Math 181 - Exam 1A - solutions
Problem 1 Assume there is a linear relationship between quantities H
and N , and you have obtained the following data.
Red alert - such an assumption must be justified if you want to
draw any conclusions from it!
8
H 2
N 20 3.8
a) Find the slope of the line for N in terms of H from these data.
b) Find the equation for N in terms of H in slope-intercept form.
c) What is N if H = 7?
d) Estimate H if N = 5.
Solution a) The slope is m = (3.8 − 20)/(8 − 2) = −16.2/6 = −2.7 (exact
answer). Expressing this as a common fraction is fine, too (luckily we don’t
have a problem with rounding here - if we had I’d recommended rounding to
1 or 2 decimals).
b) N = −2.7(H − 2) + 20 gives the slope-intercept form
N = −2.7H + 25.4
c) N = 6.5, d) common fraction H = 68/9 or
H = 15/2.7 + 2 ≈ 7.56
The word ’estimate’ here is used only because we are thinking of H as input,
N as output, so figuring out N from given input is ’usually’ more precise
than the reverse operation. Mathematically, both involve equations with an
exact solution.
Problem 2 A pinecone falls down from a tree. Suppose the height of
the pinecone at time t is given by
h(t) = −5t2 + bt + c.
t(sec) = 1 2
h(m) = 29 14
a) Tell the equations for b and c following from these data.
Suppose we have measurements
b) Find the equation of h(t).
c) At what time will the pinecone hit the ground?
Solution a) Plugging in the given values gives rise to two equations for b, c.
−5 + b + c = 29
−20 + 2b + c = 14
b) We eliminate c to get 15 − b = 15, so b = 0. Then we get c = 34 by
backward substitution. So
h = −5t2 + 34.
c) We don’t even need the quadratic formula to solve h = 0 for t:
p
t = 34/5 ≈ 2.6
The negative solution for t should be discarded.
Problem 3 The stature (height) of a human is approximately a linear
function of the length of the metacarpal bone b (both measured in cm). The
plot below shows a plot of stature vs metacarpal bone length in 9 individuals.
A computer program has determined that the regression line has equation
S = 1.6996b + 94.428
with T SS = 474, RSS = 126.70968.
a) Find the R2 -value.
b) Sketch the regression line in the plot below.
c) Would you say that the regression line fits the data well? why or why not
- explain briefly.
d) Archeologists have dug up human remains from a prehistoric settlement.
Many bones are missing, but one metacarpal bone has length 36 cm. Find
the stature of the corresponding individual as predicted by the model above.
Solution
For part a),
R2 =
T SS − RSS
≈ 0.73
T SS
(there is no good reason for giving more decimals.
b) see above.
c) You can argue both yes and no, the point is that you have to support your
statement. If you say no, you can point to the outliers and to the fact that
.73 is under 0.9. If you say yes, you can say that it’s in human biology, so
0.73 is rather high for an R2 -value there. You can add that the connection
just makes sense - taller people will have bigger bones. The ’no’ camp will
counter that the do not deny this, just that these particular points do not fit
their regression line very well.
A lot of people wrote that with more data points, we would hope for a better
fit. I have to clarify that unfortunately, this is not true. We would be more
confident that our regression line is the ’best’ in that it won’t budge anymore
after adding yet more datapoints. But the R2 -value will not budge either,
it’ll typically converge to a limit.
d) Use the regression line equation above to get S = 155.6 cm.
Problem 4 Find the following limits.
t2 + t − 4
t→5 t2 − t + 1
6u2 + 2u + 25
M = lim
u→∞
7u2
2
x + 5x − 24
N = lim
x→3
3x − 9
L = lim
Solution L = 26/21 by substitution, M = 6/7 because numerator and
denominator have equal degree, N = 11/3 by factoring out and canceling
x − 3 in numerator and denominator.
Problem 5 Suppose the concentration of a drug in the bloodstream is
C(t) in mL/L as a function of time (in seconds), given for t = 0 to t = 10 by
C(t) = −0.5t2 + 6t + 4
a) Find the average rate of change of C between t = 0 and t = 2.
b) Find the instantaneous rate of change of C at t = 2.
Solution For a), use C(0) = 4 and C(2) = 14 to get
m0,2 =
14 − 4
=5
2−0
(in mL/L).
b) You can use the substitution h = t − 2, or directly factor out t − 2 since
the problem is easy enough. The answer is m2 = 4 in PSI/sec. Here are the
steps if you decide to use h.
−0.5(t + h)2 + 6(t + h) + 4 − (−0.5t2 + 6t + 4)
h→0
h
2
−0.5(2th + h ) + 6h
= lim
h→0
h
= lim −t − 0.5h + 6 = −t + 6.
C 0 (t) = lim
h→0
And we just need to substitute t = 2 to get C 0 (2) = 4 (in PSI per second).
Of course, you could have substituted 2 for t right from the start.