Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 1.1 Direction Fields due 03/10/2021 at 11:59pm MST
1. (1 point) Match the following equations with their direc-
tion field. Clicking on each picture will give you an enlarged
view. While you can probably solve this problem by guessing,
it is useful to try to predict characteristics of the direction field
and then match them to the picture. Here are some handy char-
acteristics to start with – you will develop more as you practice.
(i) Set yequal to zero and look at how the derivative behaves
along the x-axis.
(ii) Do the same for the y-axis by setting xequal to 0
(iii) Consider the curve in the plane defined by setting y0=0
– this should correspond to the points in the picture where the
slope is zero.
(iv) Setting y0equal to a constant other than zero gives the
curve of points where the slope is that constant. These are called
isoclines, and can be used to construct the direction field picture
by hand.
1. y0=−1−2y
2. y0=3sin(x) + 1+y
3. y0=−(2x+y)
(2y)
A B C
Solution:
SOLUTION
1. matches B, because the slope is zero along the horizontal
line y=−1/2 .
2. matches A, because the slope is zero along the curve
y=−3sin(x)−1. .
3. matches C, because when y=0 (and x6=0) the slope is
infinite. Also, when x=0 (and y6=0), the slope is constant and
equal to −1
2.
Answer(s) submitted:
•b
•a
•c
(correct)
Correct Answers:
•B
•A
•C
2. (1 point) Match the following equations with their direc-
tion field. Clicking on each picture will give you an enlarged
view.
1. y0=x+2y
2. y0=xe−2x−2y
3. y0=−y(5−y)
4. y0=y3
6−y−x3
6
A B
C D
Solution:
SOLUTION
1. matches B, because the slope is zero along the line
y=−2x.
2. matches A, because the slope is zero along the curve
y=x
2e−2x. Also, when x=0, y0=−2yand therefore, on the
y-axis, the slope is positive when y<0 and negative when y>0
.
3. matches C, because the slope is zero along teh horizontal
lines y=0 and y=5.
1
4. matches D, because when y=0, y0=−x3
6and therefore,
on the x-axis, the slope is positive when x<0 and negative when
x>0.
Answer(s) submitted:
•b
•a
•c
•d
(correct)
Correct Answers:
•B
•A
•C
•D
3. (1 point)
Match the following equations with their direction field.
Clicking on each picture will give you an enlarged view. While
you can probably solve this problem by guessing, it is useful
to try to predict characteristics of the direction field and then
match them to the picture. Here are some handy characteristics
to start with – you will develop more as you practice.
A. Set yequal to zero and look at how the derivative be-
haves along the x-axis.
B. Do the same for the y-axis by setting xequal to 0
C. Consider the curve in the plane defined by setting y0=0
– this should correspond to the points in the picture
where the slope is zero.
D. Setting y0equal to a constant other than zero gives the
curve of points where the slope is that constant. These
are called isoclines, and can be used to construct the
direction field picture by hand.
1. y0=e−x+2y
2. y0=−2+x−y
3. y0=2y+x2e2x
4. y0=2sin(x) + 1+y
A B
C D
Answer(s) submitted:
•d
•b
•a
•c
(correct)
Correct Answers:
•D
•B
•A
•C
4. (1 point)
A function y(t)satisfies the differential equation
dy
dt =−y4+7y3+18y2.
(a) What are the constant solutions of this equation?
Separate your answers by commas.
.
(b) For what values of yis ystrictly increasing?
<y<and <y<.
Solution: We have
dy
dt =−y2(y2−7y−18) = −y2(y−9)(y+2)
Thus the equilibrium solutions are
y=−2,y=0,y=9
The solutions are strictly increasing when dy
dt >0, that is,
−2<y<0 and 0 <y<9.
Answer(s) submitted:
•0,-2,9
•-2
•0
•0
•9
(correct)
Correct Answers:
•-2, 0, 9
•-2
•0
•0
2
•9
5. (1 point) The graph of the function f(x)is
(the hori-
zontal axis is x.)
