Math 331
Test 1
Name________________
Directions: You may type your solutions directly on this test or rewrite them by hand, or use a combination of both. In
any case you must show your work where appropriate.
I. Common Core Modeling Schematic (5 pts each)
1. Draw the Common Core Modeling Schematic.
2. In a regression model, which part of the Common Core
Modeling Schematic applies to the R2 statistic? Explain!
R2 is used to interpret, but especially validate the linear
model for how accurately it fits the data. The closer R2 is
to one, the better the model fits the data.
II. Continuous Functions (20 pts)
1. State the definition of a function f continuous at a point
(c,f(c)).
2. By dividing and multiplying by factors, define a function
g that is equal to f(x) = x2 everywhere except at x = -2 and
x =1.
f: DC is continuous at (c,f(c)) if and only if
lim 𝑓(𝑥) = 𝑓(𝑐).
g(x)=
That is
Sketch a graph of your function below, labeling all
discontinuous points.
𝑥→𝑐
(𝑥+2)(𝑥−1)𝑥 2
(𝑥+2)(𝑥−1)
20
1. Limit exists as xc;
2. f(c) exists; and
3. 1. = 2.
15
10
5
0
-6
-4
-2
0
2
4
G is discontinuous at (1,1) and (-2,4).
3. Prove or disprove the function below is continuous at x = 0.
F(x) =
x2 for x < 0
x for x > 0
The idea is to show the three conditions in 1. (above) are true, for f when x = c = 0.
The limit as x0 of x2 is 0, and the limit as x0 of x is 0, so the limit of f(x) as x0 is 0. Also f0)= 0. So, the limit exists
and the functional value exists at x = 0, and they are equal. Therefore, f(x) is continuous at x = 0.
Math 331
Test 1
Name________________
III. Regression
1. Briefly, but carefully and accurately, outline the stepwise process leading to linear regression coefficients in a 2-D
model. State all equations and matrices accurately. (15 pts)
1. The main idea is to minimize the SSRes, which is the
sum of squares of all the residual values.
𝑛
2
2
SSres =∑𝑛𝑖=1(𝑦𝑖 − 𝑦̂)
𝑖 =∑𝑖=1(𝑦𝑖 − 𝑚𝑥𝑖 − 𝑏)
2. Take the partial derivatives of SSRes with respect to m
and b and set equal to 0 to find the local minimum of the
paraboloid.
𝛿𝑆𝑆𝑅𝑒𝑠
=2 ∑𝑛𝑖=1(𝑦𝑖
𝛿𝑚
− 𝑚𝑥𝑖 − 𝑏)2 (−𝑥𝑖 ) = 0
𝛿𝑆𝑆𝑅𝑒𝑠
=2 ∑𝑛𝑖=1(𝑦𝑖
𝛿𝑏
− 𝑚𝑥𝑖 − 𝑏)2 (−1) = 0
𝑚
3. Solve for the vector of coefficients [ ].
𝑏
∑𝑛𝑖=1(−𝑥𝑖 𝑦𝑖 + 𝑚𝑥𝑖2 + 𝑏𝑥𝑖 ) = 0
𝑛
∑(−𝑦𝑖 + 𝑚𝑥𝑖 + 𝑏)2 = 0
𝑖=1
Suppressing the subscript I, we have:
2
[∑ 𝑥
∑𝑥
∑ 𝑥 ] [𝑚] = [∑ 𝑥𝑦]
𝑏
∑𝑦
1
and, finally
2
𝑚
[ ] = [∑ 𝑥
𝑏
∑𝑥
∑𝑥 ]
1
−1
[
∑ 𝑥𝑦
]
∑𝑦
Math 331
Test 1
2. Your pilot has landed her Cessna 172 on a 1600 ft
airfield, bordered by tall trees, which sits at an altitude of
10,250 ft above sea level. She wants to depart at 2pm,
fully loaded, when the temperature is predicted to be 35
degrees Centigrade. Should you ride with her? Explain
why or why not. (10 pts)
Name________________
3. What rate of climb should the Cessna 172 pilot expect
at 10,250 ft above sea level when the air temperature is 35
degrees Celcius? (There are lots of mountains around!)
(10 pts)
From the 3-D linear model, the climb rate would be
approximately 160 ft/min.
From the 3-D linear model, the ground roll would be
approximately 4011 ft. Do not go!
4. As Earth’s population continues to grow, the solid waste generated
Year
by the population grows with it. Governments must plan for disposal
and recycling of ever growing amounts of solid waste. Planners can
use data from the past to predict future waste generation and plan for
enough facilities for disposing of and recycling the waste. Given the
following data on the waste generated in Florida from 19901994, construct a function to predict the waste that was generated there
in the year 2012? Comment on the accuracy and limitations of your
model. (20 pts)
1990
1991
1992
1993
1994
2012
State your model.
State the accuracy of your model.
y = 304.93x2 - 1E+06x + 1E+09
R² = 0.9991
Tons of Solid Waste
Generated (in thousands)
19,358
19,484
20,293
21,499
23,561
What (obviously) limits extrapolation here?
Missing data between 1994 and the extrapolation.
IV. Separation of Variables.
Derive the solution to dP/dt = kP, subject to the conditions: P(0) = 1,221 and P(2) = 5,580. Show all work! (15 pts)
dP/dt = kP
dP/P = kdt
∫
1
𝑑𝑃 = ∫ 𝑘 𝑑𝑡
Ln|P| = kt + C
Since P> 0, P(t) = ekt+C.
So, P(t) = Poekt. (This is the exponential growth model. Use the two initial conditions to find Po and k.)
P(0) = Po = 1221,
and P(2) = 1221e2k = 5580, so k ~ .760
Consequently, the growth model is P(t) = 1221e.760t.