## Wednesday, June 14, 2006

### Area under a curve

Math 400: Problem 1.1.2

How far does an object travel in 15 seconds if its velocity is given by the function v(t) = 20 + 7 cos t ft/sec? It helps to recall that we can represent the distance traveled by the area under a velocity curve. Let's look at a graph where t is the horizontal axis and y = v(t) is the vertical axis. The time units are in seconds and the velocity units are feet per second (ft/sec).

In the following graph, the red curve represents the velocity function. As we can see, the traveling object alternately slows down and speeds up (the effect of the cosine term), never going more slowly than 13 ft/sec and never going faster than 27 ft/sec. The area under the curve for time values between t = 0 and t = 15 (in blue) represents the distance traveled, if only we could figure it out. Learning how to do that is one of the goals of calculus. However, it's not too difficult to get a rough estimate.
In the second figure, I've drawn in the lines y = 13 and y = 27 to mark off the low and high points of the velocity curve. If the object were traveling only 13 ft/sec for 15 sec, then the distance it travels would be just (13 ft/sec)(15 sec) = 195 ft. That's the area of the blue rectangle that represents a lower bound on the area under the red curve. On the other hand, if the object always traveled 27 ft/sec (the upper bound on velocity), then the distance traveled would be (27 ft/sec)(15 sec) = 405 ft.

We have now figured out that the distance traveled is somewhere between 195 ft and 405 ft. That's a pretty broad range, to be sure, but we could make it better by using less extreme bounds. For example, we could try drawing triangles inside the humps of the cosine curve to estimate the area we left off while computing the lower bound. That should help. At some point, though, we would probably want something better than computing lots of different estimates. When we have the right calculus tools available, we'll be able to show that the exact distance traveled is 300 + 7 sin 15 ≈ 304.55 ft.

## Sunday, June 11, 2006

### Escher in Sacramento

Crocker Art Museum Exhibit

M. C. Escher's work will be on display at the Crocker Art Museum in Sacramento from June 10 to September 3 of this year. M. C. Escher: Rhythm of Illusion was originally organized by the Portland Art Museum and features 70 examples of the artist's surreal art. Many of Escher's drawings are familiar to college students, especially since they are so commonly displayed in the offices and classrooms of math teachers.

An article by The Sacramento Bee's art correspondent points out that Escher's original work can surprise people who are accustomed to seeing his art as represented by the many popular posters:

While reproductions of his work can sometimes seem cold and sterile, the actual prints have warmth, depth and texture that make them more sensual and personal than one might expect. One of the purposes of the show, [Crocker curator William] Breazeale says, is to demonstrate what a difference there is between reproductions and the real thing.

There is an official M. C. Escher website where interested parties can find many useful resources related to the artist, including a detailed picture gallery.