I. Recall how the integral was defined in the case of the moving car.
II. Let's look graphically at the scenario of a forward-moving car.
III. We can think about the motion of the car to see some features of the integral.
IV. Let's look at the geometric interpretation of integrals again.
V. The equation for the speed and position of a falling body can be deduced using the integral.
VI. Mean Value Theorems relate global behavior to local behavior of functions.
VII. The most important thing to remember about integrals is what they mean.
As I stated above, Starbird starts this lecture out with many graphs that I obviously choose not to reproduce here on the blog. His first example is the graph of a car with constant velocity 2 miles per minute from 0 to 3 minutes, which he integrates to find the distance, 6 miles. He points out that this also represents the area under the velocity curve.
The professor gives a more interesting example, v(t) = 2t. He reminds us that in an earlier lecture, we approximated the integral by assuming that the velocity is constant over a small interval then take the limit as the length of the intervals approaches zero. On the graph, this corresponds to finding the area of rectangles approximating the area under the curve.
Starbird now introduces the familiar notation for a definite integral. You already know what it is, but it doesn't look good in ASCII:
integral _0 ^3 2t dt
= (1) dist traveled
= (2) area
The professor explains that the integral sign is an elongated "S" -- "S" for sum. Leibniz used this symbol because the integral is defined as the limit of a sum. He also justifies 2t dt -- it's the area of a rectangle of height 2t and width dt, which is the limit as delta-t approaches zero. The limits of integration represent the starting and ending times:
Area of triangle is 1/2 * 3 * 6 = 9
So dist trav = 9 miles
Starbird now proceeds to show us that the integral from a to b plus the integral from b to c equals the integral from a to c (provided b is between a and c).
The professor reminds us that the velocity is negative if we're moving backward. Therefore the integral really gives us the change in position, not exactly the distance traveled.
Starbird now moves on to bodies falling under the influence of gravity:
integral _0 ^s -32 dt = -32s ft/sec
v(s) = -32s ft/sec
v(t) = -32t ft/sec
integral _0 ^2 -32t dt = -64 ft
v(t) = -32t ft/sec
integral _0 ^s -32t dt = -16s^2 ft
Thus we integrate the acceleration to find velocity and integrate velocity to find position.
The professor mentions that there is a Mean Value Theorem for both derivatives and integrals. I won't bother to type those up in ASCII.
Lesson 9-6 of the U of Chicago text is called "Views of Solids and Surfaces." In the modern Third Edition of the text, views of solids and surfaces appear in Lesson 9-5. (Recall that Lesson 9-4 of the new edition is "Drawing in Perspective," which is Lesson 1-5 of the old edition.)
This lesson is perfect for an architect. This is what I wrote last year about today's lesson:
Only two words are defined in this lesson -- views and elevations:
"Underneath the picture of the house are views of the front and right sides. These views are called elevations."
There's not much more for me to say about this lesson, except to add that the modern Third Edition includes some examples using isometric graph paper. This reminds me a little of the (mis)adventures my middle school students had with isometric graph paper three years ago (at the old charter school).
[2020 update: In my November 6th post, I also mentioned subbing in an Engineering Graphing Design class where the students dealt with front, side, and top views.]
And so we return to Euclid. Of course, none of his definitions or propositions have anything to do with views of solids and surfaces, so we just look at the next proposition in order:
Proposition 9
This is the three-dimensional analog of yet another theorem, previously studied in Lesson 3-4:
Transitivity of Parallelism Theorem:
In a plane, if l | | m and m | | n, then l | | n.
Some texts omit "in a plane," (after all, as Euclid is about to prove, it holds in three dimensions as well) but only prove it for a plane. The U of Chicago text is more honest and admits that it is only proving it for the plane case. Indeed, the text gives an informal argument in terms of slope -- and of course, we only define slope for lines in a (coordinate) plane.
Euclid's proof is interesting in that unlike those of the previous propositions, he does not use plane Transitivity of Parallelism to prove the spatial case.
Given:
Prove:
Statements Reasons
1. bla, bla, bla 1. Given
2. Choose G on
H on
K on
3.
(call it plane P)
4.
5.
But in this case, Euclid's original proof is simple enough to show to high school students as it is:
Let each of the straight lines AB and CD be parallel to EF, but not in the same plane with it.
I say that AB is parallel to CD.
