July 14

1. July 14 More on derivatives; Integrals

2. Questions and review Your thoughts.. Let us know if you have a question you would like us to address.

3. Discussion Here is the graph of a function. Draw a graph of its derivative.

4. Position, velocity, and acceleration: You know that if y = f(t) represents the position of an object moving along a line, the v = f '(t) is its velocity, and a = f "(t) is its acceleration. Example: For falling objects, y = is the height of the object at time t, where is the initial height (at time t=0), and is its initial velocity.

5. The meaning and uses of derivatives, in particular: • (a) The idea of linear approximation • (b) How second derivatives are related to quadratic functions • (c) Together, these two ideas help to solve max/min problems

6. Basic functions --linear and quadratic. • The derivative and second derivative provide us with a way of comparing other functions with (and approximating them by) linear and quadratic functions. • Before you can do that, though, you need to understand linear and quadratic functions.

7. Let’s review • Let's review: linear functions of one variable in the plane are determined by one point + slope (one number): • y = 4 + 3(x-2)

8. Linear functions • Linear functions occur in calculus as differential approximations to more complicated functions (or first-order Taylor polynomials): • f(x) = f(a) + f '(a) (x-a) (approximately)

9. Quadratic functions • Quadratic functions have parabolas as their graphs:

10. Quadratic functions • Quadratic functions occur as second-order Taylor polynomials: • f(x) = f(a) + f '(a)(x-a) + f "(a)(x-a)2/2! (approximately)

11. They also help us tell... • … relative maximums from relative minimums -- if f '(a) =0 the quadratic approximation reduces to • f(x) = f(a) + f "(a)(x-a)2/2! and the sign of f "(a) tells us whether x=a is a relative max (f "(a)<0) or a relative min (f "(a)>0).

12. Review - max and min problems Also, by way of review, recall that to find the maximum and minimum values of a function on any interval, we should look at three kinds of points: 1. The critical points of the function. These are the points where the derivative of the function is equal to zero. 2. The places where the derivative of the function fails to exist (sometimes these are called critical points,too). 3. The endpoints of the interval. If the interval is unbounded, this means paying attention to

13. Related Rates Recall how related rates work. This is one of the big ideas that makes calculus important: If you know how z changes when y changes (dz/dy) and how y changes when x changes (dy/dx), then you know how z changes when x changes: Remember the idea of implicit differentiation: The derivative of f(y) with respect to x is f '(y) dzdzdy dx dy dx = dy dx

14. More on related rates The idea is that "differentiating both sides of an equation with respect to x" [or any other variable] is a legal (and useful!) operation. This is best done by using examples...

15. Related Rates Greatest Hits A light is at the top of a 16-ft pole. A boy 5 ft tall walks away from the pole at a rate of 4 ft/sec. At what rate is the tip of his shadow moving when he is 18 ft from the pole? At what rate is the length of his shadow increasing? A man on a dock is pulling in a boat by means of a rope attached to the bow of the boat 1 ft above the water level and passing through a simple pulley located on the dock 8 ft above water level. If he pulls in the rope at a rate of 2 ft/sec, how fast is the boat approaching the dock when the bow of the boat is 25 ft from a point on the water directly below the pulley?

16. Greatest Hits... A weather balloon is rising vertically at a rate of 2 ft/sec. An observer is situated 100 yds from a point on the ground directly below the balloon. At what rate is the distance between the balloon and the observer changing when the altitude of the balloon is 500 ft? The ends of a water trough 8 ft long are equilateral triangles whose sides are 2 ft long. If water is being pumped into the trough at a rate of 5 cu ft/min, find the rate at which the water level is rising when the depth is 8 in. Gas is escaping from a spherical balloon at a rate of 10 cu ft/hr. At what rate is the radius chaing when the volume is 400 cu ft?

17. Integrals Start with dx -- this means "a little bit of x" or "a little change in x" If we add up a whole bunch of little changes in x, we get the "total change of x" -- A tautology question: If you add up all the changes in x as x changes from 2 to 7, what do you get? A. 0 B. 2 C. 5 D. 7 E. cannot be determined

18. We write this in integral notation as: If y = f(x), then we write dy = f '(x) dx. To add up all the "little changes in y" as x changes from 2 to 7, we should write or ... and the answer should be the total change in y as x changes from 2 to 7, in other words This is the content of the fundamental theorem of calculus!

19. The fundamental theorem of calculus… gives the connection between derivatives and integrals. It says you can calculate precisely if you can find a function whose derivative is g(x). And the result is the difference between the value of the "anti-derivative" function evaluated at b minus the same function evaluated at a.

20. Basic antiderivative formulas: except for n= -1

21. A quick example Find the value of A. 7/3 B. 0 C. 1 D. 5/3 E. 2 F. 1/3 G. 4/3 H. 2/3

22. Fundamental Theorem Workout Let Find the value of f '(1) -- the derivative of f at 1. A. 3 B. 8 C. 4 D. 0 E. 5 F. 2 G. 6 H. 1

23. Substitution In some ways, substitution is the most important technique, because every integral can be worked this way (at least in theory). The idea is to remember the chain rule: If G is a function of u and u is a function of x, then the derivative of G with respect to x is: = G'(u) u'(x) dG dx

24. For instance... could be considered as euwhere u = x2. To differentiate then, we use that the derivative of euis eu : Now we’ll turn this around...

