Parametric Equations
The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged
Discussion
analytic geometry
In previous sections, we discussed how the kinematic variables of distance, displacement, speed, velocity, and acceleration can be used to analyze motion in a straight line. During one-dimensional motion, the objects moving have only one degree of freedom. This is an idealization, of course. When I step out of my apartment on to the sidewalk there is essentially one decision I have to make …
| left or | right? |
But I'm not some sort of computer-driven robot forced by the definition of "sidewalk" to choose only between these two options. I also have the option to move forward or backward; that is, across the street or back into my apartment. This second degree of freedom makes walking a two-dimensional activity. If I walk far enough my altitude is also sure to change, so you might consider walking a three-dimensional activity, but this is merely a response to the earth's surface.
| forward or | backward? |
| left or | right? |
| up or | down? |
Flying on the other hand is truly a three-dimensional activity; especially for helicopter pilots and hummingbirds. There are always three choices available to the people and animals that fly …
or if you prefer …
| north or | south? |
| east or | west? |
| up or | down? |
These choices are mutually exclusive of one another -- a property that is known as orthogonality. Two directions are orthogonal if there is no way that motion along one of these directions could result in motion in any of the other directions. Walk east or west all you like, you'll never see any north-south change in your position. That's why walking is not normally considered a three-dimensional activity. The freedom to go "up" or "down" a hill is really just the freedom to choose between going "forward" or "backward". That one option results in upward motion while the other results in downward motion seems more a function of the surface of the earth than in any choice on my part. I really can't decide to "turn up" while walking in the same way that I can decide to "turn left".
Orthogonal directions are always perpendicular to one another. Since "perpendicular" is a completely adequate word, "orthogonal" may seem unnecessarily pretentious. The thing with dimension is that it refers to more things than just the number of independent directions. Any two measurable things that are independent of one another can be considered dimensions. In thermodynamics (the study of heat and work) pressure and volume behave like up-down and left-right in kinematics. The situation here is a bit more complicated than just saying "pressure is independent of volume", but it should be apparent that saying "pressure is perpendicular to volume" doesn't make much sense. It is correct to say that "pressure is orthogonal to volume", however.
Identifying the appropriate perpendicular directions in a kinematics problem is one of the first steps in solving it. Attach these directions (called axes) to a point in space (called the origin) and you've just created a coordinate system. There is no rule behind naming axes, but it is tradition to call the principle horizontal direction +x. Naming the other directions is open to debate, however. High school physics courses tend to concentrate on two-dimensional problems. An instructor may use +y for the perpendicular horizontal direction in one instance and then use +y for up in another. College professors tend to be more formal and usually reserve +y for the perpendicular horizontal direction and +z for up. I tend to think that if a situation is strictly two-dimensional then the other direction should be +y no matter what that direction may be. I only use +z when all three dimensions need to be considered. The thing with physics is that it doesn't really matter. The universe is isotropic. All the laws of physics are always true without modification no matter where you place the origin, how you orient the axes (as long as they're perpendicular, of course), or what you name them. Since it doesn't matter, why not make it easy on yourself? Place the origin wherever it's convenient and label the axes in whatever manner you wish. You don't even have to call them x, y, and z. Here are some alternate names for axes I have actually seen used …
a, b, and c
i, j, and k
horizontal and vertical
parallel (∥) and perpendicular (⊥)
r, θ, and φ
ζ and ξ
Since all perpendicular directions are orthogonal and since any vector quantity can be resolved into components along these directions, n-dimensional motion can be completely described by n one-dimensional algebraic expressions in n perpendicular directions (where n is any whole number greater than zero). Thus two-dimensional motion can be completely described by two one-dimensional algebraic expressions along two perpendicular directions -- usually called x and y.
