Course Information

Instructor Information

Tutorial Information

• Administratrivia (see syllabus for details)
• Some study strategies (click help button for details)
• Staircase paradox
• CS connection: measuring the complexity of an algorithm depends on knowing how quickly a function grows. For example, is it possible to get a good approximation to the size of 100! (100 factorial) without actually computing it?
• Is the function sin x continuous? Started a proof of this.

• Review of functions (a function is a set; definition of domain and range; some examples)
• Zoom in on the function f(x) = x² at (0,0), and at (1,1). Both times, if zoom in close enough, f looks like a line. Which line? We already know a point on that line, so all that remains is to figure out the slope. This turned out to be not so hard.

Don't look this up online -- try to do it on your own!

• First midterm on Monday, January 30th!
• Considered the general problem of zooming in on the point (a,a²) in the function f(x) = x². If zoom in enough on this point, f looks like a line. We figured out the equation of this line. The hardest part of the process is finding the slope, so we gave the slope a special name: the derivative of f at a.
• Considered the same problem for a more general function f. We gave a mathematical definition of the derivative of f at a, which we denoted by f'(a).
• A few examples of derivatives, including g(x) = 5x - 2 (at 2); h(x) = sin x with x in degrees (at 0); and F(x) = sin x with x in radians (at any point).
• Slick proof that sin x is continuous (including a derivation of the formula for sin x - sin y).

• Differentiability -- when does a function not look like a line, no matter how close you zoom in?
• The derivative of f(x) = x³ at every point.
• The derivative of f as a function in its own right. Second, third, and higher derivatives.

• Proof that lim_x→a f(x) = lim_h→0 f(a + h)
• Proof that if f(x) ≤ g(x), then lim_x→a f(x) ≤ lim_x→a g(x)
• Continued discussing differentiability. In particular, proved that if f is differentiable at a, then it must be continuous at a. We also saw that the converse of this statement is false.
• By using the fact that the cosine function is simply a shift of the sine function, we figured out the derivative of cos x.

• Discussed the distinction between f'(2) and d/dx (f(2)). More generally, the interpretation of constants as numbers or functions.
• Discussed the difference between f'(cx) and d/dx (f(cx)). We (almost finished) proving that d/dx (f(cx)) = c f'(cx).
• In the process of the above, we proved a useful lemma: if lim_h→0 f(h) exists, then for any constant c, lim_h→0 f(ch) = lim_h→0 f(h).

MONDAY, JAN. 30

If your last name begins with A--Y, your exam is in IC 130.
If your last name begins with Z, your exam is in IC 120.

• Let f(x) = sin x, where the sine function interprets x in degrees. We evaluated f'(0), and then more generally f'(a). (Note that f'(a) is NOT cos a!) The common theme was that when evaluating the derivative, we obtained a limit with quantities which tend to 0 at different rates. To get around this, we used two tricks. First, we multiplied by a funny form of 1 (i.e. we multiplied and divided by the same factor). Second, we employed our `scaling' lemma from before: that if G(x) tends to a limit L as x tends to 0, and c is any constant, then G(cx) tends to the same limit L as x tends to 0.
• We used these same tricks to determine the derivatives of f(cx) and, somewhat more difficult, of f(x²).
• We then took these tricks to the next level and proved the Chain Rule. Although the proof is somewhat more technical, the idea is the same as in every other proof from today's lecture.
• Finally, you applied the chain rule to find derivatives of several complicated functions.

• The Chain Rule.

• Discussed and proved the product rule
• Used this to establish that d/dx (x^a) = a x^a-1 for all a∈Q
• Discussed what 2^π might mean
• Proved the quotient rule
• Discussed maxima, minima, local maxima, local minima
• Proved that if g is defined on [a,b], has a max or min on (a,b) at c, and is differentiable at c, then g'(c) = 0.

• Continued discussing maxima and minima, by exploring examples (e.g. 3x² - 24x + 50 and x + sin x).
• Outlined a general strategy for finding all local maxima and minima of g(x) on [a,b]:
• Step 1: Determine all c in (a,b) for which g'(c) = 0
• Step 2: For each such c, determine whether g has a max, a min, or neither at c. (We discussed a couple ways of doing this: comparing values of g(c) to each other, studying the behaviour of g' near c, ...)
• Step 3: Separately consider the behaviour of g at the endpoints of the interval.

• Suppose I'm traveling, and during the time interval between 2pm and 3pm I travel 113 km. Then I definitely broke the speed limit (100 km/hr) at some point. Actually, more is true: I must have been going at 113 km/hr at some point during the trip. This last statement is a special case of what's known as the mean value theorem: that during any trip, at some point during the trip you will be travelling at precisely the same speed as your average speed during the course of the trip.

  Before discussing this in more mathematical terms, we had a short digression on average velocity vs. instantaneous velocity. More precisely: suppose I'm traveling from point A to point B in a straight line, and s(t) outputs my position (i.e. my distance from A) at time t (i.e. t hours into the trip). After some discussion, we decided that my average velocity during the time interval from t=a and t=b is [s(b) - s(a)] / [b - a], while my (instantaneous) velocity at time t=c is s'(c).

• Before getting to the formal version of the Mean Value Theorem, we spent a substantial amount of time trying to prove the following

  Conjecture: If F'(x) = 0 on some nonempty interval, then F must be constant on the interval.

