Final Term Lectures - De

8/3/2019 Final Term Lectures - De

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Orthogonal Trajectories

We have seen before (seeseparable equationsfor example) that the solutions of a differentialequation may be given by an implicit equation with a parameter something like

This is an equation describing a family of curves. Whenever we fix the parameter Cwe get

one curve and vice-versa. For example, consider the families of curves

where m and Care parameters. Clearly, we may change the names of the variables and still

have the same geometric curves. For example, the above families define the same geometric

object as

Note that the first family describes all the lines passing by the origin (0,0) while the second

the family describes all the circles centered at the origin (including the limit case when the

radius 0 which reduces to the single point (0,0)) (see the pictures below).
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and


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In this page, we will only use the variablesx andy. Any family of curves will be written as

One may ask whether any family of curves may be generated from a differential equation? In

general, the answer is no. Let us see how to proceed if the answer were to be yes. First

differentiate with respect tox, and get a new equation involving in generalx,y, , and C.Using the original equation, we may able to eliminate the parameter Cfrom the new

equation.

Example. Find the differential equation satisfied by the family

Answer. We differentiate with respect tox, to get


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Since we have

then we get

You may want to do some algebra to make the new equation easy to read. The next step is to

rewrite this equation in the explicit form

this is the desired differential equation.

Example. Find the differential equation (in the explicit form) satisfied by the family

Answer. We have already found the differential equation in the implicit form

Algebraic manipulations give

Let us reconsider the example of the two families

If we draw the two families together on the same graph we get


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As we see here something amazing happened. Indeed, it is clear that whenever one line

intersects one circle, the tangent line to the circle (at the point of intersection) and the line are

perpendicular or orthogonal. We say the two curves are orthogonal at the point of

intersection.

Definition. Consider two families of curves and . We say that and are

orthogonal whenever any curve from intersects any curve from , the two curves are

orthogonal at the point of intersection.

For example, we have seen that the familiesy = m x and are orthogonal. One

may then ask the following natural question:

Given a family of curves , is it possible to find a family of curves which is orthogonal

to ?

The answer to this question has many implications in many areas such as physics, fluid-

dynamics, etc... In general this question is very difficult. But in some cases, we may be able

to carry on the calculations and find the orthogonal family. Let us show how.


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Consider the family of curves . We assume that an associated differential equation may be

found, say

We know that for any curve from the family passing by the point (x,y), the slope of the

tangent at this point isf(x,y). Hence the slope of the line perpendicular (or orthogonal) to this

tangent is which happens to be the slope of the tangent line to the orthogonalcurve passing by the point (x,y). In other words, the family of orthogonal curves are solutions

to the differential equation

From this we see what we have to do. Indeed consider a family of curves . In order to find

the orthogonal family, we use the following practical steps

Step 1. Find the associated differential equation.

Step 2. Rewrite this differential equation in the explicit form

Step 3. Write down the differential equation associated to the orthogonal family

Step 4. Solve the new equation. The solutions are exactly the family of orthogonal

curves.

Step 5. You may be asked to give a geometric view of the two families. Also you may

be asked to find a specific curve from the orthogonal family (something like an IVP).

Example. Find the orthogonal family to the family of circles

Answer. First, we look for the differential equation satisfied by the circles. We differentiate

with respect to the variablex to get

We rewrite this equation in the explicit form


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Next we write down the equation for the orthogonal family

This is alinearas well as aseparable equation. If we use the technique oflinear equations, we

get the integrating factor

which gives

We recognize the family of lines and we confirm our earlier observation (that the two

families are indeed orthogonal).

This example is somehow easy and was given here to illustrate the technique.

Example. Find the orthogonal family to the family of circles
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Answer. We have seen before that the explicit differential equation associated to the family

of circles is

Hence the differential equation for the orthogonal family is

We recognize an homogeneous equation. Let us use the technique developed to solve this

kind of equations. Consider the new variable (or equivalentlyy =x z). Then we have

and


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Hence we have

Algebraic manipulations imply

This is aseparable equation. The constant solutions are given by

which givesz=0. The non-constant solutions are found once we separate the variables

and then we integrate

Before we perform the integration for the left-hand side, we need to use partial

decomposition technique. We have

We will leave the details to you to show thatA = 1,B=-2, and C=0. Hence we have

Hence

which is equivalent to
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where . Putting all the solutions together we get

Going back to the variabley, we get

which is equivalent to

We recognize a family of circles centered on the y-axis and the liney=0 (the x-axis which

was easy to guess, isn't it?)


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If we put both families together, we appreciate better the orthogonality of the curves (see the

picture below).


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[Differential Equations][First Order D.E.]