Now, since EF is at right angles to each of the straight lines GH and GK, therefore EF is also at right angles to the plane through GH and GK.
And EF is parallel to AB, therefore AB is also at right angles to the plane through HG and GK.
For the same reason CD is also at right angles to the plane through HG and GK. Therefore each of the straight lines AB and CD is at right angles to the plane through HG and GK.
But if two straight lines are at right angles to the same plane, then the straight lines are parallel. Therefore AB is parallel to CD.
Therefore, straight lines which are parallel to the same straight line but do not lie in the same plane with it are also parallel to each other.
Q.E.D.
Perhaps the only change we might make is replace "plane through HG and GK" with "plane P" in order to avoid repeating that cumbersome phrase.
David Joyce mentions another proof at the above link -- midpoint quadrilaterals, which are also known as Varignon quadrilaterals. The U of Chicago text only mentions midpoint quadrilaterals in an Exploration exercise (likewise in the new Third Edition, the only difference being that the new text actually mentions Varignon's name). But it's easy to prove that they are parallelograms -- just divide the quadrilateral into two triangles and apply the Midsegment Theorem of Lesson 11-5 to each.
But now Joyce points out that the theorem is true even for "space quadrilaterals." Our definition of polygon (and hence quadrilateral) states that the vertices must be coplanar. But even if we relax this requirement and allow for space quadrilaterals, Varignon's Theorem still holds. After all, even space quadrilaterals can be divided into two triangles (and each triangle lies in a plane), and so we can still apply the Midsegment Theorem to each one. The difference is that today's Proposition 9 is used to prove that opposite midsegments are parallel -- each is parallel to the the diagonal of the quadrilateral but the three lines aren't coplanar.
Notice that the final parallelogram always lies in a single plane -- this follows from last Friday's Proposition 7 that two parallel lines and a transversal are always coplanar.
Meanwhile, I've been continuing to think about a proof of Proposition 4 that is based on rotations, since this might be simpler than Euclid's proof that requires seven pairs of congruent triangles. Let's recall what we're supposed to prove -- given that line l is perpendicular to both lines m and n, prove that l must be perpendicular to the entire plane containing m and n.
Euclid begins by defining points A, B on m, C, D on n, and E, F on l. (Actually, E is where all three lines intersect.) These points have the additional property that AE = BE = CE = DE. My goal was to demonstrate that a rotation of 180 degrees about axis EF maps A and B to each other, as well as C and D to each other. Of course, E and F are fixed points of this rotation.
Let's first find A', the rotation image of A. We first notice that rotations, like all isometries, preserve angle measure, and so Angle A'EF = AEF, which is known to be 90. Also, since rotations preserve distance, AE = A'E. Finally, since the magnitude of the rotation is 180, Angle AEA' = 180. The only point satisfying all these requirements is B, so A' = B. Similarly, we have B' = A, C' = D, and D' = C.
Of course, this requires a more rigorous definition of rotation about an axis. For plane rotations about a center O, we expect A' to be a point such that AO = A'O and Angle AOA' equals the magnitude. But for rotations about an axis, we can't use the same definition unless we know where O is along the entire axis. Otherwise we won't know where to measure AO, A'O, or Angle AOA'.
The correct answer is that O is chosen along the axis such that AO is perpendicular to the axis. Since AE is perpendicular to the axis EF, E is the correct point to choose. So that's why AE = A'E and Angle AEA' must be 180. If AE weren't perpendicular to the axis, choose O to be a different point instead.
Once this is complete, then we choose line o with G and H the points of intersection. It then follows that H = G' since rotations preserve collinearity, and the proof is complete.
Or is it? I'm wondering whether I slipped and accidentally assumed what we're trying to prove. For example, can we be sure that the rotation maps o to itself unless we already know that o is perpendicular to l? Also, if we define a rotation as the composite of two reflections in intersecting planes (similar to the definition for plane rotations), then we might not be able to prove that rotations work the way I said they do unless we've already assumed that Proposition 4 is true! (Why, for example, do we choose O such that AO is perpendicular to the axis?)
Making the proof work requires much more thought than I'm willing to take now, considering that we're not actually trying to teach Proposition 4 in a classroom.
Here is the worksheet for today: Students can fill out the front of the worksheet with examples of both a 3D figure (such as a pyramid) and a block-building (similar to #2 on the back) with the respective front, right, and top views of each.
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