Math 181 - Practice exam 1 - solutions
The real exam will be shorter, but there will be plenty of space on the
exam paper.
Problem 1 Stride length of humans S is roughly a linear function of
height H. Given the following data points, find the equation for S in terms
of H which gives a line passing through both points.
H = 65 72
(8 marks)
S = 38 45
We are looking for a, b in the equation S = aH + b. The given datapoints
mean
38 = 65a + b
45 = 72a + b
which gives 7 = 7a or a = 1 after eliminating b, and then b = −27, hence
S = H − 27.
Note that this will probably not be a very accurate model since it is based
on only two data points.
Problem 2 Solve the following linear system.
3a + 5b − 2c = 21
a + b + 3c = −1
a − 2b + c = 0
(8 marks)
There are many ways to do this! Here is just one example. Eliminate c using
the third equation, by doing I + 2 · III, II − 3III. So we have the three
equations
5a + b = 21
−2a + 7b = −1
a − 2b + c = 0
Then eliminate eg b from the first two equations by subtracting 7 times I
from II, which gives −37a = −148 or a = 4. Backwards substitution then
gives b = 1, c = −2.
Problem 3 Some large birds can only take flight by jumping off a cliff.
Suppose such a bird jumps at time t = 0, and as long as it has not yet started
to flap its wings, its height above water at time t is a quadratic function of
the form
h(t) = −16t2 + bt + c.
1 2.5
t(sec) =
h(feet) = 182 95
a) Tell the equations for b and c following from these data.
b) Find the equation of h(t).
c) At what time will the bird hit the water if it never starts flapping its
wings?.
d) The bird needs until t = 3 to extend its wings and start flying. Even
then, it will still go lower another 30 feet before gaining height. What is the
bird’s margin of safety (the lowest point on its path)? Note - after t = 3, the
bird’s height is no longer given by your formula for h(t). (4 + 4 + 5 + 4 =
17 marks)
Suppose we have measurements
a) Plugging in the given values gives rise to two equations for b, c.
−16 + b + c = 182
−100 + 2.5b + c = 95
b) We eliminate c to get 1.5b = −3, so b = −2. Then we get c = 200 by
backward substitution (note that c = h(0) is the height of the cliff which was
not given). So
h = −16t2 − 2t + 200.
c) From the quadratic formula, h = 0 at time
√
−2 − 4 + 12, 800
≈ 3.47
t=
−32
The negative solution for t should be discarded.
d) At time t = 3 the bird has height h = −144 − 6 + 200 = 50 feet above
water. Plunging a further 30 feet, it will sink to 20 feet above water before
rising again. Note that we do not know when the lowest point will be reached
- there will be a different equation governing the height after time t = 3.
Problem 4 Test subjects are given varying doses of a cold medicine.
Then they have to make basketball free throws for 30 seconds. The number
F of free throws is plotted against the dosis x of medicine which they received
(F on the vertical axis, x on the horizontal axis). A computer program has
determined that the regression line has equation
F = −0.245x + 3.429
a) Make a table of these data points.
b) Sketch the regression line in the plot above.
c) Find the numbers T SS and RSS of this regression model. Round
answers to 3 digits after the decimal point.
d) Find the number R2 of this regression model. Round the answer to 2
digits after the decimal point. Would you say that the regression line fits the
data well? Explain your answer very briefly.
e) Find the predicted F -value of the regression line for x = 5.
( 2 + 2 + 4 + 4 + 4 = 16 points)
x 2 4 5 6 8 10
F 3 3 1 1 2 1
c) I get T SS = 4, and from the computer RSS = 1.551.
d) R2 = 0.61. You can make a case for every answer! The fit is not very
good, just because R2 is rather low. Compared to engineering and physics,
these points just don’t lie on a straight line at all! But considering we are
dealing with just a small number data points, and what’s worse, coming
from humans, we cannot expect anything better. For social sciences, this fit
is rather good!
For part a),
e) y = −0.245 · 5 + 3.429 = 2.204.
Problem 5 Find the following limits.
(3+3+3=9 marks)
x−4
6x + 8
t −4
M = lim−
t→4 t − 4
3u2 + 5u − 10, 000
N = lim
u→∞
7 + 5u
L = lim
x→4 x2 −
2
L = 1/2, M = −∞, N = ∞.
Problem 6 Suppose air pressure P (t) in an autoclave is given in PSI as
a function of time (in seconds) by
P (t) = −0.1t2 + 5t + 14.
a) Find the average rate of change of P between t = 0 and t = 3.
b) Find the instantaneous rate of change of P at t = 3.
For a), use P (0) = 14 and P (3) = 28.1 to get
m0,3 =
14 − 28.1
= 4.7
0−3
(in PSI/sec).
b) You can use the substitution h = t − 3, or directly factor out t − 3 since
the problem is easy enough. The answer is ma = 4.4 in PSI/sec. Here are
the steps if you decide to use h.
−0.1(t + h)2 + 5(t + h) + 14 − (−0.1t2 + 5t + 14)
h→0
h
2
−0.1(2th + h ) + 5h
= lim
h→0
h
= lim −0.2t − 0.1h + 5 = −0.2t + 5.
P 0 (t) = lim
h→0
And we just need to substitute t = 3 to get P 0 (3) = 4.4 (in PSI per second).
Of course, you could have substituted 3 for t right from the start – but
honestly, would that have made the problem easier?