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Applications of FirstOrder EquationsOrthogonal trajectories. The term orthogonal meansperpendicular, and trajectory means

path or cruve. Orthogonal trajectories, therefore, are two families of curves that always

intersect perpendicularly. A pair of intersecting curves will be perpendicular if the product of

their slopes is 1, that is, if the slope of one is the negative reciprocal of the slope of theother. Since the slope of a curve is given by the derivative, two familes of curves 1(x,y, c) =

0 and 2(x,y, c) = 0 (where c is a parameter) will be orthogonal wherever they intersect if

Example 1: The electrostatic field created by a positive point charge is pictured as a

collection of straight lines which radiate away from the charge (Figure1). Using the fact that

the equipotentials (surfaces of constant electric potential) are orthogonal the electric field

lines, determine the geometry of the equipotenitials of a point charge.

Figure 1

If the origin of anxy coordinate system is placed at the charge, then the electric field lines canbe described by the family

The first step in determining the orthogonal trajectories is to obtain an expression for theslope of the curves in this family that does notinvolve the parameter c. In the present case,
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The differential equation describing the orthogonal trajectories is therefore

since the right-hand side of (**) is the negative reciprocal of the right-hand side of (*). Because this

equation is separable, the solution can proceed as follows:

where c2

= 2 c.

The equipotential lines (that is, the intersection of the equipotential surfaces with any plane

containing the charge) are therefore the family of circlesx

2

+y

2

= c

2

centered at the origin.The equipotential and electric field lines for a point charge are shown in Figure2.

Figure 2

Example 2: Determine the orthogonal trajectories of the family of circlesx2

+ (y c)2

= c2

tangent to thex axis at the origin.

The first step is to determine an expression for the slope of the curves in this family that doesnot involve the parameter c. By implicit differentiation,


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To eliminate c, note that

The expression for dy/dx may now be written in the form

Therefore, the differential equation describing the orthogonal trajectories is

since the right-hand side of (**) is the negative reciprocal of the right-hand side of (*).

If equation (**) is written in the form

note that it is not exact (since My= 2 ybut Nx= 2 y). However, because

is a function ofxalone, the differential equation has


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as an integrating factor. After multiplying through by =x2

, the differential equation describing the

desired family of orthogonal trajectories becomes

which is now exact (because My= 2x2

y= Nx). Since

and

the solution of the differential equation is

(The reason the constant was written as 2 c rather than as c will be apparent in the following

calculation.) With a little algebra, the equation for this family may be rewritten:

This shows that the orthogonal trajectories of the circles tangent to thex axis at the origin are

the circles tangent to they axis at the origin! See Figure3.


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Figure 3

Radioactive decay. Some nuclei are energetically unstable and can spontaneously transform

into more stable forms by various processes known collectively as radioactive decay. The

rate at which a particular radioactive sample will decay depends on the identity of the sample.

Tables have been compiled which list the half-lives of various radioisotopes. The half-life is

the amount of time required for one-half the nuclei in a sample of the isotope to decay;

therefore, the shorter the half-life, the more rapid the decay rate.

The rate at which a sample decays is proportional to the amount of the sample present.

Therefore, ifx(t) denotes the amount of a radioactive substance present at time t, then

(The rate dx/dtis negative, sincex is decreasing.) The positive constant kis called the rate

constant for the particular radioisotope. The solution of this separable first-order equation is

wherexodenotes the amount of substance present at time t= 0. The graph of this equation (Figure4

) is known as the exponential decay curve:


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Figure 4

The relationship between the half-life (denoted T1/2) and the rate constant kcan easily be

found. Since, by definition,x = x6 at t= T1/2, (*) becomes

Because the half-life and rate constant are inversely proportional, the shorter the half-life, the

greater the rate constant, and, consequently, the more rapid the decay.

Radiocarbon dating is a process used by anthropologists and archaeologists to estimate the

age of organic matter (such as wood or bone). The vast majority of carbon on earth is

nonradioactive carbon-12 (12

C). However, cosmic rays cause the formation ofcarbon-14

(14

C), a radioactive isotope of carbon which becomes incorporated into living plants (and

therefore into animals) through the intake of radioactive carbon dioxide (14

CO2). When the

plant or animal dies, it ceases its intake of carbon-14, and the amount present at the time of

death begins to decrease (since the14

C decays and is not replenished). Since the half-life of14

C

is known to be 5730 years, by measuring the concentration of14

C in a sample, its age can be

determined.

Example 3: A fragment of bone is discovered to contain 20% of the usual14

C concentration.

Estimate the age of the bone.

The relative amount of14

C in the bone has decreased to 20% of its original value (that is, the

value when the animal was alive). Thus, the problem is to calculate the value oftat whichx(

t) = 0.20xo (wherex = the amount of14

C present). Since


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the exponential decay equation (*) says

Newton's Law of Cooling. When a hot object is placed in a cool room, the object dissipates

heat to the surroundings, and its temperature decreases. Newton's Law of Cooling states thatthe rate at which the object's temperature decreases is proportional to the difference between

the temperature of the object and the ambient temperature. At the beginning of the colling

process, the difference between these temperatures is greatest, so this is when the rate of

temperature decrease is greatest. However, as the object cools, the temperature difference

gets smaller, and the cooling rate decreases; thus, the object cools more and more slowly as

time passes. To formulate this process mathematically, let T( t) denote the temperature of the

object at time tand let Ts denote the (essentially constant) temperature of the surroundings.

