Precalculus I Class 01/14
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:
196 a + 14 b + c = 80
3481 a + 59 b + c = 50
16129 a + 127 b + c = 20
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:
3285 a + 45 b = -30
15933 a + 113 b = -60.
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
-371205·a - 5085·b = 3390
716985·a + 5085·b = -2700
Adding the two equations we get
345780 a = 690
Solving we get
a = 690 / 345780 = .001995.
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
3285 * .001995 + 45 b = -30, which gives us
6.554 + 45 b = -30 so that
45 b = -36.554 and
b = -36.554 / 45 = -.8123.
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
196*.001995 + 14 * -.8123 + c = 80 so that
c = 90.9816
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.