lecture10_3.4
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Math 220: Lecture 10§ 3.4
Professor Nicholls
Department of Mathematics, Statistics,and Computer Science
University of Illinois at Chicago
Math 220: Lecture 10
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 1 / 21
Chapter 3: Mathematical Models and NumericalMethods Involving First Order Equations
As we have commented, DE are very effective modeling tools andfirst order ODEs are no exceptions:
Compartmental Analysis [§ 3.2]: Mixing and populations!Heating and Cooling of Buildings [§ 3.3, not in this class].Newtonian Mechanics [§ 3.4, today].Electrical Circuits [§ 3.5, not in this class].
Numerical Methods are a crucial tool in the study of DE and thereare many improvements to Euler’s Method. For first order ODEthese include:
Improved Euler’s Method [§ 3.6].Taylor and Runge–Kutta Methods [§ 3.7].
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 2 / 21
Newtonian Mechanics
Newtonian mechanics focuses on the motion of objects and theeffects of forces.It is appropriate for very large and very slow objects.Question: How slow? Answer: Fast means close to the speed oflight (Relativity).Question: How large? Answer: Small (typically) means on thescale of nanometers (Quantum Mechanics).Clearly, nearly all of our daily experience is well within the realm ofapplicability of Newtonian Mechanics so it is incredibly useful. Wemust use care to always work in an Intertial Reference Frame.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 3 / 21
Newtonian Mechanics: Modeling
The key to our analysis is Newton’s Second Law: The rate ofchange of the momentum is equal to the sum of the forces on anobject:
dpdt
= F (t , x ,p).
Recall that momentum and velocity are “equivalent” up to theconstant mass, m:
p(t) = mv(t).
Acceleration is the time rate of change of velocity so
a(t) =dvdt
=1m
dpdt
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 4 / 21
Newtonian Mechanics: Modeling, cont.
Typically one must use the position, x(t), velocity x ′(t), andacceleration x ′′(t) as unknowns.However, is this section it suffices to use velocity v(t) andacceleration v ′(t). Thus we obtain a first–order ODE!In situations where the forces are independent of position we have
mv ′ = F (t , v).
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 5 / 21
Example # 1
Example: (NSS 3.4 # 5): An object of mass 5 kg is given an initialdownward velocity of 50 m/sec and then allowed to fall under theinfluence of gravity. Assume that the force in newtons due to airresistance is −10v , where v is the velocity of the object in m/sec.Determine the equation of motion of the object. If the object is initially100 m above the ground, determine when the object will strike theground.Remark: We will use the book’s convention that down is the positivedirection.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 6 / 21
Example # 1, cont.
Solution: We use Newton’s second law with
F (t , v) = 5(9.81)− 10v , m = 5, v(0) = 50.
Thus, we must solve the IVP
5v ′ = 5(9.81)− 10v , v(0) = 50.
We begin by solving the ODE which is linear:
v ′ + 2v = 9.81.
The Integrating Factor is
µ(t) = eR
2 dt = e2t .
Using this, our solution is
v(t) = e−2t{∫ t
e2s9.81 ds + C}.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 7 / 21
Example # 1, cont.
Solution: Recall that our solution formula is
v(t) = e−2t{∫ t
e2s9.81 ds + C}
so that
v(t) = e−2t{
(9.81/2)e2t + C}
= (9.81/2) + Ce−2t .
Now, the initial condition gives
50 = v(0) = (9.81/2) + C,
soC = 50− (9.81/2),
and
v(t) = (9.81/2) + [50− (9.81/2)]e−2t = 4.905 + 45.095e−2t .
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 8 / 21
Example # 1, cont.
Solution: Finally, to recover the position we simply integrate thevelocity:
x(t) =
∫ tv(s) ds + K
=
∫(4.905 + 45.095e−2s) ds + K
= 4.905t − (45.095/2)e−2t + K
= 4.905t − 22.5475e−2t + K .
Using the initial condition, x(0) = 0, we find
0 = 4.905(0)− 22.5475(1) + K ,
so K = 22.5475, and,
x(t) = 4.905t − 22.5475e−2t + 22.5475.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 9 / 21
Example # 1, cont.
Finally, we wish to find the T such that x(T ) = 100:
100 = x(T ) = 4.905T + 22.5475(1− e−2T )
≈ 4.905T + 22.5475,
soT ≈ (100− 22.5475)/4.905 ≈ 15.79.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 10 / 21
Comment: Solving Mechanics Problems
We point out that the “canonical” problem we wish to solve inNewtonian Mechanics is
mv ′ = mg − bv , v(0) = v0.
This is a linear ODEv ′ + (b/m)v = g,
which is linear with Integrating Factor
µ(t) = eR t (b/m) ds = e(b/m)t .
With this, the exact solution is
v(t) = e−(b/m)t{∫ t
e(b/m)sg ds + C}.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 11 / 21
Comment: Solving Mechanics Problems, cont.