When a situation is essentially two-dimensional, the position of an object can be adequately described by two numbers (its xand y coordinates), the velocity of an object by two numbers (the xand y components of its velocity), and its acceleration by two numbers (the x and y components of its acceleration). From the Pythagorean theorem, the magnitudes of these quantities are related by the following expressions …
| r2 = | x2 | + | y2 |
| v2 = | vx2 | + | vy2 |
| a2 = | ax2 | + | ay2 |
The relation is more completely described in vector notation where the vectors with a hat (^) over them are unit vectors along the coordinate axes …
| r = | x | î | + | y | ĵ |
| v = | vx | î | + | vy | ĵ |
| a = | ax | î | + | ay | ĵ |
Since kinematic vectors at right angles are independent of each other, associated with each direction are functions for displacement, average velocity, and average acceleration …
| x = | x(t) | y = | x(t) | |
| vx = | Δx | vy = | Δy | |
| Δt | Δt | |||
| ax = | Δvx | ay = | Δvy | |
| Δt | Δt |
Similar functions can be derived for instantaneous velocity and instantaneous acceleration …
| x = | x(t) | y = | x(t) | |||||
| vx = | lim | Δx | vy = | lim | Δy | |||
| Δt → 0 | Δt | Δt → 0 | Δt | |||||
| ax = | lim | Δvx | ay = | lim | Δvy | |||
| Δt → 0 | Δt | Δt → 0 | Δt |
Or, in the language of calculus …
| x = | x(t) | y = | x(t) | |||||
| vx = | dx | vy = | dy | |||||
| dt | dt | |||||||
| ax = | dvx | = | d2x | ay = | dvy | = | d2y | |
| dt | dt2 | dt | dt2 |
Likewise, three-dimensional motion can be completely described by three one-dimensional algebraic expressions along three mutually perpendicular directions usually called x, y, and z. Since the universe has three spatial dimensions, the position of an object can be completely described by three numbers (its x, y, and z coordinates), the velocity of an object by three numbers (the x, y, and z components of its velocity), and its acceleration by three numbers (the x, y, and z components of its acceleration). From the Pythagorean theorem, the magnitudes of these quantities are related by the following expressions …
| r2 = | x2 | + | y2 | + | z2 |
| v2 = | vx2 | + | vy2 | + | vz2 |
| a2 = | ax2 | + | ay2 | + | az2 |
The relation is more completely described in vector notation where the vectors with a hat (^) over them are unit vectors along the coordinate axes …
| r = | x | ˆi | + | y | ˆj | + | z | ˆk |
| v = | vx | ˆi | + | vy | ˆj | + | vz | ˆk |
| a = | ax | ˆi | + | ay | ˆj | + | az | ˆk |
Since kinematic vectors at right angles are independent of each other, associated with each direction are functions for displacement, average velocity, and average acceleration …
| x = | x(t) | y = | x(t) | z = | z(t) | |||
| vx = | Δx | vy = | Δy | vz = | Δz | |||
| Δt | Δt | Δt | ||||||
| ax = | Δvx | ay = | Δvy | az = | Δvz | |||
| Δt | Δt | Δt |
Similar functions can be derived for instantaneous velocity and instantaneous acceleration …
| x = | x(t) | y = | x(t) | z = | z(t) | ||||||||
| vx = | lim | Δx | vy = | lim | Δy | vz = | lim | Δz | |||||
| Δt → 0 | Δt | Δt → 0 | Δt | Δt → 0 | Δt | ||||||||
| ax = | lim | Δvx | ay = | lim | Δvy | az = | lim | Δvz | |||||
| Δt → 0 | Δt | Δt → 0 | Δt | Δt → 0 | Δt |
Or, in the language of calculus …
| x = | x(t) | y = | x(t) | z = | z(t) | ||||||||
| vx = | dx | vy = | dy | vz = | dz | ||||||||
| dt | dt | dt | |||||||||||
| ax = | dvx | = | d2x | ay = | dvy | = | d2y | az = | dvz | = | d2z | ||
| dt | dt2 | dt | dt2 | dt | dt2 |
This application of algebra to geometry is known by several names, analytic geometry being perhaps the most common with coordinate geometry running a close second. Since the person generally credited with the discovery of this subject was the French philosopher and mathematician René Descartes (1596-1650), it is often also called cartesian geometry in his honor. While Descartes was the first to publish his thoughts on this subject, it was almost certainly discovered independently at about the same time by another French mathematician, Pierre de Fermat (1601-1665).
Equations in analytic geometry correspond to curves and surfaces. A typical curve is described by a function that generates a value on the y-axis from every value on the x-axis. A typical surface is described by a function that generates a value on the z-axis from every pair of x-y coordinates. Kinematic equations are described in a way that is somewhat different. The position of a moving object changes with time. Because the x, y, and z values depend on an additional parameter (time) that is not a part of the coordinate system, kinematic equations are also known as parametric equations.
Albert Einstein (1879-1955) turned physics on its head by removing time from the list of parameters and adding it to the list of coordinates. This was a central proposition in his famous theory of relativity; a topic that will be discussed in more detail later in this book. According to Einstein the universe is not three-dimensional with three coordinates -- x, y, and z -- it's four-dimensional with four coordinates -- x, y, z, and t. Objects appear to us to move only because we are swept up in the flow of time. If we could see time the same way we see length then objects in motion would appear as rigid curves fixed in space -- a four-dimensional space that included time, called space-time. In a sense, relativity abolished motion.
