Math 251
Final Exam - Fall 2015 - Solutions
Note: The wording of the problems here is not identical to that on the exam as administered.
The total number of points was 100, so the point values are about half of what similar questions
would be on a midterm.
Essentially no proofreading has been done!
1. (8 points; point values of parts as shown) A small spacecraft takes off from the surface
of a planet, reaches a maximum height, and then crashes. Its position at time t is given by
y(t) = 9t2 − 4t3 , where y(t) is measured in kilometers (km) above the surface and t is measured
in minutes (min). Answer the following questions, being careful to give correct units when
called for.
(a) (2 points) Find the speed of the spacecraft at time t = 1.
Solution: Calculate y 0 (t) = 18t − 12t2 , so the speed is y 0 (1) = 18 − 12 = 6. Thus, the answer
is 6km/min. (The units are required .)
(b) (2 points) How long will it take for the spacecraft to reach its maximum height?
Solution: First find the maximum height, which occurs when y 0 (t) = 0. We have y 0 (t) =
18t − 12t2 , so solve:
18t − 12t2 = 0
6t(2 − 2t) = 0
3
t = 0 or t = .
2
We are told that the spacecraft starts out by rising, reaches a maximum height, and then
crashes, to the answer is after 32 min. (This can also be checked by applying the methods we
have learned to the function y(t) = 9t2 − 4t3 .)
(c) (4 points) At time t = 2, what is the acceleration of the spacecraft? Would a person in
the craft feel himself speeding up or slowing down at this moment? Explain.
Solution: Calculate y 0 (t) = 18t − 12t2 , so y 00 (t) = 18 − 24t, and y 00 (2) − 18 − 48 = −30. The
numerical answer is therefore −30km/min2 . (The units are required .)
The acceleration is negative, and the velocity is y 0 (2) = 18·2−12·22 = −12, so the spacecraft
is falling fast as time passes. Thus, the person feels himself speeding up towards the ground.
Physically, strictly speaking, this is “slowing down”, since position is measured in distance
above the ground.
2. (6 points) The derivative of the function f (x) = (x3 + 3x2 + 3x + 5)e−x is given by f 0 (x) =
−(x + 2)(x − 1)2 e−x . Find the critical numbers of f and for each one determine if it is a local
minimum, local maximum, or neither.
1
Solution: We find critical numbers by solving f 0 (x) = 0. This happens when x + 2 = 0, when
(x − 1)2 = 0, and when e−x = 0. Since e−x is never zero, the critical numbers are at x = −2
and x = 1.
To decide which critical numbers are local minimums or maximums, we consider the factors
of f 0 (x):
• On (−∞, −2), −1 < 0, x + 2 < 0, (x − 1)2 > 0, and e−x > 0, so f 0 (x) > 0 and f is
nondecreasing.
• On (−2, 1), −1 < 0, x + 2 > 0, (x − 1)2 > 0, and e−x > 0, so f 0 (x) < 0 and f is
nonincreasing.
• On (1, ∞), −1 < 0, x + 2 > 0, (x − 1)2 > 0, and e−x > 0, so f 0 (x) < 0 and f is
nonincreasing.
So f has a local maximum at x = −2, but neither a local maximum nor a local minimum at
x = 1.
Note: The second derivative test doesn’t help much here. A calculation gives f 00 (x) =
(x − 1)(x2 − 2x − 5)e−x . So f 00 (−2) = −9e2 < 0, showing that there is a local maximum at
x = −2, but f 00 (1) = 0, which doesn’t help.
3. (8 points) A pendulum swings back and forth on the surface of the earth. The relation
between
√ the period T and the length l of the pendulum can be modelled by the equation
T = kl, where T is measured in seconds, l is measured in meters, and k = 4 s2 /m. Answer
the following questions, being careful to include units when appropriate.
(a) (1 point) When l is 25 meters, compute the period of the pendulum.