25. To do an integral problem... For a problem like we suspect that the x4should be considered as u and then x3 dx is equal to du/4. And so:

26. In general... In substitution, you 1. Separate the integrand into factors 2.Figure out which factor is the most complicated 3. Ask whether the other factors are the derivative of some (compositional) part of the complicated one. This provides the clue as to what to set u equal to.

27. Here’s another one: -- the complicated factor is clearly the denominator (partly by virtue of being in the denominator!) and the rest (x dx) is a constant times the differential of x -- but it's a good idea to try and make u substitute for as much of the complicated factor as possible. And if you think about it, x dx is a constant times the differential of 2x +5! So we let u = 2x +5, then du = 4 x dx, in other words x dx = du / 4 . So we can substitute: 2 2 2

28. Now you try a couple... A) 0 B) 1/2 C) 1 D) p/2 E)

29. p/4 Find sec x sin(tan x) dx 2 0 A) p/2 B) 1-p/4 C) sin 1 D) 1 - cos 1 E) p/2 - sin 1 F) p/4 + cos 1 G) 1 + 3p/4 H) 1 + tan 1

30. Integrals and Areas A problem that was around long before the invention of calculus is to find the area of a general plane region (with curved sides). And a method of solution that goes all the way back to Archimedes is to divide the region up into lots of little regions, so that you can find the area of almost all of the little regions, and so that the total area of the ones you can't measure is very small.

31. By Newton's time, people realized that it would be sufficient to handle regions that had three straight sides and one curved side (or two or one straight side -- the important thing is that all but one side is straight). Essentially all regions can be divided up into such regions. Ameba

32. These all-but-one-side-straight regions look like areas under the graphs of functions. And there is a standard strategy for calculating (at least approximately) such areas. For instance, to calculate the area between the graph of y = 4x - x2 and the x axis, we draw it and subdivide it as follows:

33. Since the green pieces are all rectangles, their areas are easy to calculate. The blue parts under the curve are relatively small, so if we add up the areas of the rectangles, we won't be far from the area under the curve. For the record, the total area of all the green rectangles is: 246 25 whereas the actual area under the curve is: Also for the record, 246/25 = 9.84 while 32/3 is about 10.6667.

34. We can improve the approximation by dividing into more rectangles: Now there are 60 boxes instead of 20, and their total area is: which is about 10.397. Getting better. We can in fact take the limit as the number of rectangles goes to infinity, which will give the same value as the integral. This was Newton's and Leibniz's great discovery -- derivatives and integrals are related and they are related to the area problem. Area 60 boxes 7018 675

35. Limits of Riemann sums A kind of limit that comes up occasionally is an integral described as the limit of a Riemann sum. One way to recognize these is that they are generally expressed as , where the “something” depends on n as well as on i.

36. Green graph Again, recall that one way to look at integrals is as areas under graphs, and we approximate these areas as sums of areas of rectangles. This is a picture of the“right endpoint” approximation to the integral of a function.

37. approximating

38. Example...

39. solution

40. Area between two curves: A standard kind of problem is to find the area above one curve and below another (or to the left of one curve and to the right of another). This is easy using integrals. Note that the "area between a curve and the axis" is a special case of this problem where one of the curves simply has the equation y = 0 (or perhaps x=0 )

41. Solving area problems 1. Graph the equations if possible 2. Find points of intersection of the curves to determine limits of integration, if none are given 3. Integrate the top curve's function minus the bottom curve's (or right curve minus left curve).

42. Example: Find the area between the graphs of y=sin(x) and y=x(p-x)

43. It’s easy to see that the curves intersect on the x-axis, and the values of x are 0 and p. The parabola is on top, so we integrate: And this is the area between the two curves.

44. An Area Question: Find the area of the region bounded by the curves y=4x2 and y=x2+3. A. 1/2 B. 1 C. 3/2 D. 2 E. 5/2 F. 3 G.7/2 H. 4

45. Position, velocity, and acceleration: Since velocity is the derivative of position and acceleration is the derivative of velocity, Velocity is the integral of acceleration, and position is the integral of velocity. (Of course, you must know starting values of position and/or velocity to determine the constant of integration.)

46. Example... An object moves in a force field so that its acceleration at time t is a(t) = t -t+12 (meters per second squared). Assuming the object is moving at a speed of 5 meters per second at time t=0, determine how far it travels in the first 10 seconds. 2

47. Solution... First we determine the velocity, by integrating the acceleration. Because v(0) = 5, we can write the velocity v(t) as 5 + a definite integral, as follows: The distance the object moves in the first 10 seconds is the total change in position. In other words, it is the integral of dx as t goes from 0 to 10. But dx = v(t) dt. So we can write: (distance traveled between t=0 and t=10) = = = 3950/3 = 1316.666... meters .

48. Good night! See you next week (or on the web!)