Given the differential equation x0(t) = f(x(t)).
List the constant (equilibrium) solutions to this differential
equation in increasing order and indicate whether or not these
equations are stable, semi-stable, or unstable.
?
?
?
?
Solution: The equilibrium solutions can be determined by
solving f(x) = 0. From the graph we can see that f(x) = 0
when x=−3.5,x=−1.5,x=−0.5 and x=1.5.
Since f(x)<0 for x<−3.5 and f(x)>0 for −3.5<x<−1.5,
the equilibrium solution x=−3.5 is unstable.
Since f(x)has the same sign for −3.5<x<−1.5 and
for −1.5<x>−0.5, the equilibrium solution x=−1.5 is
semistable.
Since f(x)>0 for −1.5<x<−0.5 and f(x)<0 for −0.5<
x<−1.5, the equilibrium solution x=−0.5 is stable.
Since f(x)<0 for x<1.5 and f(x)>0 for x>1.5, the equi-
librium solution x=1.5 is unstable.
Answer(s) submitted:
•-3.5
•unstable
•-1.5
•semi-stable
•-.5
•stable
•1.5
•unstable
(correct)
Correct Answers:
•-3.5
•UNSTABLE
•-1.5
•SEMI-STABLE
•-0.5
•STABLE
•1.5
•UNSTABLE
6. (1 point)
Given the differential equation x0= (x+2.5)(x+1.5)3x2(x−1).
List the constant (i.e. equilibrium) solutions to this differential
equation in increasing order and indicate whether or not these
solutions are stable, semi-stable, or unstable. Confirm your an-
swer by plotting the slope field using MATLAB.
?
?
?
?
Answer(s) submitted:
•-2.5
•unstable
•-1.5
•stable
•0
•semi-stable
•1
•unstable
(correct)
Correct Answers:
•-2.5
•UNSTABLE
•-1.5
•STABLE
•0
•SEMI-STABLE
•1
•UNSTABLE
7. (1 point) Consider the direction field below for a differen-
tial equation y0=y(y−3)(y−5).Use the graph and equation to
answer the following questions.
1. Find the equilibrium solutions for the differential equation.
Answer (separate by commas): y=
3
2. If the initial condition is y(0) = c,for what values of cis
lim
t→∞
y(t)finite?
Answer (as an interval):
Note: You can click on the graph to enlarge the image.
Answer(s) submitted:
•0,3,5
•0
(score 0.5)
Correct Answers:
•0, 3, 5
•[0,5]
8. (1 point)
The slope field for y0=0.6(1+y)(3−y)is shown below
On a print out of the slope field, draw solution curves through
each of the three marked points.
(a) As x→∞(As needed, enter ∞in your answers as Inf):
For the solution through the top-left point: y→
For the solution through the origin: y→
For the solution through the bottom-right point: y→
(b) What are the equilibrium solutions of this differential
equation?
(Enter your answers as a list, e.g., 3,5,ordered smallest to
largest.)
(c) Are these stable or unstable (enter Sor Ufor each, as a
list in the same order as your answers for the equilibrium
points)?
Solution:
SOLUTION
The slope field with solution curves is shown in the figure
below.
(a) From the slope field and sketched solutions, we can see
that as x→∞the solution curve through the top-left initial
condition approaches an equilibrium solution. This must be
where the derivative goes to zero, so it must be when y0=
0.6(1+y)(3−y) = 0, or y=3 or y=−1. Because this equilib-
rium is positive, it must be that y→3.
Again, from the slope field and sketched solutions, we can
see that as x→∞the solution curve through the origin ap-
proaches the same equilibrium as before. Thus y→3.
Finally, the solution curve through the bottom-right point
clearly diverges. There are no equilibria below y=−1, so it
must be that y→ −∞.
(b) The equilibrium solutions were found in (a), and are
y=−1 and y=3.
