Lecture10_3.4

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].


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.


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


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).


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.


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}.


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 .


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.


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.


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}.


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).


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.


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.


2v ′ = 2(9.81)− v/20, v(0) = −200.


v ′ + (1/40)v = 9.81.


µ(t) = eR

(1/40) dt = et/40.


v(t) = e−t/40{∫ t

es/409.81 ds + C}.


Example # 2, cont.


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.


−200 = v(0) = 392.4 + C,

soC = −592.4,

andv(t) = 392.4− 592.4e−t/40.


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.


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.


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.


Example # 3, cont.

Solution: We use Newton’s second law with

F (t , v) = mg − (1/40)mg − 10v , m = 100, v(0) = 0.


100v ′ = 100(9.81)− (1/40)100(9.81)− 10v , v(0) = 0.


v ′ + 0.1v = 0.975(9.81).


µ(t) = eR

0.1 dt = e0.1t .


v(t) = e−0.1t{∫ t

e0.1s(0.975)9.81 ds + C}.


Example # 3, cont.


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 .


0 = v(0) = 95.6475 + C,

soC = −95.6475,

andv(t) = 95.6475

(1− e−0.1t

).


Example # 3, cont.

Finally, we seek T such that v(T ) = 70:

70 = 95.6475(

1− e−0.1T)

=⇒ T ≈ 13.162.


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