Making Connection With Rational Function and Equation.

OH mY GOD!!!! tHE Due date is today!! I haven't post the last sub chapter yet. Relax, take a deep breath.....PANIC!!! Ms. Joanne and Ms. Malar, please forgive me.....

This chapter is quite short. We can see an example of everyday life related to rational function. An example from the text book is, "Most of us know that getting closer in a concert means a better view, but it is also means more exposure to potential damage sound levels. The intensity, I, increase by what is known as the inverse square law, or reciprocal of the square of the distance, d, from the sound source, so that I=k/d^2. This law also applies to gravitational force and light intensity.", text book page 186. 

The question:

The intensity of sound, in watts per square metre, varies inversely as the square of the distance, in metres, from the source of the sound. The intensity of the sound from the loudspeaker at the distance of 2 m is 0.001W/m^2. 

a) Detemine a function to represent this relationship.

b) Graph this function.

c) What is the effect of halving the distance from the source of the sound.

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The solutions: 

a) So, I=k/d^2, k is constant.

substitute I=0.001 and d=2 into the equation to find k.

I=k/d^2

0.001=k/(2)^2

0.001=k/4

k=0.001 x 4

k=0.004

So, I=0.004/d^2

b) The graph: 

The graph will looks like this, but the scale is different according to the equations. The y-axis represent the Intensity and the x-axis represent the distance. The equation of the graph should be written on the graph: I=0.004/d^2.

c) substitute 1/2d for d.

I= 0.004/(1/2)d^2

I= 0.004/(d^2/4)

I=4 x 0.004/d^2

Therefore if the distance is halved, the sound is four times as intense. In other word, the further you get to the stage of the concert the high the exposure to potentially damaging sound levels.

I have finish the question!!!Hope you all understand. I also hope you all can answer a similar question as this one, as most of the question is using the same method. 

oNE, Last thing, the special case: 

The special case is that there is discontinuous at a point where if we calculated manually we will say that it is the restriction and the vertical asymptote. But if we graph the function using the gc, we will see there is no discontinuous in the graph. If we want to identify which point is the discontinuous point we just have to substitute the value then see the result. If the the result does not approaches negative or positive infinity instead the results is a whole number, thus that value is the point where the discontinuous is. In other word, we can just say there is a hole at the point (x,y). This is an example of a graph with a hole.

That is all for chapter 3.5 and unfortunate it is the end for this chapter. Later I will see you all again in with chapter 7 ahead. I am looking forward to the next chapter that i will post in a few months time. Bye!!!