Miscellaneous notes
- Instantaneous velocity is always tangential to the curve.
- Acceleration has both tangential and normal/radial/centripetal components.
- A function is a mathematical relation that maps a single output value onto
a single input value. Many curves in the cartesian plane can be described
by a function y =ƒ(x). Some curves can't. One way around this might be to use an inverse function
to draw the curve x =ƒ−1(y). Sometimes even this doesn't work.
Try drawing a circle, for instance.
The equation r2 =x2 +y2
works but it can't be manipulated into a function. The rearranged equation
y = ±√(r2 − x2) doesn't cut it. It maps one value of x onto two values of y.
I want a honest function, not some mutant equation
that would drive me to seek the professional help of a mathematician. I
need something so simple a physicist can handle it, or even better still,
an equation so simple a computer can handle it. One number in, one number
out. Well … not exactly one number. What I really want is one number in and one
location out, one place on the cartesian plane, one (x, y) coordinate pair.
I can do this by parameterizing the curve -- turning the x and y coordinates
into separate functions of a third variable, the parametric variable, usually
identified with the letter t because the best physical meaning to assign
to it is time. I don't say "best" cavalierly. I say best because I mean
best -- best for us, the physicists and those who study physics. Parameterizing
a curve in time gives it a physical reality.
- This pair of equations will show you where the object is in space at any moment in time. If I was good, I could rig this computer to show you the way this thing wanders around in a little movie.
- Wow! A movie. Why that's a thousand pictures and a picture's worth a thousand words. That makes a movie worth a million words. Thank you physics man. Now I have a million words -- Um -- What's a word worth?
- Um. Ida no.
- The simplest parametric equations for an object moving on a circular path
at a constant speed are …
x = r cos t y = r sin t - More stuff is needed for this chain of thought.
Summary
- N-dimensional motion can be completely described by n one-dimensional
algebraic expressions along n mutually perpendicular directions
(where n is any whole number greater than zero).
- Two-dimensional motion can be completely described by two, one-dimensional algebraic expressions along two perpendicular directions.
- Three-dimensional motion can be completely described by three, one-dimensional algebraic expressions along three mutually perpendicular directions.
| r = | x | ˆi | + | y | ˆj | + | z | ˆk | r2 = | x2 | + | y2 | + | z2 | |
| v = | vx | ˆi | + | vy | ˆj | + | vz | ˆk | v2 = | vx2 | + | vy2 | + | vz2 | |
| a = | ax | ˆi | + | ay | ˆj | + | az | ˆk | a2 = | ax2 | + | ay2 | + | az2 | |
| x = | x(t) | y = | x(t) | z = | z(t) | ||||||||
| vx = | Δx | vy = | Δy | vz = | Δz | ||||||||
| Δt | Δt | Δt | |||||||||||
| ax = | Δvx | ay = | Δvy | az = | Δvz | ||||||||
| Δt | Δt | Δt | |||||||||||
| x = | x(t) | y = | x(t) | z = | z(t) | ||||||||
| vx = | lim | Δx | vy = | lim | Δy | vz = | lim | Δz | |||||
| Δt → 0 | Δt | Δt → 0 | Δt | Δt → 0 | Δt | ||||||||
| ax = | lim | Δvx | ay = | lim | Δvy | az = | lim | Δvz | |||||
| Δt → 0 | Δt | Δt → 0 | Δt | Δt → 0 | Δt | ||||||||
| x = | x(t) | y = | x(t) | z = | z(t) | ||||||||
| vx = | dx | vy = | dy | vz = | dz | ||||||||
| dt | dt | dt | |||||||||||
| ax = | dvx | = | d2x | ay = | dvy | = | d2y | az = | dvz | = | d2z | ||
| dt | dt2 | dt | dt2 | dt | dt2 | ||||||||
Problems
practice
- An alien is flying her spaceship at half the speed of light in the positive x direction
when the autopilot begins accelerating the ship uniformly in the negative y direction
at 2.34 m/s2 (0.2 times the acceleration due to gravity on the
alien's home planet, the name of which is impossible to write in human symbols). Determine the resultant
displacement and velocity of the spacecraft when the acceleration ceases 137 earth days later.