Solution: At l = 25, we get T =
(The units are required .)
p
(4 s2 /m) · (25 m) =
√
100 s2 = 10 s. So the period is 10 s.
(b) (4 points) Using part (a) as your base value, give a linear approximation for the period
of the pendulum when the length l is changed to 27 meters.
√
√
Solution: It is better to use function notation: T (l) = 4l = 2 l = 2t1/2 .
Now we do the linear approximation. We have T 0 (l) = 2 21 t−1/2 = t−1/2 . So
1
2
T (27) ≈ T (25) + T 0 (25)(27 − 25) = 10 + √ (2) = 10 + .
5
25
So the estimated is 10.4 s. (The units are required .)
(c) (3 points) Suppose you do a linear approximation for l = 205 meters, based on the
starting point l = 200. Is the linear approximation greater or less than the actual period?
Explain (perhaps using a picture).
Solution: We have T 00 (l) = − 21 t−3/2 . This is negative for all l in the domain, so the graph
of T is concave down, and hence below its tangent lines. Therefore the linear approximation is
greater than the actual period.
(Picture missing.)
4. (4 points) Let f and g be functions such that:
g(1) = 2,
g 0 (1) = 1,
f (1) = 2,
and f 0 (1) = 6
and
g(2) = 1, g 0 (2) = 5, f (2) = 3, and f 0 (2) = −1.
Let h(x) = f (g(x)). Find h0 (2). (You will not need to use all the information provided.)
Solution: Using the chain rule,
h0 (2) = f 0 (g(2))g 0 (2) = f 0 (1) · 5 = 6 · 5 = 30.
5. (11 points; point values of parts as shown) Air Krill runs daily flights from the Eugene
Airport to the Falkland Islands. When the company charges $400 per ticket, 24 people fly.
Market research has shown each $10 that the fare is increased, 1 fewer person flies. Conversely,
for each $10 the fare is lowered, 1 more person flies.
(a) (4 points) Let c denote the price charged for each ticket. (“c” stands for “cost”.) Let
T (c) be the number of tickets sold at that price, and let R(c) be the revenue at that price.
Write down formulas for T (c) and R(c).
Solution: We have
1
T (c) = 24 + −
10
Therefore
(c − 400) = 64 −
c
.
10
c
c2
R(c) = cT (c) = c 64 −
= 64c − .
10
10
(b) (3 points) How much should Air Krill charge per ticket to maximize total revenue?
Solution: We want to maximize R(c) We need to determine the appropriate interval. We
c
≥ 0,
need to have c ≥ 0. Also, the number of passengers must be nonnegative, so 64 − 10
2
c
so c ≤ 640. So we need to maximize the function R(c) = 64c − 10 over the interval [0, 640].
Compute:
c
R0 (c) = 64 − .
5
This is zero when c = 320.
To verify that this is a maximum, the second derivative test is easiest. Calculate: R00 (c) = − 15 ,
which is negative everywhere, so any critical number is a global maximum. Therefore revenue
is maximized at a ticket cost of $320.
Next easiest is the first derivative test. Factor:
320 − c
R0 (c) =
,
5
which is positive for c < 320 and negative for c > 320. So any critical number is a global
maximum. Therefore revenue is maximized at a ticket cost of $320.
For the closed interval method, compare:
R(0) = 0,
R(320) = 64(320) −
640 · 320 − 320 · 320
3202
3202
=
=
> 0,
10
10
10
and
6402
= 0.
10
Therefore revenue is maximized at a ticket cost of $320.
R(640) = 64(640) −
(c) (2 points) Suppose that the Air Krill airplanes can only seat 30 people. Write down an
inequality involving T (c) that represents this, and deduce a corresponding inequality for c.
Solution: The statement says T (c) ≤ 30. Thus:
c
64 −
≤ 30
10
c
≥ 34
10
c ≥ 340.
(d) (2 points) Does the condition in (c) change the answer for (b), and if so what is the new
answer? Explain.