(c) By inspection of the slope field, we can see that the equi-
librium solution y=−1 is unstable and that y=3 is stable.
Answer(s) submitted:
•3
•3
•-inf
•-1,3
•U,S
(correct)
Correct Answers:
•3
•3
•-infinity
•-1, 3
•U, S
9. (1 point)
Consider the slope field shown.
(a) For the solution that satisfies y(0) = 0, sketch the
solution curve and estimate the following:
y(1)≈and y(−1)≈
(b) For the solution that satisfies y(0) = 1, sketch the
solution curve and estimate the following:
y(0.5)≈and y(−1)≈
(c) For the solution that satisfies y(0) = −1, sketch the
solution curve and estimate the following:
y(1)≈and y(−1)≈
Answer(s) submitted:
4
•.75
•.5
•1.75
•.75
•-2
•0
(correct)
Correct Answers:
•0.718282
•0.367879
•1.79744
•0.735759
•-2
•0
10. (1 point)
Consider the two slope fields shown, in figures 1 and 2 below.
figure 1 figure 2
On a print-out of these slope fields, sketch for each three so-
lution curves to the differential equations that generated them.
Then complete the following statements:
For the slope field in figure 1, a solution passing through the
point (2,-1) has a
• ?
• positive
• negative
• zero
• undefined
slope.
For the slope field in figure 1, a solution passing through the
point (-3,-3) has a
• ?
• positive
• negative
• zero
• undefined
slope.
For the slope field in figure 2, a solution passing through the
point (2,-1) has a
• ?
• positive
• negative
• zero
• undefined
slope.
For the slope field in figure 2, a solution passing through the
point (0,-4) has a
• ?
• positive
• negative
• zero
• undefined
slope.
Solution:
SOLUTION
There are many possible solution curves that could be drawn.
For the slope field in figure 1, a solution passing through the
point (2,-1) has a positive slope.
For the slope field in figure 1, a solution passing through the
point (-3,-3) has a negative slope.
For the slope field in figure 2, a solution passing through the
point (2,-1) has a positive slope.
For the slope field in figure 2, a solution passing through the
point (0,-4) has a negative slope.
Answer(s) submitted:
•positive
•negative
•positive
•negative
(correct)
Correct Answers:
•positive
•negative
•positive
•negative
11. (1 point)
The slope field for the equation y0=x+yis shown below
On a print out of this slope field, sketch the solutions that
pass through the points
(i) (0,0);
(ii) (-3,1); and
(iii) (-1,0).
From your sketch, what is the equation of the solution to
the differential equation that passes through (-1,0)? (Verify that
your solution is correct by substituting it into the differential
equation.)
y=
5
Solution:
SOLUTION
The solution curves are shown in the graph below (that
through the origin in red, through (-3,1) in black, and that
through (-1,0) in blue).
We can guess from this that the solution through (-1,0) is
y=−x−1. Plugging this into the differential equation confirms
this.
Answer(s) submitted:
•-x-1
(correct)
Correct Answers:
•-x-1
12. (1 point)
Supposed a body of mass 4 kg is falling in the atmosphere
near sea level.
Let v(t)m/s be the velocity of the body at time tin seconds.
Assume that vis positive in the downward direction - that is,
when the object is falling. We assume that the forces acting
on the body are the force of gravity and a retarding force of
air resistance with direction opposite to the direction of motion
and with magnitude cv(t)where c=0.4kg
s. The gravitational
constant is g=9.8m/s2.
a) Find a differential equation for the velocity v:
dv
dt =
(b) Find the equilibrium solution of the differential equation,
that is, find the limiting velocity.
limiting velocity = m/s
Solution: (a)
The force of gravity is given by Fg=mg, while the air resistance
is Fr=−cv.
