it's a right triangle -- the only angle that matters is the 6-degree angle between the two known sides.
This is what I wrote two years ago about today's lesson:
Question 19 of the SBAC Practice Exam is on solving equations:
Consider this solution to a problem:
Problem: -4(6 - y) + 4 = -4
Step 1: -24 - 4y + 4 = -4
Step 2: -20 - 4y = -4
Step 3: -4y = 16
Step 4: y = -4
In the first response box, enter the number of the step where the mistake is made.
In the second response box, enter the correct solution to the problem.
This is a strong first semester Algebra I problem. The original problem has lots of negative signs, and so the obvious error to search for is a sign error. And sure enough, there is a sign error right away -- -4 times -y should be 4y, not -4y. So the step that contains the mistake is step 1.
The solution is easy to correct. All we have to do is change all the signs on y:
Problem: -4(6 - y) + 4 = -4
Step 1: -24 + 4y + 4 = -4
Step 2: -20 + 4y = -4
Step 3: 4y = 16
Step 4: y = 4
Both the girl and the guy from the Pre-Calc class correctly answer Step 1 for this question. But only the girl correctly identifies the solution y = 4. The guy just writes -4 = y for the solution again, even though it's implied that the solution is not -4.
Question 20 of the SBAC Practice Exam is on solving equations:
Consider a sequence whose first five terms are: -1.75, -0.5, 0.75, 2, 3.25
Which function (with domain all integers n > 1) could be used to define and continue this sequence?
A) f (n) = (7/4)(n - 1) - 5/4
B) f (n) = (5/4)(n - 1) - 7/4
C) f (n) = (7/4)n - 5/4
D) f (n) = (5/4)n - 7/4
Because this is an arithmetic sequence, this is also considered to be first semester Algebra I. The first thing we notice is that the terms are listed as decimals, but the choices are all fractions. So the students must convert between decimals and fractions. There is an embedded calculator available, but that assumes that the students know how to use it to make the conversion. I used the calculator (powered by Desmos!) to divide to convert fractions to decimals. Some calculators can convert decimals to fractions, but I don't see that option on this calculator. So the first step would be to divide to convert 5/4 (a major 3rd!) and 7/4 (a barbershop 7th!) to fractions.
We find out that the first term -1.75 is -7/4 and the common difference is -1.25 or 5/4. But we can't use the first term unless we use the (n - 1) version of the formula, f (n) = f (1) + (n - 1)d. Plugging in to this formula, we obtain f (n) = -7/4 + (n - 1)5/4, which is rewritten as f (n) = (5/4)(n - 1) - 7/4. So the correct answer is B).
And as for the girl and the guy from the Pre-Calc class, neither one answers this question -- since for some reason, it didn't print on the SBAC Practice packet! At least this packet is much better than what I had at the old charter school two years ago -- where none of the SBAC Practice problems could print.
SBAC Practice Exam Question 20
Commentary: Distributing a negative sign correctly appears in Lesson 6-9 of the U of Chicago Algebra I text. But the method of finding explicit formulas in Lesson 6-3 -- finding the phantom "zeroth term," doesn't work for Question 10. Instead, students must know and be able to use the formula f (n) = f (1) + (n - 1)d, which most likely appears in later texts based on the Common Core, such as Glencoe and Edgenuity.
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