Solution: We must now maximize R(c) over the interval [340, 640]. The function has no
critical numbers in this interval, so the maximum must occur at an endpoint.
The easiest method is to observe, as in the second approach to part (b), that R0 (c) < 0 on
[340, 640], so the maximum must be at c = 340.
The alternative method is to compare values at the endpoints:
640 · 340 − 340 · 340
300 · 320
3402
=
=
> 0,
10
10
10
and R(640) = 0 (as in part (b)). Since R(340) > R(640), revenue is maximized at c = 340.
R(340) = 64(340) −
6. (3 points/part) Find the exact values of the following limits (possibly including ∞ or −∞),
or explain why they do not exist. Show supporting work.
(a) lim
x→0
x4
.
e7x4 − 1
0
Solution 1: The limit has the indeterminate form “ ”. Therefore we may use L’Hospital’s
0
d 7x4 4
Rule. The chain rule gives
e
= 24x3 e7x , so
dx
4x3
x4
,
lim 7x4
= lim
x→0 e
− 1 x→0 28x3 e7x4
if the second limit exists. To evaluate the second limit, cancel 4x3 from the numerator and
denominator, getting:
1
4x3
1
1
1
=
lim
=
=
=
.
4
4
4
x→0 28x3 e7x
x→0 7e7x
7e0
7
7e7·0
lim
Therefore
x4
1
= .
4
7x
x→0 e
7
−1
lim
0
Note: If you don’t cancel the factors 4x3 , the limit still has the indeterminate form “ ” You
0
can therefore use L’Hospital’s Rule again. To get an answer this way, you will in fact need to
use L’Hospital’s Rule a total of 3 more times, which will involve computing the 4th derivative
4
of e7x . This is quite messy, so messy that I will not give details here.
Solution 2: We note that x4 → 0+ as x → 0. Substituting h = x4 , we therefore get
x4
h
1
= lim+ 7h
=
4
7x
x→0 e
− 1 h→0 e − 1
limh→0+
lim
e7h −1
h
.
With f (t) = e7t , the denominator in this last expression is
lim+
h→0
0
f (h) − f (0)
e7h − 1
= lim+
= f 0 (0).
h→0
h
h
7t
We know f (t) = 7e , so the limit is 7e0 = 7. Therefore
x4
1
.
=
4
x→0 e7x − 1
7
lim
e7h − 1
using L’Hospital’s Rule (once).
h→0
h
Note: We could also evaluate lim
(b) lim
x→1 x2
x−2
.
−x−6
Solution: Both the numerator and denominator are continuous at 1, and the denominator is
not zero there. Therefore the limit can be evaluated by simply substituting x = 1. That is,
x−2
1−2
1
lim
= 2
= .
x→1 x2 − x − 6
1 −1−6
6
Note that L’Hospital’s Rule doesn’t apply, since the limit does not have an indeterminate
form. If you try to use it anyway, you get
1
= 1,
lim
x→1 2x − 1
which is the wrong answer.
(c) lim
x→0
x2
.
cos(2x) − 1
0
Solution: The limit has the indeterminate form “ ”. Therefore we may use L’Hospital’s
0
d
Rule. The chain rule gives
(cos(2x) − 1) = −2 sin(2x), so
dx
x2
2x
lim
= lim
,
x→0 cos(2x) − 1
x→0 −2 sin(2x)
0
if the second limit exists. This limit also has the indeterminate form “ ”. Therefore we may
0
use L’Hospital’s Rule again. Using the chain rule again, we get
2
2x
= lim
,
lim
x→0 −4 cos(2x)
x→0 −2 sin(2x)
if the second limit exists. Clearly
2
1
2
=
=− .
lim
x→0 −4 cos(2x)
−4 cos(2 · 0)
2
So
x2
1
lim
− .
x→0 cos(2x) − 1
2
3x + 1
.
x→∞ 5x2 − 9
(d) lim
Solution: This has the indeterminate form “ ∞
”, so work is needed. We factor out x2 from
∞
both the numerator and denominator, and then use the limit laws:
3
x2 x3 + x12
+ x12
3x + 1
x
lim
= lim 2
= lim
x→∞ 5x2 − 9
x→∞ x
x→∞ 5 − 92
5 − x92
x
limx→∞ x3 + limx→∞
=
5 − limx→∞ x92
1
x2
=
0+0
= 0.
