The pedals on a bicycle have a maximum height of 30cm above the ground and a minimum distance of 8cm above the ground. Jeng pedals at a rate of 20 cycles per minute.
a) What is the period, in seconds for this function?
b) At t = 0, Jeng's right foot is closest to the ground.
i) Write 2 equations that represent the height of her right foot above the ground; 1 sine; 1 cosine.
ii) For how long per cycle is Jeng's right foot 20cm, or higher, above the ground?
To start of, part A of the question was fairly straight forward. To find the period, we take 1 minute and convert it to seconds which is 60 seconds. And to find out how long each cycle took, we divided the time, 60 seconds, by the number of cycles per 60 seconds, which is 20 cycles.
And this results in,
3 seconds per cycle.
For part B, i, we're asked to make up two equations, one for sine, and one for cosine. For some people it might be easy to make the equations, but sometimes it's easier to determine the equations for the function by making a graph of it first.
NOTE: This graph has some errors. For one, when drawing a graph don't forget to label your x and y axis and draw the arrows. For two, don't forget to add arrow(s) or dot(s) to the end of the function depending on the context of the original questions. An example can be time, because time does not go backwards there would be a dot at the point where the function touches the y-axis and an arrow on the other end indicating that the graph continues with time. For three, you should always number at least the 4 points (for your max, min, and where it touches the sinusoidal axis). Usually it's labeled as π/2, π, 3π/2 and 2π. In this question we should label it 0.75, 1.5, 2.25 and 3.
From the graph, it's a lot easier to determine the equations that go with it.
Making a list also helps come up with the equations.
It's usually easier to do it in the DABC order. So starting with D, we can easily find it by finding the middle between the max and min value. In this case it's 19. Since the sign is always positive for D, we can put 19 beside D for both sine and cosine. For A, we simply find the distance from either the max or min, depending on what you prefer, to the sinusoidal axis. For this question A is 11. The sign for A we don't know yet, but we'll come back to that afterwards. The period, which we calculated earlier (3), is the same for both sine and cosine so we can place that where B is. For C, we're looking for a phase shift. For cosine we know that the max or min value starts at the y-axis.. therefore C for cosine is 0. Knowing this, we can figure out the sign of A, which in this case is negative (-) because cosine is starting at it's min. For sine, we know that there has been a phase shift of 0.75 and it's moved to the right so C for sine is +0.75. We can tell from the graph that A for sine is positive so parameter A is +11.
Part B, ii, we can do on our calculator. First we take the equation we made and put it into Y1. Then for Y2 we put in a 20.
Now that we've done that, we find the intersection of Y2 and the function -11cos((2π/3)x)+19.
The first intersection was 0.7935 and the second intersection was 2.2065. To find the amount of time that Jeng's foot is 20cm or higher above the ground we subtract 0.7935 from 2.2065 and we end up with 1.4131. 1.4131 is the amount of time that Jeng's foot is 20cm or higher above the ground. Wasn't that easy?
Well we started another question in class today, but I'll leave that for Danny to do tomorrow since we're not even close to being finished it. Homework will be posted on the blog if it hasn't already and during tomorrow's morning class we'll be completing the question we started today. Hope you enjoy the rest of early dismissal!. (: Tootles.