By Newton’s second law, the net force is equal to mdv
dt . Thus
mdv
dt =Fg−Fr
mdv
dt =mg −cv
dv
dt =g−c
mv
Substituing the given values of g,mand cyields the differential
equation for the velocity:
dv
dt =9.8−0.4
4v
(b)
The limiting velocity can be found by solving dv
dt =0, that is,
9.8−0.4
4v=0. This yields the value 98m/s.
Answer(s) submitted:
•9.8-(((.4)/4))v
•98
(correct)
Correct Answers:
•9.8 - 0.4*v/4
•98
13. (1 point)
Supposed a body of mass 38 kg is falling in the atmosphere
near sea level.
Let v(t)m/s be the velocity of the body at time tin seconds.
Assume that vis positive in the downward direction - that is,
when the object is falling. We assume that the forces acting on
the body are the force of gravity and a retarding force of air
resistance with direction opposite to the direction of motion and
with magnitude proportional to the square of the velocity. Let c
be the constant of proportionality.
The gravitational constant is g=9.8m/s2.
a) Find a differential equation for the velocity v:
dv
dt =
(b) Determine the limiting velocity after a long time. Your
answer should be an expression in c.
limiting velocity = m/s
(c) Find the drag coefficient, c, so that the limiting velocity
is 47 m/s.
c= kg/m
Solution: (a)
The force of gravity is given by Fg=mg, while the air resistance
is Fr=−cv2.
By Newton’s second law, the net force is equal to mdv
dt . Thus
mdv
dt =Fg−Fr
mdv
dt =mg −cv2
dv
dt =g−c
mv2
Substituing the given values of gand myields the differential
equation for the velocity:
6
dv
dt =9.8−c
38 v2
(b)
The limiting velocity can be found by solving dv
dt =0, that is,
9.8−c
38 v2=0. Solving the equation gives v=±q372.4
c. Since
vis positive in the downward direction, the limiting velocity is
given by
r372.4
c
(c)
We solve the equation 47 =q372.4
c. This yields c=
0.16858306926211.
Answer(s) submitted:
•9.8-(((c/(38)))vˆ(2))
•sqrt(((372.4)/c))
•.16858
(correct)
Correct Answers:
•9.8 - c*vˆ2/38
•sqrt(9.8*38/c)
•0.16858306926211
14. (1 point) A pond contains 2640 L of pure water and an
uknown amount of an undesirable chemical. Water contaninig
0.06 kg of this chemical per liter flows into the pond at a rate of
8 L/h. The mixture flows out at the same rate, so the amount of
water in the pond remains constant. Assume that the chemical
is uniformly distributed throughout the pond.
Let Q(t)be the amount of chemical (in kg) in the pond at time t
hours.
(a) Write a differential equation for the amount of chemical
in the pond?
at any time time (enter Q for Q(t):
dQ
dt =
(b) How much chemical will be in the pond after a long time?
Q∞=(kg)
(c) Does the limiting value in part (b) depend on the amount
that was present initially? [?/yes/no]
Solution: (a) First note that, since the incoming and outgo-
ing flows of water are the same, the amount of water in the pool
remains constant at 2640 L. We have
dQ
dt =rate in −rate out
where ”rate in” and ”rate out” refer to the rates at which the
chemical enters and exits the tank, respectively. The rate at
which the chemical enters the pond is given by
rate in = (8 L/h)(0.06 kg/L) = 0.48 kg/h
The concentation of chemical in the pond is Q
2640 kg/L, so the
rate of flow out is
rate out = (8 L/h)Q
2640 kg/L=1
330 Qkg/h
Thus we obtain the differential equation
dQ
dt =0.48 −1
330 Q
where each term has the units kg/min.
(b) The limiting value can be found by solving dQ
dt =0, that
is, 0.48 −1
330 Q=0. This gives
Q∞=158.4
(c) The limiting value is independent of the initial amount of
chemical in the pond.
Answer(s) submitted:
•.48-(((.48Q)/(158.4)))
•158.4
•no
(correct)
Correct Answers:
•0.48-8*Q/2640
•158.4
•no
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