5−0
Alternate solution: Here is a different way to arrange essentially the same calculation:
1
3
(3x + 1)
+ x12
3x + 1
2
x
x
lim
=
lim
=
lim
1
x→∞ 5x2 − 9
x→∞
(5x2 − 9) x→∞ 5 − x92
x2
=
limx→∞ x3 + limx→∞
5 − limx→∞ x92
1
x2
=
0+0
= 0.
5−0
Second alternate solution. (This solution can only been used after L’Hopital’s rule has been
treated in the course.)
We have limx→∞ (3x + 1) = ∞ and limx→∞ (5x2 − 9) = ∞. The limit above therefore has the
∞
indeterminate form “ ∞
”, so more work is needed. In this case, L’Hospital’s Rule applies. We
get:
3x + 1
3
lim
= lim
,
2
x→∞ 5x − 9
x→∞ 5x
provided the last limit exists. This limit is zero because limx→∞ 3 = 3 and limx→∞ 5x = ∞.
Therefore
3x + 1
lim
= 0.
x→∞ 5x2 − 9
7. (7 points) At noon Horton is 10 miles due west of Gertrude. Horton walks north and
Gertrude walks west; neither walks at a steady rate. Both have devices that can continuously
measure the distance between them. After one hour, Horton has walked 3 miles and Gertrude
has walked 6 miles. Horton’s device tells him that he is at that moment walking at 5 miles per
hour, and the distance between him and Gertrude is increasing at a rate of 2 miles per hour.
How fast is Gertrude walking at this moment? Include units in your answer.
Solution: Note: There is no picture in this file.
Let h(t) be the distance (in miles) Horton has walked north at time t (with time measured
in hours), and let g(t) be the distance (in miles) Gertrude has walked west at time t. Let t = 0
represent noon on the day in question. Let z(t) be the distance (also in miles) between Horton
and Gertrude at time t. Then the information given says that:
h0 (1) = 5,
h(1) = 3,
g(1) = 6,
and z 0 (1) = 2.
We want to find g 0 (1). At time t, So the Pythagorean Theorem gives z(t)2 = h(t)2 +[10−g(t)]2 .
Differentiating, we get
2z(t)z 0 (t) = 2h(t)h0 (t) + 2[10 − g(t)][−g 0 (t)].
(Don’t forget to use the chain rule!) Put t = 1 and (for simplicity) divide by 2:
z(1)z 0 (1) = h(1)h0 (1) + [10 − g(1)][−g 0 (1)].
Next, substitute values. (Note that this can only be done after differentiating!) We need
p
p
z(1) = h(0)2 + [10 − g(t)]2 = 32 + (10 − 6)2 = 5,
and we then get
5 · 2 = 3 · 5 + (10 − 6) · (−g 0 (1)).
So
10 = 15 − 4g 0 (1),
and
5
10 − 15
= .
−4
4
miles per hour. (The units are necessary!)
g 0 (1) =
So Gertrude is walking west at
5
4
Here are descriptions of some alternatives. First, one could solve for g(t). This is messy
(involving square roots), and the details are omitted.
Second, one could measure distances differently. If one imagines the Horton’s location at
noon to be the origin, with Horton on the y-axis and Gertrude on the x-axis, then it is natural
to let y(t) be how far Horton has gone north and let x(t) be how far east Gertrude is from that
point. With these conventions, the distance between them is given by z(t)2 = x(t)2 + y(t)2 ,
and the information given says that:
x(0) = 4,
y(0) = 3,
y 0 (0) = 5,
and z 0 (0) = 2.
