class 050907

If you are having trouble solving the system of equations, you should complete the exercise solving a system of quadratic equations and submit your solution using the form.

Note questions submitted for today's class at questions 050907.

1.  If our depth vs. clock time model is y = .02 t^2 - 3 t + 90, with y in cm when x is in seconds, then

• What equation tells us the depth at clock time 25 seconds, and how do we solve this equation?

• What equation tells when the depth will be 60 cm, and how do we solve this equation?

2. The best quadratic function describing illumination y in W / m^2 as a function of distance x from the Sun in AU is approximately y = 44 x^2 - 600 x + 1900.

• What three points did you use to get your model?

• What function model did you end up with?

• How well did your model fit the data?  Be as specific as possible.

• How did you find the range of distances that would correspond to the illumination range 25 W / m^2 to 100 W / m^2, and what were your results?  If you didn't get these results, how would you use the best-fit quadratic equation given above to find them?

The figure below shows the best quadratic fit and the best power-function fit to the data given for the illumination vs. distance problem.  It should be clear that the quadratic fit isn't good at all, while the power-function fit is excellent.

We can approximate the power-function fit using the form y = A x^p.  Substituting the second and seventh points (2, 264) and (7, 20) we get the equations

• 264 = A * 2^p
•  20 = A * 7^p.

Dividing the first by the second we get

• 13.2 = (2/7)^p.

Trial and error tells us that p lies between -2 and -2.1.

Using p = -2.05 in the first of the original equations we get

• 264 = A * 2^(-2.05) so that
• A = 264 / (2^(-2.05)) = 1093.

So our model would be y = 1093 * x^(-2.05).

This differs a bit from the model shown below, but being based on only two points this should be expected.