Precalculus I Class 01/14


http://164.106.222.236/ 

dsmith@vhcc.edu

dasmith@naxs.com

Using the points (14, 80), (59, 50) and (127, 20), each of the form (t, y), we plugged the y coordinate of each into the form y = a t^2 + b t + c to get the three equations:

Subtracting the first equation from the second, the subtracting the first equation from the third we obtained a system of 2 equations from which we have eliminated c:

Now continue the solution according to the following instructions:

Multiply the first of these equations by -113 and multiply the second by 45. 

The equations become

Adding the two equations we get

Solving we get 

Substitute this variable back into either of the original 2-equation system and solve for the variable that remains.

Substituting into  3285 a + 45 b = -30 we obtain

Substitute the values of the two variables back into any of the original 3-equation system and solve for the variable that remains.

Using the first equation 196 a + 14 b + c = 80 we get

Substitute your values of a, b and c back into the form y = a t^2 + b t + c of the quadratic.

Plug in t = 27 and evaluate y.  Compare with our actual observations of depth vs. clock time.

Our solutions are 

a = 0.001995, b = -0.8123, c = 90.9816

giving us the model

y = 0.001995 t^2 + -0.8123 t + 90.9816, or

y = 0.001995 t^2 - 0.8123 t + 90.9816.

This compares with the model y = 0.0019 t^2 - 0.7907 t + 90.183 given by our spreadsheet, which is based on all data points rather than just the 3 we have used here.

At t = 27 we obtain

y = 0.001995 * 27^2 -0.8123 * 27 + 90.9816 = 70.5,

representing a depth of 70.5 cm when t = 27 sec.  This compares nicely with our observations, where we saw that t = 27 when y = 70.  

Using the model find the clock time t at which the depth is 60, and compare with our observations.

Using y = 0.001995 t^2 -0.8123 t + 90.9816 we note that depth is represented by y so our equation becomes

60 = 0.001995 t^2 -0.8123 t + 90.9816.

This is a quadratic equation, which we solve using the quadratic formula.  The quadratic formula tells us that the equation a t^2 + b t + c = 0 when and only when t = [- b +- sqrt( b^2 - 4 a c) ] / (2 a).

Our equation is not in the form a t^2 + b t + c = 0, but it can easily be put into this form if we subtract 60 from both sides:

60 - 60 = 0.001995 t^2 -0.8123 t + 90.9816 - 60

0.001995 t^2 -0.8123 t + 30.9816 = 0.

We obtain 

t = [ - (-.8123) +- sqrt( (-.8123)^2 - 4 * .001995 * 30.9816)  ] / (2 * .001995) ] 

 =  [ - (-.8123) +- sqrt( .4126) ] / (.003990) ] 

 =  [ - (-.8123) +- .6423 ] / (.003990) ] 

Using the + of the +- we obtain 1.455 / .003990 = 364.6.

Using the - of the +- we obtain  .1700 / .003990 = 42.59.

This indicates that the depth is 60 when t = 42.59 or when t = 364.6.

Our original observations indicate that the depth is 60 cm when t = 42 sec, which is reasonably close to the 43.59 sec of our observation.  When we consider that our observations of the second hand of the clock could easily be off by a second, this provides good confirmation of our model.

This also shows us why we really didn't need all those decimal places in our model.  Our observations themselves were only accurate to 2 significant figures.  A 2-significant-figure model can be obtained from our solution y = 0.001995 t^2 -0.8123 t + 90.9816 by simply rounding everything to 2 decimal places:

y = .0020 t^2 - .81 t + 91.