Lecture 1 (Feb 7)
After a brief interlude on continued fractions and how awesome they are, we discussed the concept of average temperature over the course of a day. We approximated it by measuring the temperature at a bunch of times throughout the day and averaging them, and realized that the more times we sample the temperature throughout the day, the closer we get to an ideal notion of average temperature. We then worked out a precise algebraic formula that represents this. Finally, we noticed a funny connection to area (which doesn't seem to have anything to do with average temperature at first glance). We'll discuss this more, and begin our review of derivatives, next time.
Lecture 2 (Feb 10)
In the previous class we saw that the average temperature is related to the area of a region defined by the graph of temperature. This foreshadows integral calculus, which deals primarily with (i) reinterpreting natural phenomena in terms of area, and (ii) providing techniques to compute area. We then moved on to start our `review' of differential calculus, albeit from a different perspective than you've seen it. We started with a philosophical paradox: what does it mean to say that a car is moving at 30 miles per hour at 3pm? After all, at a given moment in time, the car doesn't move at all! We realized the best way to resolve this seeming contradiction is to think about where the car was just before 3pm and where it was just after 3pm; dividing the distance it traveled over that tiny time window by the amoung of time it took gives the velocity of the car. The tinier the time window becomes, the better an idea we get about the actual speed precisely at 3pm! We then represented this idea pictorially: drawing a graph of the location (position) of the car at different times, and then zooming in on 3pm in this graph. If we zoom in enough, the graph will begin to look like a line, and we saw that the slope of this line is precisely the velocity of the car. More generally, given any function f, if we zoom in really far on the point (3,f(3)), the graph will start to resemble a line, and the slope of that line is called the derivative of f at 3, denoted f'(3).
Lecture 3 (Feb 12)
Last time we discussed the geometric definition of f'(3): zoom in on the point (3,f(3)) until the function looks like a line, and set f'(3) to be the slope of this line. (You have likely seen a related geometric definition before, in terms of the tangent line. But was is the tangent line? How do you find it? The answer is: you zoom in until the function looks like a line, and that line is the tangent line!)
Next we used the limit defintion to compute the derivative of the function x2 at x = 3. Reviewing our computation more carefully, we saw that the derivative of x2 at any input x is 2x. Of course this is a formula that's already familiar to you; the purpose of the above was to explain where that formula comes from. A different notation for the derivative is d⁄dx: d⁄dx f(x) means the derivative of f(x). For example, d⁄dx x2 = 2x. The advantage of this notation is that x2' is awkward to write (and you shouldn't write it ever). The notation does have a serious disadvantage, however, which is best illustrated by example. Let f(x) = x2. Then d⁄dx f(3) = d⁄dx 9 = 0. In general, it's difficult to use this notation to express the derivative of f at a specific point. The last part of our class was dedicated to deriving the product rule: a rule for how to differentiate the function f(x) g(x). By definition, the derivative (say, at 3) of this function is the limit as h approaches 0 of (f(3+h) g(3+h) - f(3)g(3))/h. But examination of the limit definition of the quantities f'(3) and g'(3) demonstrate that f(3+h) ≈ f(3) + f'(3) h and g(3+h) ≈ g(3) + g'(3) h, with both of these approximations getting better and better as h approaches 0. Plugging these approximations into the limit definition of d⁄dx (f(x)g(x)) at x = 3, we found that this equals f(3) g'(3) + g(3) f'(3). Replacing the number 3 by x throughout, we find that we've recovered the product rule.Lecture 4 (Feb 17)
Last time we approximated values of f(x) near x = 3 in terms of the derivative, and used this to discover the product rule. After reviewing this, we used the product rule to find the derivative of 1/x. (The idea: 1 = x * 1/x. Differentiate both sides and use product rule.) We then used the product rule to differentiate x3, and more generally, to differentiate xn for any positive whole number n. This gave us the power rule. Using the same trick as for 1/x, we saw how we could also use the product rule to differentiate x-n, and obtained the same power rule formula as before.
Lecture 5 (Feb 19)
We used the approximation from before (approximating the value of f(x) near x = 3 in terms of the derivative of f) to discover the chain rule: the derivative of f(g(x)) is f'(g(x))g'(x). This formula is useful, but it's not always obvious how to apply it in practice. One way of thinking about it is: given a complicated-looking function, try to say it out loud as a single function of "something". For example, f(x)3 looks like "something cubed", where the "something" is f(x). (If you instead tried to say it looked like "f of something", what's the "something"? You can't say it, so that means you're not looking at it in the best way.) Chain rule then asserts that you take the derivative as if you were simply differentiating that simple version, but then multiplying by the derivative of the "something". For example, since f(x)3 looks like "something cubed", we take the derivative in the usual way using the power rule: 3*(something)2. But then we multiply this by the derivative of the something. In other words, the derivative of f(x)3 is 3*f(x)2*f'(x). We then practiced chain rule, for example finding the derivative of √x by realizing that (√x)2 = x, so we could differentiate both sides.
Next we turned to trigonomety. Everyone in the room "knew" what the derivative of sin x was. But it turns out that almost no one knew what sin x means! After some discussion, we gave a definition of cosine and sine: if you start at the point (1,0) and walk a total distance of θ along the boundary of the unit circle counterclockwise, the point you end up at is (cos x, sin x). We finished with a discussion of angle measurement, and why measuring angles in degrees is rubbish.
Lecture 6 (Feb 21)
After a presentation by Aniah, we practiced a bit of chain rule, in particular working out the derivative of 1/g(x) from the homework. We then practiced computing some sines and cosines. We used the idea of similar triangles to derive the familiar SOHCAHTOA rule. We finished by deriving an alternative formula for the angle of a triangle: half the product of any two sides with the sine of the angle between them.
Lecture 7 (Feb 24)
Using the Pythagorean Theorem, we saw that sin2 θ + cos2 θ = 1. But why is the Pythagorean Theorem true? We explored a variant on a proof invented by James Garfield (class of 1856). The idea was to join two copies of a given right triangle together, and then to complete the picture to form a rectangle; by computing the area in two different ways, we deduce the Pythagorean Theorem. We then employed a similar strategy to deduce a formula for sin(x+y). Finally, we used this formula to show that the derivative of sin x (where x is in radians) is cos x.
