     Calculus Animations,Graphics and Lecture Notes Parametric Equations 2-space    Home About Kelly Liakos Sponsors Calculus 1- Limits and Derivatives Calculus 1 - The Second Derivative Calculus 1 and 3 Formula Sheets Chain Rule Computer Lab Assignments Conic Sections Differential Equations Directional Derivatives/Gradient Double Integrals Equillibrium Solutions 1st order DEs 1st order Diff Eqs -Motion Flux Integrals and Surface Integrals Infinite Sequences Infinie Series Level Curves and Level Surfaces Line Integrals Optimization and Related Rates Optimization for Functions of 2 Variables Parametric Equations 2-space Parametric Equations 3-space Partial Derivatives Polar Coordinate System Polar Coordinates- Derivatives and Integrals PreCalculus Riemann Sums and the Fundamental Theorem of Calculus 2d order Diff EQS-Motion 2d Partial Derivatives Supplemental Exercises and Solutions Tangent Planes/ Differential for f(x,y) Trigonometry Trigonometry Applications Triple Integrals Unit Tangent Vectors/Unit Normal Vectors Vectors in General Vector Valued Functions Visualizing Limits of Functions of 2 Variables Work Links Special Topics Scratch Paper 2 Poems  Parametric Equations 2-Space The idea of parametric equations is to describe a curve in the plane by using 2 equationsx = x(t) and y = y(t) instead of describing a curve by y = f (x). The advantage is that If we describe a curve by y = f(x) we have the path of a particle but have no idea of direction along the path or speed.In the first 3 animations we see how a curve is generated from the parametric equationsx(t) = t + 1  and y(t) = t2  simply by plotting points over ever shorter time intervals.Obviously we develop a better method than plotting points shortly. One Second Intervals Half Second Intervals Intervals of 1/10 of a second So how do we graph a curve defined parametrically? There are 3 basoc steps1. Eliminate t to obtain the path. In our previous example x= t+1  y= t2  it follows t = x-1 and therefore y = (x-1)2. So the particle is restricted to this parabola.(Note If you are using Mathcad you would simply put x(t) on the horizontal and y(t) on the vertical eliminating this step altogether)2. Locate the initial point--In our eg x(0) = 1 and y(0) = 03. Analyze the derivatives over smooth intervals to determine all subsequent motion.What do we mean by smooth? Pretty much what intuitively we would think it is,If you're driving down the road and don't hit any red lights or bounce off a tree you have a more or less a smooth trip.Smoothness:1.  If x(t) and y(t) are differentiable over an interval and not simultaneously 0 the trajectory is smooth. (This is the no stop light case)2. The trajectory is not smooth at any pt where either x(t) or y(t) is not differentiable. (In practice hopefully you don't bounce off of a tree instantaneously changing direction).The importance of smoothness is that if the trajectory is smooth then the orientation along the path doesn't change so we can determine subsequent motion from the consideration of the derivatives initially.In our example x ' (t) = 1 therefore x is always increasing, Further y ' (t) = 2t >0 for t>0 therforey is always increasing. Therefore the particle starts at (1,0) and moves up and to the right along the parabola for all time. Example 1 Trajectory Example 2 Circular motionLet x(t) = cos(t)   y(t) = sin(t)1. it is easy to verify x2 + y2 = 1. Therefore the paricle travels on a circle of radius 1.2. The initial pt is (1,0)3. x '(t) = -sin(t) and y' (t) =cos(t) so the derivatives are not simultaneously 0Just after t =0 x' <0 and y' >0 So since we have smoothness the partice travels counter clockwise for all time.Note the direction the particle travels changes at each pt but the orientation does not.Note the movement of the particles and the derivatives in the following animation Circular Trajectory

Be Careful -- you may wonder why we just don't consider the position of the particle at 2 times and deduce subsequent motion from that.

For the same reason wagon wheels turn backward in old movies.

Consider the following animation in which circular motion viewed from different time intervals shows in one case the particle moving counterclockwise and in another it appears to be moving counterclockwise.

Consideration of the instantaneous rates of change eliminates this paradox

A non smooth example

Consider x(t) =cos2(t) and y(t) =cos(t).

1. By inspection we see y = x2.

2. Initially the particle is at (1,1)

3. x'(t) = -2cos(t)sin(t) = -sin(2t)   y' (t) =-sin(t)

We see the derivatives are simultaneously 0 ar integer multiple of π

We'll analyze the derivatives on the intervals 0 to π and then π to 2π.

Just after t = 0 the derivaties <0 so the particle is moving to the left and down. At π/2 the particle starts moving up but is still moving to the left. At t =π the particle comes to a stop and just after π starts moving down and to the right returning to (1,1) at t=2π.
Noe at 3π/2 the particle starts moving up but the motion is still to the right.

This motion then repaeats over every 2π interval.

Non Smooth Example-Animation

Here's one for you to try x(t) =sin(t)  y(t) =sin(t)

An Example with pts of non-differentiability

Let x(t) = |cos(t)|  and y(t) = sin(t)

We still have circular motion,however at odd multiples of π/2 x '(t) does not exist and instaneously changes signs,therefore the particle "bounces off the wall" at these times

Nondifferentiable-Animation

Speed

Consider the 2 sets of parametric eqns :

x(t) = t   y(t) = t    and  x(t) =et - 1 and y(t) = et - 1

1. Both travel on y = x
2. Both start at (0,0)
3. Both are smooth and represent a particle moving up and to the right.

The difference is speed

The best way to see this is in terms of vectors
let  r(t) = x(t) i + y(t) j then v(t) = x'(t)i +y'(t)j and since the speed ds/dt  = ||v(t)|| it follows the speed is
√(x'2 +y'2)

Therefore for our first particle the speed ds/dt = √2 and for the second is et√2

It follows the distance traveled is then ∫√(x'2 +y'2)dt integrated from t1 to t2 .

See animation below

Same Path Different Speeds

In this example we'll consider a planet-moon system.

Suppose a planet orbits a star every 365 days at a distance of 9.3 million miles.

Then x(t) = 9.3cos(2π/365t)   y(t) = 9.3sin(2π/365t)

At the same time a moon orbits the planet every 28 days at a distance of 2.5 million miles

Then  x(t) = 9.3cos(2π/365t) + 2.5cos(2π/28t)         y(t) = 9.3sin(2π/365) + 2.5sin(2π/28t)

See the animation below

EarthMoon

In this example we consider a mosquito walking on a rotating disk.

In the first animation the mosquito approaches the center at a constant rate.

In the second animation the mosquito has become disoriented. Leaving the center

It alternates between  moving toward the edge and back to the center.

Notes- Mosquito on a Rotating Disk

Animation - Mosquito on a Disk

Animation - Disoriented Mosquito on a Disk

A cycloid is the curve traced out by a fixed point on a circle as the circle rolls along the x-axis. The following notes and animation develops and demonstraes the parametric equations of the cycloid.

Notes - The Cycloid

Animation - The Cycloid

A hypocycloid is the curve traced out by a fixed point on a circle that rotates inside a larger circle--you can think of taking the cycloid in the previous example and folding  a segment of the x-axis  into a circle

The following are animations of various hypocycloids.

For an excellent discussion see the work of Nick Whitman:

Animation - Hypocycloid Ratio of 3 to 1

Animation - Hypocycloid - Ratio of 5 to 1

Animation - Hypocycloid - Ratio of 11 to 2.7

The following links are to the pages on supplemental exercises for parametric equations in 2 space and a discussion of parametric equations in 3 space respectively

The Smart Bunny-A very short story by Kurt Vonnegut Jr.  