Chapter 24 of Morris Kline's Mathematics and the Physical World is called "Differential Equations -- the Heart of Analysis." Differential equations are the next step beyond Calculus.
"Nature is pleased with simplicity, and affects not the pomp of superfluous causes." -- Isaac Newton.
"Man's expectation that he can understand the multifarious workings of nature has rested upon the belief that nature is designed not only rationally but simply."
And so physicists wanted to derive as many facts from as few principles as possible. But to do this, they needed to solve differential equations.
Kline's first example of a differential equation is:
d = 32t
But this is clearly not a differential equation. Kline calls this an equation about velocity, so he clearly means to write d-dot here. In ASCII, we use the prime symbol instead:
d' = 32t
A solution of this differential equation is easily found:
d = 16t^2
So notice that every time we calculate an integral, we're actually solving a differential equation. Kline gives a more complicated differential equation:
v' = -96,000/(r^2)
From the solution of this, Kline calculates the escape velocity of a spacecraft to reach the moon. The solution of this equation gives 6.87 miles per second, and he writes that it would take a spacecraft about four days, 21 hours to reach the moon. (Kline wrote this book before the moon landing, but notice that Apollo 11 really did take a little over four days to reach the moon.)
Question 24 of the PARCC Practice Exam is on the circumference of a circle:
24. A landscaper is designing a display of flowers for an area in a public park. The flower seeds will be planted at points that lie on a circle that has a diameter of 8 feet. The point where any seed is planted must be at least 2 feet away from the seeds on either side of it.
What is the maximum number of flower seeds that can be planted using the design?
After planting the flower seeds, the landscaper has 20 seeds left over. The landscaper wants to plant all of the remaining seeds in another circle so that the seeds are 2 feet apart. To the nearest tenth of a foot, what is the diameter of the smallest circle that the landscaper can use to plant all of the remaining seeds?
This question clearly asks students to use the circle circumference formula, C = pi d. So for Part A, the circumference of the circle of diameter 8 feet is 8pi feet. Since we must allow for 2' per seed, we conclude that there must be at most 4pi seeds. Now 4pi is about 12.6, but there isn't quite enough room for the 13th seed (13 seeds would require a diameter of 8.3'), so there are only 12 seeds.
For part B, we see that 20 seeds require 40', so the diameter is 40/pi feet. We calculate that 40/pi is about 12.73', which we round to the nearest tenth as 12.7'.
For the second time in the last three questions, students may have some problem with rounding. Now according to the key provided by PARCC, 12.6', 12.7', and 12.8' are all acceptable answers. But for the number of seeds, only 12 is the correct answer. We see that PARCC is especially picky with rounding when it comes to real countable objects such as seeds, as opposed to measurements. There isn't enough room for the thirteenth seed, so the answer is twelve seeds. On the other hand, six bags of candy wasn't enough to fill the spherical candy jar -- we needed seven bags.
In each case, students need to read the question to know which way to round. For today's question, each seed needs at least two feet of clearance -- we're allowed to give each seed extra clearance, but we can't shortchange the amount of clearance by having too many seeds. Last week, we were asked to fill the sphere -- we're allowed to overfill it, but we couldn't leave any extra space. On the other hand, last week's question could have asked, "What's the most number of bags that can fit in the sphere?" To that question, the answer would have been six bags, because the seventh bag can't fit.
I also wonder whether students might be confused as to what to do to solve the problem. Some students might try to find the area of the circle or something like that, or they might find the circumference, but the fail to divide it by two since each seed need 2' clearance. Notice that once we knew what we had to find, the calculation was trivial.
Some readers might notice that the pre-rounded answers to Part A and Part B are almost the same, each about 12.6. But observe that we're calculating different things in each case -- the exact answer for Part A is 4pi, and for Part B, it is 40/pi. These are nearly equal because pi is so close to sqrt(10) -- remember our discussion about how sqrt(10) Day would be just two days after Pi Day.
I decided that since we never actually completed any questions from Lesson 8-8 of the U of Chicago text, I included some on this worksheet. But now I realize that some of the "review" questions from this section may be too difficult for the students. One of the questions asks for the area of an equilateral triangle -- this is a "review" question from Lesson 8-8, but in reality students can't answer until they learn about 30-60-90 questions in Lesson 14-1! (Of course, we have finished Lesson 14-1, so we have no such handicap.)
Another question asks to compare the perimeters of two triangles. Even though students might be able to eyeball the answer, a rigorous proof is quite tricky. Students must notice that one pair of sides form opposite sides of a parallelogram, and they have a second pair in common. The third pair of sides can be compared using the SAS Inequality (Hinge Theorem).
PARCC Practice EOY Question 24
U of Chicago Correspondence: Lesson 8-8. Arc Measure and Arc Length
Key Theorem: Circle Circumference Formula
If a circle has circumference C and diameter d, then C = pi d.
Common Core Standard:
Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.
Commentary: Some questions from Lesson 8-8 are included.