Differentiation gives 2z(t)z 0 (t) = 2x(t)x0 (t) + 2y(t)y 0 (t). The algebra will be a bit different
through the rest of the calculation, but the final result will be the same.
Finally, you could do everything in physicists’ notation. I will only show the first version.
The equation relating the quantities is z 2 = h2 + [10 − g]2 . Differentiating (using the chain rule,
because everything is a function of t!), we get
dz
dh
dg
2z
= 2h + 2(10 − g) −
,
dt
dt
dt
so
dh
dg
dz
= h − (10 − g) .
dt
dt
dt
Substituting values (implicitly putting t = 1, and using z = 5 at t = 1, as above):
dg
5 · 2 = 3 · 5 − (10 − 6) · .
dt
dg
5
So
= .
dt
4
z
8. (6 points) Find the equation of the tangent line to the curve x2 + xy + y 2 = 3 at the point
(1, 1).
Solution: Let’s write it with y as an explicit function y(x) of x:
x2 + xy(x) + y(x)2 = 3.
Differentiate both sides with respect to x, using the chain rule on the last term on the left:
2x + y(x) + xy 0 (x) + 2y(x)y 0 (x) = 0.
Now substitute x = 1 and y(x) = 1:
2 · 1 + 1 + 1 · y 0 (x) + 2 · 1 · y 0 (x) = 0.
3 + 3y 0 (x) = 0
y 0 (x) = −1.
This is the slope of the tangent line. Therefore the equation of the tangent line is
y − 1 = (−1)(x − 1),
which is simplified to y = −x + 2.
dy
For those who prefer the other notation, here it is written with
. Differentiate with respect
dx
to x as before:
dy
dy
= 0.
2x + y + x + 2y
dx
dx
Now substitute x = 1 and y = 1:
dy
dy
2·1+1+1·
+2·1·
= 0.
dx
dx
dy
3+3
=0
dx
dy
= −1.
dx
This is the slope of the tangent line. Therefore the equation of the tangent line is
y − 1 = (−1)(x − 1),
which is simplified to y = −x + 2.
9. (4 points/part) In each part below, find the derivative of the given function.
(a) f (x) = sin(x2 ) + (sin(x))2 .
Solution: Use the chain rule on both terms, getting
d 2
d
f 0 (x) = cos(x2 )
x ) + 2 sin(x)
sin(x)) = 2x cos(x2 ) + 2 sin(x) cos(x).
dx
dx
(b) g(x) = tan(5x) + ln(7)x3 + π 2 .
Solution: Use the chain rule on the first term, getting
g 0 (x) = sec2 (5x) · 5 + ln(7) · 3x2 = 5 sec2 (5x) + 3 ln(7)x2 .
(The expression π 2 is a constant, so its derivative is zero.)
√
(c) q(t) = e3t t.
Solution: Rewrite as q(t) = e3t t1/2 . Use the product rule, using the chain rule on the first
factor:
d 3t 1/2
d 1/2
q 0 (t) =
e )t + e3t
t ) = 3e3t t1/2 + e3t 12 t−1/2 .
dt
dt
You can (but don’t have to) rearrange this more nicely as
q 0 (t) = t−1/2 e3t 3t + 12 .
(d) h(s) =
7s + 3
.
3s − 7
Solution: Use the quotient rule, getting:
d
d
7s + 3 (3s − 7) − (7s + 3) ds
3s − 7
7(3s − 7) − (7s + 3) · 3
−58
0
ds
h (s) =
=
=
.
2
2
(3s − 7)
(3s − 7)
(3s − 7)2
10. (2 points/part) The picture below is the graph of the DERIVATIVE y = f 0 (x) for a certain
function f . CAUTION: You are given the graph of the derivative f 0 (x), not the graph of
f (x), but you are asked questions about f (x). The points referred to in parts (a) and (b) are
marked on the graph with dots.
