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Chapter All

# MA 26200 Chapter All: MA 262 Solution Manual

Department
Mathematics
Course Code
MA 26200
Professor
Sa Barreto Antonio
Chapter
All

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Solutions to Section 1.1
True-False Review:
1. FALSE. A derivative must involve some derivative of the function y=f(x), not necessarily the ﬁrst
derivative.
2. TRUE. The initial conditions accompanying a dierential equation consist of the values of y, y0,... at
t= 0.
3. TRUE. If we deﬁne positive velocity to be oriented downward, then
dv
dt =g,
where gis the acceleration due to gravity.
4. TRUE. We can justify this mathematically by starting from a(t)=g, and integrating twice to get
v(t)=gt +c, and then s(t)=1
2gt2+ct +d, which is a quadratic equation.
5. FALSE. The restoring force is directed in the direction opposite to the displacement from the equilibrium
position.
6. TRUE. According to Newton’s Law of Cooling, the rate of cooling is proportional to the dierence
between the object’s temperature and the medium’s temperature. Since that dierence is greater for the
object at 100Fthan the object at 90F, the object whose temperature is 100Fhas a greater rate of
cooling.
7. FALSE. The temperature of the object is given by T(t)=Tm+cekt, where Tmis the temperature
of the medium, and cand kare constants. Since ekt 6= 0, we see that T(t)6=Tmfor all times t. The
temperature of the object approaches the temperature of the surrounding medium, but never equals it.
8. TRUE. Since the temperature of the coee is falling, the temperature dierence between the coee and
the room is higher initially, during the ﬁrst hour, than it is later, when the temperature of the coee has
9. FALSE. The slopes of the two curves are negative reciprocals of each other.
10. TRUE. If the original family of parallel lines have slopes kfor k6= 0, then the family of orthogonal tra-
jectories are parallel lines with slope 1
k. If the original family of parallel lines are vertical (resp. horizontal),
then the family of orthogonal trajectories are horizontal (resp. vertical) parallel lines.
11. FALSE. The family of orthogonal trajectories for a family of circles centered at the origin is the family
of lines passing through the origin.
Problems:
1. Starting from the dierential equation d2y
dt2=g, where gis the acceleration of gravity and yis the unknown
position function, we integrate twice to obtain the general equations for the velocity and the position of the
object:
dy
dt =gt +c1and y(t)=gt2
2+c1t+c2,
where c1,c
2are constants of integration. Now we impose the initial conditions: y(0) = 0 implies that c2= 0,

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and dy
dt (0) = 0 implies that c1=0.Hence, the solution to the initial-value problem is
y(t)=gt2
2.
The object hits the ground at the time t0for which y(t0) = 100. Hence 100 = gt2
0
2, so that t0=q200
g4.52
s, where we have taken g=9.8ms
2.
2. Starting from the dierential equation d2y
dt2=g, where gis the acceleration of gravity and yis the unknown
position function, we integrate twice to obtain the general equations for the velocity and the position of the
ball, respectively: dy
dt =gt +cand y(t)=1
2gt2+ct +d,
where c, d are constants of integration. Setting y= 0 to be at the top of the boy’s head (and positive
direction downward), we know that y(0) = 0. Since the object hits the ground 8 seconds later, we have that
y(8) = 5 (since the ground lies at the position y= 5). From the values of y(0) and y(8), we ﬁnd that d=0
and 5 = 32g+8c. Therefore, c=532g
8.
(a) The ball reaches its maximum height at the moment when y0(t) = 0. That is, gt +c= 0. Therefore,
t=c
g=32g5
8g3.98 s.
(b) To ﬁnd the maximum height of the tennis ball, we compute
y(3.98) ⇡253.51 feet.
So the ball is 253.51 feet above the top of the boy’s head, which is 258.51 feet above the ground.
3. Starting from the dierential equation d2y
dt2=g, where gis the acceleration of gravity and yis the unknown
position function, we integrate twice to obtain the general equations for the velocity and the position of the
rocket, respectively: dy
dt =gt +cand y(t)=1
2gt2+ct +d,
where c, d are constants of integration. Setting y= 0 to be at ground level, we know that y(0) = 0. Thus,
d= 0.
(a) The rocket reaches maximum height at the moment when y0(t) = 0. That is, gt +c= 0. Therefore, the
time that the rocket achieves its maximum height is t=c
g. At this time, y(t)=90 (the negative sign
accounts for the fact that the positive direction is chosen to be downward). Hence,
90 = yc
g=1
2gc
g2
+cc
g=c2
2gc2
g=c2
2g.
Solving this for c, we ﬁnd that c=±p180g. However, since crepresents the initial velocity of the rocket,
and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose
c=p180g⇡42.02 ms1, and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 42.02 ms1.

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(b) The time that the rocket reaches its maximum height is t=c
g⇡42.02
9.81 =4.28 s.
4. Starting from the dierential equation d2y
dt2=g, where gis the acceleration of gravity and yis the unknown
position function, we integrate twice to obtain the general equations for the velocity and the position of the
rocket, respectively:
dy
dt =gt +cand y(t)=1
2gt2+ct +d,
where c, d are constants of integration. Setting y= 0 to be at the level of the platform (with positive
direction downward), we know that y(0) = 0. Thus, d= 0.
(a) The rocket reaches maximum height at the moment when y0(t) = 0. That is, gt +c= 0. Therefore, the
time that the rocket achieves its maximum height is t=c
g. At this time, y(t)=85 (this is 85 m above
the platform, or 90 m above the ground). Hence,
85 = yc
g=1
2gc
g2
+cc
g=c2
2gc2
g=c2
2g.
Solving this for c, we ﬁnd that c=±p170g. However, since crepresents the initial velocity of the rocket,
and the initial velocity is negative (relative to the fact that the positive direction is downward), we choose
c=p170g⇡40.84 ms1, and thus the initial speed at which the rocket must be launched for optimal
viewing is approximately 40.84 ms1.
(b) The time that the rocket reaches its maximum height is t=c
g⇡40.84
9.81 =4.16 s.
5. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
d2y
dt2=g, y(0) = 0,dy
dt (0) = 2,
where gis the acceleration of gravity and yis the unknown position function. We integrate this dierential
equation twice to obtain the general equations for the velocity and the position of the object:
dy
dt =gt +c1and y(t)=gt2
2+c1t+c2.
Now we impose the initial conditions: since y(0) = 0, we have c2= 0. Moreover, since dy
dt (0) = 2, we have
c1=2. Hence the solution to the initial-value problem is y(t)=gt2
22t. We are given that y(10) = h.
Consequently, h=g(10)2
22·10 =)h= 10(5g2) 470 m where we have taken g=9.8ms
2.
6. If y(t) denotes the displacement of the object from its initial position at time t, the motion of the object
can be described by the initial-value problem
d2y
dt2=g, y(0) = 0,dy
dt (0) = v0.