  We were unable to prove this conjecture. The main difficulty we encountered was that F'(x) is written in terms of a certain limit, and it is not at all obvious how to deduce anything precise about the behaviour of F from the knowledge of this limit. Having convinced ourselves that this problem is much harder than it looks, we moved on to the Mean Value Theorem, which turns out to be the key to proving the conjecture.

• Mean Value Theorem: suppose F(x) is differentiable on the open interval (a,b) and is continuous at both endpoints (i.e. F is continuous on [a,b]). Then there exists c in (a,b) such that F'(c) = [F(b) - F(a)] / [b - a]. The example above gives a physical interpretation of this theorem, which is a good way to remember the statement. We noted that continuity at the endpoints is a necessary hypothesis, since otherwise the function defined by F(x) = 1 on (0,1] and F(0) = 0 would be a counterexample.
• Before proving the MVT, we used it to resolve our earlier conjecture. Indeed, suppose F'(x) = 0 for all x in some nonempty interval I. Given any a,b in I, the MVT guarantees the existence of a c between a and b such that F'(c) = [F(b) - F(a)] / [b - a]. But F'(c) = 0 by hypothesis, whence F(a) = F(b). Since a and b were arbitrary, we see that F must be constant.
• As a second application of MVT, we proved that if
  F'(x) > 0 for all x in a nonempty interval I, then F is increasing on I. (Recall: F is increasing means that
  F(b) > F(a) whenever b > a.) The proof is quite similar to the first corollary, so you should try to reconstruct it on your own.
• Philosophically, the point of the MVT is this. The definition of F'(x) involves a limit of a certain quotient. Using the MVT, we can remove the lim symbol... at a cost (we cannot control where the derivative is taken).
• Properly motivated, we now prove the MVT. We began with a special case called Rolle's theorem: if h(x) is differentiable on (a,b) and continuous on the endpoints, and if h(a) = h(b), then there exists some c in (a,b) such that h'(c) = 0. The proof was not too hard. By Theorem 7-2, h(x) has a max and a min on [a,b]. There are two cases: either h(x) has a max or a min in (a,b), or else the max and the min both occur at the endpoints. Both cases easily lead to the conclusion of the theorem. In the former case, we know the derivative vanishes at maxes and mins; in the latter case, we saw that h(x) must equal h(a) for all x in [a,b].
• Next, we proved the MVT by using a trick. We're given F(x) which is differentiable on (a,b) and continuous at the endpoints. We cannot apply Rolle's theorem to F because F(a) might not equal F(b). Instead, we construct a function h(x) which mimics the behaviour of F, but satisfies h(a) = h(b). How do we do this? Write
  h(x) = F(x) + (...) (x-a), so that h(a) = F(a). We wish to determine (...) in such a way that h(b) = h(a), i.e. we want h(b) = F(a). A little thought shows that letting (...) = F(a) - F(b) gives us what we want -- that h(x) satisfies the hypotheses of Rolle's theorem. It follows that there exists a point c such that h'(c) = 0. Writing out h'(c) in terms of F and F' yields the MVT.
• Finally, we briefly discussed Cantor's Ternary Function, which is a crazy continuous function C : [0,1] → [0,1] satisfying C(0) = 0, C(1) = 1, and C'(x) = 0 whenever C'(x) is defined. (You should convince yourself that this is pretty weird.)

We sketched the graph of F(x) = (1+x^2)^-1 by using derivatives. This involved the following steps:
• Find all places where F'(x) = 0. (Just x=0.)
• For each of these, determined whether F has a max, a min, or neither there. (Since F'(x) > 0 for x < 0, and F'(x) < 0 for x > 0, we must have that F is increasing to the left of 0, and decreasing to the right of 0, i.e. F has a max at 0.)
• Determined when F is positive and negative. (In this case, always positive.)
• Determined the long-term behaviour of F, i.e. the limits as x → ± ∞ (in this case, both limits tend to 0).
• Sketch!
• Next we discussed the Second Derivative Test; click here for a pdf.
• Finally, we talked about L'Hôpital's Rule, discussed in this pdf.

FRIDAY, MAR. 9

• Reviewed the calculation from last time: the integral from 0 to 5 of the function f(x) = x.
• Trick for quickly guessing the integral from -2 to 2 of the same function f(x).
• Integrated the function g from 0 to 9, where g(x) = 2 everywhere except at x=1, and g(1) = 5.
• New notation for integrals (to help tell what's the variable of integration).
• Covered four basic properties of integrals:
• The integral on [a,b] equals the integral on [a,c] plus the integral on [c,b];
• The integral of the sum is the sum of the integrals;
• The integral of Cf(x) equals C times the integral of f(x); and
• If f is bounded between m and M on the interval [a,b], then the integral of f on [a,b] is bounded between m(b-a) and M(b-a).
• Discovered the fundamental theorem of calculus by looking at a picture.

• Stated and proved the "first form" of the fundamental theorem of calculus
• Observed that a modification of the proof shows that the function F(x), defined to be the integral from a to x of f, is continuous even if f is not.
• Proved that if f is continuous and integrable on [a,b], and if g' = f, then the integral from a to b of f equals g(b) - g(a).

If you want a challenge, try 7(vi).

SUNDAY, MAR. 25

MONDAY, APR. 23

IC 130