6
5
4
3
2
1
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
-1
-2
-3
-4
-5
-6
For reference, the formula I used is f 0 (x) =
1
(x
195
− 7)x(x + 4)(x + 7).
For reference in the solutions below, here is the graph of the function f whose derivative is
shown above and which satisfies f (6) = −17. (Any other function with the given derivative
differs from this one by a constant, as we saw in Section 4.2 of the book.) You can check, by
drawing tangent lines, that the graph of the derivative of this function does indeed have the
shape shown in the graph above. Also shown on the graph below are the tangent lines at the
points in parts (a) and (b), and all the points (including those in Part (c)) are marked on the
graph with dots.
4
-8
-7
-6
-5
-4
-3
-2
-1
1
2
3
4
5
6
7
8
-4
-8
-12
-16
-20
-24
This function has the formula
f (x) =
1 5
1 4
49 3
98
833
x +
x −
x −
x+
.
975
195
585
195
325
(a) Is f increasing, decreasing, or nearly flat at x = −1, or is there not enough information
provided to determine this? Why?
Solution: You can read off the graph of f 0 that f 0 (−1) ≈ 0.7 > 0. (The exact value is
≈ 0.738462.) Therefore f is increasing at x = −1. The tangent line to the graph of f at
x = −1 is shown on the graph above, and you can see that its slope is positive.
48
65
(b) Is f concave up or concave down at x = 3.5, or does f (nearly) have an inflection point
at x = 3.5, or is there not enough information provided to determine this? Why?
Solution: The graph shown is of the derivative f 0 of f . Therefore f 00 (3.5) is the slope (first
derivative) of the graph shown at 3.5. Clearly it is negative. Thus f 00 (3.5) < 0. Therefore f is
concave down at x = 3.5. The tangent line to the graph of f at x = 3.5 is shown on the graph
above, and you can see it is above the graph of f .
(c) At which values of x in [−8, 8] (the interval shown) does f have a local minimum? Explain.
Solution: Local minimums occur at critical numbers, that is, numbers c such that f 0 (c) = 0
or f 0 (c) does not exist. There are no numbers c such that f 0 (c) does not exist. We can read off
the graph of f 0 that f 0 (c) = 0 for c = −7, c = −4, c = 0, and c = 7. We can test which of these
are local minimums in either of two ways: the first derivative test or the second derivative test.
Here is the first derivative test. (This explanation is much more detailed than was required
on the exam.) At c = −7, the values shown for f 0 (x) on the graph of f 0 , for x < −7 but
x close to −7, are positive, while values shown for f 0 (x), for x > −7 but x close to −7, are
negative. So f is increasing before we get to −7, and decreasing afterwards; therefore, f has a
local maximum at c = −7.
Similarly, f has a local maximum at c = 0.
At c = −4, the values shown for f 0 (x) on the graph of f 0 , for x < −4 but x close to −4,
are negative, while values shown for f 0 (x), for x > −4 but x close to −4, are positive. So f is
decreasing before we get to −4, and increasing afterwards; therefore, f has a local minimum
at c = −4.
Similarly, f has a local minimum at c = 7.
Thus, the answer to the problem is x = −4 and x = 7.
Here is the second derivative test. (Again, this explanation is much more detailed than was
required on the exam.) Since the graph shows f 0 , the second derivative f 00 (x), which is the
derivative of f 0 at x, is the slope of the tangent line to the graph shown at x. The tangent lines
at x = −7 and x = 0 clearly have negative slopes, while those at x = −4 and x = 7 clearly have
positive slopes. Thus, the second derivative test tells us that f has local minimums at x = −4
and x = 7 (critical numbers where the second derivative is positive). (Also, there are local
maximums at x = −7 and x = 0, since these are critical numbers where the second derivative
is negative.)
You can see from the graph of f provided above that f does indeed have local minimums
(marked with dots) at x = −4 and x = 7, and local maximums at x = −7 and x = 0.