Newton's Law of Cooling then says

Since Ts < T(that is, since the room is cooler than the object), Tdecreases, so the rate ofchange of its temperature, dT/dt, is necessarily negative. The solution of this separable

differential equation proceeds as follows:

Example 4: A cup of coffee (temperature = 190F) is placed in a room whose temperature is

70F. After five minutes, the temperature of the coffee has dropped to 160F. How many

more minutes must elapse before the temperature of the coffee is 130F?

Assuming that the coffee obeys Newton's Law of Cooling, its temperature Tas a function of

time is given by equation (*) with Ts= 70:


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Because T(0) = 190, the value of the constant of integration ( c) can be evaluated:

Furthermore, since information about the cooling rate is provided ( T= 160 at time t= 5minutes), the cooling constant kcan be determined:

Therefore, the temperature of the coffee tminutes after it is placed in the room is

Now, setting T= 130 and solving for tyields

This is the total amount of time after the coffee is initially placed in the room for its

temperature to drop to 130F. Therefore, after waiting five minutes for the coffee to cool

from 190F to 160F, it is necessary to then wait an additional seven minutes for it to cool

down to 130F.


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Skydiving. Once a sky diver jumps from an airplane, there are two forces that determine her

motion: the pull of the earth's gravity and the opposing force of air resistance. At high speeds,

the strength of the air resistance force (the drag force) can be expressed as kv2, where v is the

speed with which the sky diver descends and kis a proportionality constant determined by

such factors as the diver's cross-sectional area and the viscosity of the air. Once the parachute

opens, the descent speed decreases greatly, and the strength of the air resistance force is givenby Kv.

Newton's Second Law states that if a net force Fnet acts on an object of mass m, the object will

experience an accelaration a given by the simple equation

Since the acceleration is the time derivative of the velocity, this law can be expressed in the

form

In the case of a sky diver initially falling without a parachute, the drag force is Fdrag = kv2, and

the equation of motion (*) becomes

or more simply,

where b = k/m. [The letter g denotes the value of the gravitational acceleration, and mg is the force

due to gravity acting on the mass m (that is, mg is its weight). Near the surface of the earth, g is

approximately 9.8 meters per second2.] Once the sky diver's descent speed reaches v=

, the preceding equation says dv/ dt= 0; that is, vstays constant. This occurs when

the speed is great enough for the force of air resistance to balance the weight of the sky diver; the

net force and (consequetly) the acceleration drop to zero. This constant descent velocity is known as

the terminal velocity. For a sky diver falling in the spread-eagle position without a parachute, the

value of the proportionality constant kin the drag equation Fdrag = kv2

is approximately kg/m.

Therefore, if the sky diver has a total mass of 70kg (which corresponds to a weight of about 150

pounds), her terminal velocity is


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or approximately 120 miles per hour.

Once the parachute opens, the air resistance force becomes Fair resist = Kv, and the equation of

motion (*) becomes

or more simply,

where B = K/m. Once the parachutist's descent speed slows to v= g/B = mg/K, the preceding

equation says dv/dt= 0; that is, vstays constant. This occurs when the speed is low enough for the

weight of the sky diver to balance the force of air resistance; the net force and (consequently) the

acceleration reach zero. Again, this constant descent velocity is known as the terminal velocity. For a

sky diver falling with a parachute, the value of the proportionality constant Kin the equation Fair resist

= Kvis approximately 110 kg/s. Therefore, if the sky diver has a total mass of 70 kg, the terminal

velocity (with the parachute open) is only

which is about 14 miles per hour. Since it is safer to hit the ground while falling at a rate of 14 miles

per hour rather than at 120 miles per hour, sky divers use parachutes.

Example 5: After a free-falling sky diver of mass m reaches a constant velocity ofv1, her

parachute opens, and the resulting air resistance force has strength Kv. Derive an equation for

the speed of the sky diver tseconds after the parachute opens.

Once the parachute opens, the equation of motion is

where B = K/m. The parameter that will arise from the solution of this first-order differential

equation will be determined by the initial condition v(0) = v1 (since the sky diver's velocity is v1 at the

moment the parachute opens, and the clock is reset to t= 0 at this instant). This separable

equation is solved as follows:


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Now, since v(0) = v1gBv1 = c, the desired equation for the sky diver's speed tsecondsafter the parachute opens is

Note that as time passes (that is, as tincreases), the term e( K/m)t

goes to zero, so (as expected)

the parachutist's speed v slows to mg/K, which is the terminal speed with the parachute open.

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Final Term Lectures - De