Continuing,
v(t) = e−(b/m)t{∫ t
e(b/m)sg ds + C}
= e−(b/m)t{
((mg)/b)e(b/m)t + C}
= (mg)/b + Ce−(b/m)t ,
which is the “limiting velocity” with an exponentially decayingcorrection. Using the initial condition
v0 = v(0) = (mg)/b + C,
we findv(t) = (mg)/b + [v0 − (mg)/b]e−(b/m)t .
Finally, if we set the initial position to be x(0) = 0 we find
x(t) = (mgt)/b + (m/b)[v0 − (mg)/b](
1− e−(b/m)t).
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 12 / 21
Example # 2
Example: (NSS 3.4 # 12): A shell of mass 2 kg is shot upward with aninitial velocity of 200 m/sec. The magnitude of the force on the shelldue to air resistance is |v | /20. When will the shell reach its maximumheight above the ground? What is the maximum height?Remark: We will use the book’s convention that down is the positivedirection.Remark: Notice that when the velocity is negative (upwards), the dragforce is positive (downwards), so we can use F = −v/20.Remark: Also, when the velocity is positive (downwards), the dragforce is negative (upwards), so, again, we can use F = −v/20.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 13 / 21
Example # 2, cont.
Solution: While the shell moves upwards we use Newton’s secondlaw with
F (t , v) = 2(9.81)− v/20, m = 2, v(0) = −200.
Thus, we must solve the IVP
2v ′ = 2(9.81)− v/20, v(0) = −200.
We begin by solving the ODE which is linear:
v ′ + (1/40)v = 9.81.
The Integrating Factor is
µ(t) = eR
(1/40) dt = et/40.
Using this, our solution is
v(t) = e−t/40{∫ t
es/409.81 ds + C}.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 14 / 21
Example # 2, cont.
Solution: Recall that our solution formula is
v(t) = e−t/40{∫ t
es/409.81 ds + C}
so that
v(t) = e−t/40{
(40)9.81et/40 + C}
= 392.4 + Ce−t/40.
Now, the initial condition gives
−200 = v(0) = 392.4 + C,
soC = −592.4,
andv(t) = 392.4− 592.4e−t/40.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 15 / 21
Example # 2, cont.
Solution: Finally, the shell reaches it maximum when the velocityequals zero. Seek T such that
0 = v(T ) = 392.4− 592.4e−T/40,
or392.4 = 592.4e−T/40,
so thatT = −40 ln(392.4/592.4) ≈ 16.47.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 16 / 21
Example # 2, cont.
Solution: The height of the shell at this time can be found byintegrating the velocity to give the displacement:
x(t) =
∫ tv(s) ds + K
=
∫ t(392.4− 592.4e−s/40) ds + K
= 392.4t + 40(592.4)e−t/40 + K .
Setting the initial height to be x = 0 we can solve for K :
0 = x(0) = 392.4(0) + 40(592.4) + K ,
giving K = −40(592.4) = −23,696. Finally,
x(16.47) = 392.4(16.47) + 23,696(
e−16.47/40 − 1)≈ −1534.81
which is 1534 meters above the initial position since downwards ispositive.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 17 / 21
Example # 3
Example: (NSS 3.4 # 9): An object of mass 100 kg is released fromrest from a boat into the water and allowed to sink. While gravity ispulling the object down, a buoyancy force of 1/40 times the weight ofthe object is pushing the object up (weight = mg). If we assume thatwater resistance exerts a force on the object that is proportional to thevelocity of the object, with proportionality constant 10 N-sec/m, find theequation of motion of the object. After how many seconds will thevelocity of the object be 70 m/sec.Remark: We will use the book’s convention that down is the positivedirection.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 18 / 21
Example # 3, cont.
Solution: We use Newton’s second law with
F (t , v) = mg − (1/40)mg − 10v , m = 100, v(0) = 0.
Thus, we must solve the IVP
100v ′ = 100(9.81)− (1/40)100(9.81)− 10v , v(0) = 0.
We begin by solving the ODE which is linear:
v ′ + 0.1v = 0.975(9.81).
The Integrating Factor is
µ(t) = eR
0.1 dt = e0.1t .
Using this, our solution is
v(t) = e−0.1t{∫ t
e0.1s(0.975)9.81 ds + C}.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 19 / 21
Example # 3, cont.
Solution: Recall that our solution formula is
v(t) = e−0.1t{∫ t
e0.1s(0.975)9.81 ds + C}
so that
v(t) = e−0.1t{
0.975(9.81/0.1)e0.1t + C}
= 95.6475 + Ce−0.1t .
Now, the initial condition gives
0 = v(0) = 95.6475 + C,
soC = −95.6475,
andv(t) = 95.6475
(1− e−0.1t
).
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 20 / 21
Example # 3, cont.
Finally, we seek T such that v(T ) = 70:
70 = 95.6475(
1− e−0.1T)
=⇒ T ≈ 13.162.
Professor Nicholls (UIC) Math 220: Lecture 10 Math 220: Lecture 10 21 / 21