(d) Is f (0) positive, negative, or zero, or is there not enough information to determine this?
Explain.
Solution: Since we only know f 0 , we only know f up to a constant. (For example, if g(x) =
f (x) + 117 for all x, then g 0 (x) = f 0 (x) for all x.) So we can’t determine anything about f (0)
by itself (only how it is related to the values of f at other points).
11. (8 points) You want to build a box with a square base. Let b be the length of one side of the
base, and let h be the height, both measured in meters. Given the constraints that 10 ≤ b ≤ 25
and 2b + h = 60, what are the biggest and smallest volumes for such a box?
Solution: Since the base is square with side length b, its area is b2 . Therefore its volume
is V = b2 h. So we are supposed to minimize and maximize b2 h subject to the constraints
2b + h = 60 and 10 ≤ b ≤ 25.
We eliminate one variable from the formula for V using the first constraint:
2b + h = 60
h = 60 − 2b
V = b2 h = b2 (60 − 2b).
So we must minimize and maximize the function V (b) = b2 (60 − 2b) on the interval [10, 25].
We find critical numbers. To differentiate easily, rewrite
V (b) = 60b2 − 2b3 .
Differentiate, and factor so we can solve V 0 (b) = 0:
V 0 (b) = 120b − 6b2 = 6b(20 − b).
This is zero when b = 0 or b = 20. We reject b = 0 since it is not in the interval [10, 25]. (I
must see you do this!)
Now compare values:
V (10) = 102 (60 − 2 · 10) = 4000,
V (20) = 202 (60 − 2 · 20) = 400 · 20 = 8000,
and
V (25) = 252 (60 − 2 · 25) = 625 · 10 = 6250.
So the maximum volume is 8000m3 and the minimum volume is 4000m3 . (The units are
necessary.)
One can use the first or second derivative test to decide whether the critical number is a
maximum or minimum. But this is wasted effort, since one eventually needs to find V (10),
V (20), and V (25) anyway.
12. (3 points/part) The function P (t) = (6t + 1)ek(t−1) models the population of a colony of
bacteria at time t, where P is measured in hundreds of bacteria and t is measured in hours.
Observations indicate that after one hour there are 700 bacteria, and at that time the colony
is growing at a rate of 200 bacteria per hour.
(a) Find k.
Solution: We are told that P (1) = 7 and P 0 (1) = 2. (Caution! Read the problem! P (t)
is measured in hundreds of bacteria!) So
7 = P (1) = (6 · 1 + 1)ek(1−1) = 7e0 = 7.
This equation doesn’t tell us anything: the statement that at t = 1 there are 700 bacteria is
redundant.
The statement about P 0 (1) is more useful. Using the product and chain rules, we get
d
P 0 (t) = 6ek(t−1) + (6t + 1)ek(t−1)
k(t − 1) = 6ek(t−1) + (6t + 1)ek(t−1) · k.
dt
Substituting t = 1 gives
P 0 (1) = 6ek(1−1) + (6 · 1 + 1)ek(1−1) k = 6 + 7k.
This is supposed to be 2, so 2 = 6 + 7k, that is, k = − 47 .
(b) What happens to the population of bacteria in the long run, as t → ∞? Explain.
Solution: Using the information from part (a), we can rewrite the formula as
P (t) = (6t + 1)e(−4/7)(t−1) .
We are asked about limt→∞ P (t). Write this as
6t + 1
lim (4/7)(t−1) .
t→∞ e
∞
In this form, it has the indeterminate form
. Therefore we may use L’Hospital’s Rule. Thus
∞
6t + 1
6
lim (4/7)(t−1) = lim 4 (4/7)(t−1) ,
t→∞ e
t→∞ e
7
if the second limit exists. But this limit is zero, since limt→∞ e(4/7)(t−1) = ∞. So
6t + 1
lim
= 0.
t→∞ e(4/7)(t−1)
The population of bacteria goes to zero, that is, the bacteria die out.