MAT 171 Test 2. February 2014. Jennings.
SHOW YOUR WORK!
1. Find the derivative of
(a) (5 points) f (x) = x5
Solution:
f 0 (x) = 5x4
(b) (5 points) g(x) = 5x7 − 3x5 + 2x − 29
Solution:
g 0 (x) = (5)(7x6 ) − (3)(5x4 ) + 2 = 35x6 − 15x4 + 2
(c) (5 points) h(x) = (2x4 + 3x2 − 1)4
Solution: Chain rule:
h0 (x) = 4(2x4 + 3x2 − 1)3 (2)(4x3 ) + (3)(2x)
= 4(2x4 + 3x2 − 1)3 (8x3 + 6x)
(d) (5 points) k(x) =
√
1
x+ √
x
Solution:
k(x) = x1/2 + x−1/2
1
1
k 0 (x) = x−1/2 − x−3/2
2
2
2. (5 points) Let F (x) = 3x . Here is a table that gives the slopes of secant lines through the points on the
graph whose x coordinates are 1 and 1 + h.
h
0.1
0.01
0.001
0.0001
−0.0001
−0.001
−0.01
−0.1
(F (1 + h) − F (1))/h
3.48369522101714
3.31400758135607
3.29764795261189
3.29601791497769
3.29565583029101
3.29402710536542
3.27779874834877
3.12124620477713
Use the table to estimate the slope F 0 (1) of the line that is tangent to the graph of y = F (x) at the
point where x = 1. Your answer should have at least five significant digits.
Solution: By definition, F 0 (1) is the limit of the slopes of secant lines
F 0 (1) = lim
h→0
F (1 + h) − F (1)
h
so to get the best approximation from the table choose the value(s) of h that are closest to 0. There
are two: h = 0.0001 and h = −0.0001. The corresponding slopes of secant lines are in the right-hand
column:
(F (1 + h) − F (1))/h
h
0.0001
3.29601791497769
−0.0001 3.29565583029101
Any number between the two slopes in this table would be a pretty good estimate of the derivative.
The best estimate based on the information in the table would be the average of the two slopes
1
(3.29565583029101 + 3.29601791497769)
2
≈ 3.2958
F 0 (1) ≈
(1)
(2)
3. (10 points) If w pounds per acre of nitrogen fertilizer are spread on a corn field the yield is
210 −
4150
w + 30
bushels of corn per acre.
Corn sells for $5 per bushel and fertilizer costs $0.70 per pound. All other costs of growing and harvesting
the corn total $450 per acre, and don’t depend on the amount of fertilizer that is used.
How much nitrogen fertilizer per acre should the farmer use to maximize his net dollar return per acre?
Solution: For each acre the dollar return is
(return) = (earnings from corn sales) − (costs)
For each acre the earnings from corn sales are
(earnings from corn sales) = (bushels of corn per acre)(dollars per bushel)
4150 = 210 −
($5)
w + 30
and the costs are
(costs) = (fertilizer cost) − (other costs)
= (pounds of fertilizer)(cost per pound) − (other costs)
= (w)($0.70) + $450.
Thus the return per acre is
4150 ($5) − $0.70w + $450
w + 30
20750
= 1050 −
− 0.70w − 450 dollars.
w + 30
R = 210 −
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To find the maximum, computer the derivative R0 (w) then solve the equation 0 = R0 (w). First
re-write the formula for R in a way that makes it easy to compute the derivative.
R = 1050 − 20750(w + 30)−1 − 0.70w − 450
then differentiate
R0 (w) = −20750(−1)(w + 30)−2 − 0.70
20750
− 0.70
=
(w + 30)2
and solve the equation 0 = R0 (w)
20750
− 0.70
(w + 30)2
20750
0.70 =
(w + 30)2
0=
0.70(w + 30)2 = 20750
20750
(w + 30)2 =
r 0.70
20750
w + 30 = ±
0.70
r
20750
w=±
− 30
0.70
It doesn’t make sense to use a negative amount of fertilizer so the right answer must be the positive
one
r
20750
w=
− 30 ≈ 142.17 pounds of fertilizer per acre.
0.70
I didn’t ask this, but how do we know this is a maximum, not a minimum? One way to find out
would be to look at the graph of R(w):
R
300
200
100
w
100
200
300
400
-100
4. (10 points) At the beginning of an experiment there were one million bacteria in a bottle. The growth
rate of the bacteria is 4.2% per hour.
(a) Find the difference equation (the text calls this a dymanic equation) that tells how much bacteria
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population grows in a typical hour, based on the population at the beginning of that hour.
Solution: Let B(t) be the number of bacteria t hours after the experiment begins. The growth
rate is 4.2% = 0.042 per hour so
B(t + 1) − B(t) = 0.042B(t)
(b) Find the closed-form formula (hint: it’s an exponential equation) that predicts the bacteria population t hours after the experiment begins.
Solution:
B(t + 1) = B(t) + 0.042B(t) = (1 + 0.042)B(t) = (1.042)B(t)
so
B(1) = (1.042)B(0) = (1.042)(106 )
B(2) = (1.042)B(1) = (1.042)(1.042)(106 ) = (1.0422 )(106 )
B(3) = (1.042)B(2) = (1.042)(1.0422 )(106 ) = (1.0423 )(106 )
B(4) = (1.042)B(3) = (1.042)(1.0423 )(106 ) = (1.0424 )(106 )
and so on. Thus the formula must be
B(t) = (1.042t )(106 ).
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