Monday, 27 August 2012

Viva Voce - Balram Sharma

Part I: My Viva Voce Video


Name: Balram Sharma
Class: S2-03
Scenario #2
Elapsed Time: 4:08 minutes

Part II: My Reflections

a) What are my key learning points?

My problem-solving skills have surely been built upon through this performance task. I have become more proficient in video editing and presentation skills, as well, and after having completed this performance task, I feel more comfortable using technology to present the solutions to mathematics questions, and even the answers to questions in other subjects. 

b) What have I understood about trigonometry and Pythagoras' theorem as an applied learner?

In my opinion, an applied learner is one who learns effectively from real-world experiences, and can successfully apply the concepts learned through these experiences to his or her academics, and vice-versa. From this performance task, I have learned more about trigonometry and Pythagoras' theorem, and I have also learned how to apply these concepts, as well as some of the other concepts I have learnt at SST (including many ICT, ADMT and Oral Communications skills), in the real world.

Part III: The Script

The question is, "A spider sits at the middle point, P, of my living room wall. It sees a fly at corner C. Find the shortest distance the spider must travel to reach point C". Okay, that sounds simple enough. Let's start by drawing a diagram. I've labelled the points A, E, D, H, C and G and I've also added P, as well as Y and X. Now, Y is in the middle of the line DH, and X is in the middle of the line AD. So, first of all, let me draw the most obvious route, which takes us through PY, YD and DC. That's PYDC. Next, we're going to try something a little shorter, PDC. That's PD, and DC, simple as that. Let's not forget point X. How about using that? So, we could go, PX, and XC? How about that? But then, there must be something shorter than that. So, let's create a point, and call it M. And let's put it smack in the middle of line DX. So we could go, PM, and MC. That seems quite a bit shorter. But all these routes mean nothing if we don't do the math. Let's start with PYDC. That's PY, which is one meter, and YD, which is two meters, and then DC, which is 5 meters. That means PYDC is an eight meter long route. Then, we'll start on PDC. That's PD, which is the square root of five meters, and DC, which is five meters, according to the diagram. So that leads us to PDC, which is 7. 236067977 meters, or 7.24 meters when rounded off to three significant figures. Of course, we can't forget PXC, which is PX, two meters, plus XC, which is the square root of twenty six meters. So, that means PXC is a 7.099019 meter long route, or 7.10 meters to three significant figures. Finally, let's go on to PMC, which is a little more complicated. That's PM, which is 2.061552813 meters, and MC, which is 5.024937811 meters. That means PMC is 7.086490265 meters long, or 7.10 meters long when rounded off to three significant figures. Therefore, the shortest route is PMC. Of course, that's in exact figures. Rounded off to three significant figures, it's a tie between PXC and PMC. Now we're going to look at the assumptions we've made. First of all, the sum of two sides of a triangle must be greater than the length of the third side. Secondly, the spider must only be able to move in straight lines which connect two points, and not in curves or irregular lines. Also, the spider must not be able to retreat along any route. Finally, we're going to assume that the spider can't be able to swing from surface to surface. In conclusion, Route 4, that is, PMC, is the shortest route, and of course, we have working to back that claim up. Thank you for watching this video.

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