Start your solution by listing the given quantities along each axis and the time in convenient units … t = 137 × 24 × 3600 = 1.184 × 107 s x0 = 0 m y0 = 0 m vx0 = 0.050(3 × 108 m/s) = 1.5 × 107 m/s vy0 = 0 m/s ax = 0 m/s2 ay = −2.34 m/s2 Apply the one-dimensional equations of motion for constant acceleration in each direction to get the
components of the displacement and velocity.x = x0 + vx0Δt + ½ axΔt2 y = y0 + vy0Δt + ½ ayΔt2 x = vx0Δt y = ½ ayΔt2 x = (1.5 × 107 m/s)(1.184 × 107 s) y = ½ (−2.34 m/s2)(1.184 × 107 s)2 x = +1.776 × 1014 m y = −1.639 × 1014 m vx = vx0 + axΔt vy = vy0 + ayΔt vx = vx0 vy = ayΔt vx = 1.5 × 107 m/s vy = (−2.34 m/s2)(1.184 × 107 s) vy = −2.770 × 107 m/s Add the components to get the resultant displacement and velocity. r = √[ x2 + y2 ] v = √[ vx2 + vy2 ] r = √[ (+1.776 × 1014 m)2 + (−1.639 × 1014 m)2 ] v = √[ (1.5 × 107 m/s)2 + (−2.770 × 107 m/s)2 ] r = 2.42 × 1014 m v = 3.15 × 107 m/s tan θ = y = −1.639 × 1014 m x +1.776 × 1014 m tan θ = vy = −2.770 × 107 m/s vx 1.5 × 107 m/s θ = −42.7° θ = −61.6° r = 2.42 × 1014 m at −42.7° v = 3.15 × 107 m/s at −61.6°
[magnify]
[magnify]
[magnify] - An alien spacecraft accidentally flies into a plasma cloud (a collection of ionized
gas). This disrupts the ship's guidance system, which makes the velocity varying
according to the following parametric equations.
To make the calculations simpler for us humans, the aliens have adapted this problem to our standards. The equations use m/s for velocity, seconds for time, and radians for angular measure. In addition, the initial coordinates of the ship were (0,0); that is, the ship started acting this way when it was located at the origin.vx = 0.4 − cos (t / 100) vy = 0.0 + sin (t / 100)
Determine …- the displacement as a function of time,
- the path of the ship for the first 2000 s,
- the direction of the ship at the beginning and end of this interval, and
- the maximum and minimum speed of the ship.
Solutions …
- This is a problem that requires calculus. By definition, velocity is
the first derivative of displacement with respect to time. From the fundamental
theorem of calculus then, displacement is the time integral of velocity.
All that remains is to adjust the constants for the initial conditions: x0 = y0 = 0.x = ∫ vx dt y = ∫ vy dt x = ∫(0.4 − cos (t / 100)) dt y = ∫(0.0 + sin (t / 100)) dt x = 0.4t − 100 sin (t / 100) + cx y =− 100 cos (t / 100) + cy
Therefore0 = 0.4(0) − 100 sin (0 / 100) + cx 0 = −100 cos (0 / 100) + cy 0 = cx 0 = −100 + cy cx = 0 cy = 100 x = 0.4t − 100 sin (t / 100) y = 100 − 100 cos (t / 100) - The ship is moving forward in the x direction, as the first term shows,
but also oscillating, as the second term shows. The coefficient of the
second term is much larger than the first, so we can be quite certain
that there will be times when the ship will reverse direction, but overall
the motion will be forward. It's sort of a "two steps forward, one
step back" kind of kind of shuffle.
At the same time, the ship is oscillating in the y direction. The constant term is basically unimportant. The ship starts at a particular location and then dodges side to side without ever making any headway in this direction.
Combining these two motions results in a path that looks something like a doodle from a bored physics student's notebook. Don't try to sketch it. Let a computer or graphing calculator draw it for you.
- To determine the direction of motion, evaluate the velocity equations
at the appropriate times.
At the beginning of the interval, the velocity was entirely in the negative x direction.vx(t) = 0.4 − cos (t / 100) vy(t) = 0.0 + sin (t / 100) vx(0) = 0.4 − cos (0 / 100) vy(0) = 0.0 + sin (0 / 100) vx(0) = 0.4 − 1.0 vy(0) = 0.0 + 0.0 vx(0) = − 0.6 m/s vy(0) = 0.0 m/s
At the end of the interval, the velocity was entirely in the positive x direction.vx(t) = 0.4 − cos (t / 100) vy(t) = 0.0 + sin (t / 100) vx(2000) = 0.4 − cos (2000 / 100) vy(2000) = 0.0 + sin (2000 / 100) vx(2000) = 0.4 − 0.4 vy(2000) = 0.0 + 0.9 vx(2000) = 0.0 m/s vy(2000) = + 0.9 m/s - Speed is the magnitude of velocity and is found by adding the components
using the Pythagorean theorem.
