First-Order Equations in Practice: Decay, Cooling, Circuits and Trajectories
Chapter One Hundred Four
Syllabus topic Module 2, "2.1 Equation of the first order and of the first degree"
Pages 231 to 233 of 303
In one line
Four applications, each solved with a method from this section, including the orthogonal trajectories that MU's own reading list treats as part of the topic.
Where this chapter stands
GUIDELINES section 2.3 of this book's house rules: not the driver of the syllabus, but worth having. None of what follows is in MU's printed labels. It is here because a method learned with no idea what it is for is forgotten, and because orthogonal trajectories appear in every one of the three books on her reading list.
One: radioactive decay and half life
The model, from the chapter on where equations come from:
dN/dt = -k N
N = C e^(-k t)
With N equal to N0 at time zero, C is N0. The half life T is the time for N to halve, so e to the minus kT equals one half, giving T equal to log 2 over k.
log(2)/k = log(2)/k
Worked with numbers: if a quantity falls to 90 per cent of its value in 10 years, then e to the minus 10k is 0.9, so k is minus log(0.9) over 10, about 0.01054 per year, and the half life is log 2 over that, about 65.8 years.
log(2)/(-log(9/10)/10) = 10 log(2)/log(10/9)
The same arithmetic answers carbon dating, drug clearance and the decay of a signal in a lossy medium.
Two: Newton's law of cooling
dT/dt = -k(T - 20)
T = 20 + C e^(-k t)
Worked: a body at 100 degrees cools to 60 in 10 minutes in a room at 20. How long to reach 30?
At t = 0, T is 100, so C is 80. At t = 10, T is 60, so 40 equals 80 e to the minus 10k, giving e to the minus 10k equal to one half, so k is log 2 over 10.
For T = 30: 10 equals 80 e to the minus kt, so e to the minus kt is one eighth, so kt is log 8, which is 3 log 2. Since k is log 2 over 10, t is 30 minutes.
3 log(2)/(log(2)/10) = 30
The neatness is not an accident: the temperature difference halved in 10 minutes, so it takes 30 minutes to halve three times, from 80 to 40 to 20 to 10. Recognising that saves all the algebra.
Three: a charging capacitor
dV/dt = (E - V)/2
V = E + C e^(-t/2)
With V(0) = 0 the constant is minus E, so V is E(1 minus e to the minus t over 2). The time constant is the 2 in the exponent, and at t equal to one time constant the voltage has reached 1 minus one over e of its final value, about 63 per cent.
First-Order Equations in Practice: Decay, Cooling, Circuits and Trajectories
1 - e^(-1) = 1 - 1/e
That number, 0.632, is the one every electronics student memorises, and this is where it comes from.
Four: orthogonal trajectories
This is the application MU's reading list treats as part of the topic, and it is set.
The problem. Given a family of curves, find the family that crosses every one of them at right angles.
The method. Three steps.
- Find the differential equation of the given family, by eliminating its constant, as the earlier chapter did.
- Replace dy/dx by minus one over dy/dx, because perpendicular slopes are negative reciprocals.
- Solve the new equation. Its solutions are the orthogonal trajectories.
Worked: the family of circles about the origin.
Their equation, from the earlier chapter, is dy/dx equal to minus x over y.
dy/dx = -x/y
x^2 + y^2 = C
Replacing dy/dx by its negative reciprocal gives dy/dx equal to plus y over x.
dy/dx = y/x
y = C x
So the orthogonal trajectories of the circles about the origin are the straight lines through the origin, which is geometrically obvious and is the reassuring first example.
Worked: the family of parabolas y = C x squared.
Eliminating C: differentiating gives dy/dx equal to 2Cx, and C is y over x squared, so dy/dx equals 2y over x.
dy/dx = 2y/x
y = C x^2
Replacing dy/dx by minus one over it: minus x over (2y) equals dy/dx.
dy/dx = -x/(2y)
x^2/2 + y^2 = C
So the orthogonal family is a set of ellipses, with their long axis along the x axis. Each parabola crosses each ellipse at right angles.
Worked: the family of hyperbolas xy = C.
Differentiating: y plus x dy/dx = 0, so dy/dx equals minus y over x.
Replacing by the negative reciprocal: dy/dx equals x over y.
dy/dx = x/y
y^2 - x^2 = C
So the orthogonal trajectories of one family of rectangular hyperbolas is the other family of rectangular hyperbolas, rotated by 45 degrees. That symmetry is pleasing and is a standard question.
Where you meet orthogonal trajectories outside an examination
Electric field lines cross lines of equal potential at right angles. Lines of steepest descent cross contour lines at right angles. Streamlines cross lines of equal pressure at right angles in a certain kind of flow. In every case, one family is given and the other is what you want, and this is the calculation.
Check yourself
| Given family | Its equation | Orthogonal trajectories |
|---|---|---|
| circles about the origin | dy/dx = -x/y | lines through the origin |
| lines through the origin | dy/dx = y/x | circles about the origin |
| parabolas y = Cx^2 | dy/dx = 2y/x | ellipses x^2 + 2y^2 = C |
| hyperbolas xy = C | dy/dx = -y/x | hyperbolas y^2 - x^2 = C |
First-Order Equations in Practice: Decay, Cooling, Circuits and Trajectories
dy/dx = -x/(2y)
x^2 + 2 y^2 = C
The last block is the third row's answer written without the halves, which is the tidier form and the same family.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.