The cosine term is what matters here. It has its maximum value of +1 when t/100 is 0, 2π, 4π and other even multiples of π and its minimum value of −1 when t / 100 is π, 3π, 5π and other odd multiples of π. To determine the range of the ship's speed, simply substitute ±1 for the cosine term.v2 = (vx)2 + (vy)2 v2 = (0.4 − cos (t / 100))2 + (0.0 + sin (t / 100))2 v2 = 0.16 − 0.8 cos (t / 100) + cos2 (t / 100) + sin2 (t / 100) v2 = 0.16 − 0.8 cos (t / 100) + 1.00 v2 = 1.16 − 0.8 cos (t / 100) v2 = 1.16 − 0.8 cos (t / 100) vmin = √(1.16 −1.00) = √ 0.16 = 0.40 m/s vmax = √(1.16 +1.00) = √ 2.16 = 1.47 m/s
- The parametric equations below are used for generating an interesting family of curves called lissajous figures.
Where …x = A sin (at + φ) y = B sin (bt) - x and y are coordinates on a plane,
- t is the parameter,
- A and B are amplitudes,
- a and b are angular frequencies, and
- φ is the phase angle
Since sine is a circular function, the range of parameter values should be thought of in terms of laps around the unit circle. We will use radians for all angular measures, so be sure your calculator mode or computer preferences are set appropriately.−1.5 < x < +1.5 −1.5 < y < +1.5 - Let the amplitudes and angular frequencies equal one (A = B = a = b = 1).
Set the parameter range to 0 < t < 2π with
a reasonable size increment (something that your calculator or computer
can complete in under thirty seconds). Draw the lissajous figures for
various phase angles. Try simple fractions of a complete circle like
φ = 0, ⅙π, ¼π, ½π, ⅔π, 1π, 1½π.
- What is the domain of x and y in all cases?
- What effect does phase angle have on the lissajous figure?
- Let the amplitudes equal one (A = B = 1).
Let the phase angle equal a quarter lap around the unit circle (φ = ½π).
Draw the lissajous figures for various angular frequencies.You will need to increase the maximum parameter
value to 4π, 6π, 20π or higher depending on your choice of frequencies.
- Try different whole number values to start with and do not choose the same value for each frequency (a, b ∈ ℤ+ and a ≠ b). Start with small numbers like 1, 2, 3, 4, 5. Try larger numbers if you have the patience. Be sure to vary each frequency. How is the appearance of the lissajous figure affected by your choice of angular frequencies?
- Set one of the frequencies to 1 and the other to an irrational number like √2 or π. How is the behavior of this lissajous figure different from those of all your previous trials?
- Answer it.
- Write something completely different.
- Answer it.
numerical
- The parametric equations below are used for generating an interesting family
of curves that are informally called spirograph
curves in honor of the mechanical drawing toy first manufactured by
the Kenner Products toy company in 1965.
hypocycloid epicycloid x = (A − B) cos t + B cos ⎛
⎝A − B t ⎞
⎠B y = (A − B) sin t − B sin ⎛
⎝A − B t ⎞
⎠B x = (A + B) cos t − B cos ⎛
⎝A + B t ⎞
⎠B y = (A + B) sin t − B sin ⎛
⎝A + B t ⎞
⎠B - Some simple parametric curves to try. The symbols are as follows: x and y are coordinates, a and b are constants, and t is the parameter.
a. circle x = a cos t y = a sin t b. ellipse x = a cos t y = b sin t c. cycloid x = at − b sin t y = a − b cos t d. deltoid x = 2a cos t + a cos 2t y = 2a sin t − a sin 2t e. astroid x = a cos3 t y = a sin3 t f. nephroid x = ½ a (3 cos t − cos 3t) y = ½ a (3 sin t − sin 3t) g. folium of descartes x = (3at) / (1 + t3) y = (3at2) / (1 + t3) h. involute of circle x = a cos t + at sin t y = a sin t − at cos t i. serpentine x = a cot t y = b (sin t) (cos t) j. witch of agnesi x = a cot t y = b sin2 t k. pusuit curve, tractrix x = t − a tanh (t / a) y = a sech (t / a)
Resources
- no resources for this topic
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