1
Parametric Equations and Curves
There are a great many curves that we can’t even write down as a single equation in terms of only x and y.
So, to deal with some of these problems we introduce parametric equations. Instead of defining y in terms of x
(y = y(x)) we define both x and y in terms of a third variable called a parameter as
x = f(t)
y = g(t)
Each value of t defines a point
(x, y) = (f(t), g(t))
The collection of points that we get by letting t be all possible values is the graph of the parametric equations and is called the
parametric curve. Parametric curves have a direction of motion. The direction of motion is given by increasing t. So, when
plotting parametric curves we also include arrows that show the direction of motion.
Sketching a parametric curve is not always an easy thing to do.
1. method: Elimination of the parameter from the set of the equations and substitution:
example: draw the graph: x = t2 + t
y = 2t – 1
–1≤t≤ 1
t = ½ (y+1)
into x to give: x = ¼ y2 + y + ¾
this is a parabola that opens to the right.
t=–1
x=0
y=–3
t=1
x=2
y=1
direction of motion (arrows): from t = – 1 to t = 1
2. method: Using trigonometry to eliminate a parameter
example: draw the parametric curve for the following set of parametric equations. Clearly indicate direction of motion.
x = 5 cos t
y = 2 sin t
0 ≤ t ≤ 2π
Instead solving one of these equations for t and plugging this into the other (very ugly) we do this:
𝑥2
25
+
𝑦2
4
= sin2 t + cos2 t = 1
𝑒𝑙𝑖𝑝𝑠𝑒:
for direction of motion we have to go back to parameter t:
𝑥2
25
+
𝑦2
4
=1
t
0
π/2
π
3π/2
2π
x
5
0
-5
0
5
y
0
2
0
–2
0
example: draw the parametric curve for the following set of parametric equations. Clearly indicate direction of motion.
x = 5 cos 3t
y = 2 sin 3t
0 ≤ t ≤ 2π
we replaced the t with 3t.
𝑒𝑙𝑖𝑝𝑠𝑒:
As we’re going around faster, while we have the same ellipse that
we got in the previous example we’ll trace out the curve three
times, each in the following ranges: 0 ≤ t ≤ 2π/3, 2π/3 ≤ t ≤ 4π/3
and 4π/3 ≤ t ≤ 2π
circle: x = r cos t
y = r sin t
𝑥2
25
+
𝑦2
4
=1
2
3. method : there are some equations that you can’t do anything else but use calculator
– as I have only TI – 84 + (silver edition) , I know only this one:
TI – 84 plus
CLEAR RAM follow these key strokes: "2nd", "+", "7","1","2"
Graphing parametric equations
graph the parametric equations
x = 5cos3t, y = 5sin3t
1.
2.
3.
Hit the MODE key.
Arrow down to where it says Func and then use the right arrow to choose Par.
Hit ENTER The calculator is now in parametric equations mode.
4.
5.
6.
7.
8.
9.
10.
Hit the Y= key.
In the X1T slot, type 5(cos(
Hit X,T,q,n key: you get 5(cos(T
Hit two closing parentheses, )) to get: 5(cos(T)) .
Hit MATH 3 to get cubed. You should now have X1T=5(cos(T))3.
Hit the down arrow to go to Y1T and in that space type 5(sin(T))3
Finally, hit GRAPH.
If you did everything right, the result looks a bit like a diamond, with all four edges bowed inward.
You can also specify the T values that the calculator begins and ends with; for instance, you may limit the graph to
0<T<3. This would not change the viewing window, but it would only draw part of the graph. When you change T,
hit GRAPH again.
SPECIAL PLANE CURVES
CYCLOID A famous curve that was named by Galileo in 1599. This is a curve traced by a point P on a circle of radius a
rolling along x axis..
Equations in parametric form: x = a ( - sin )
y = a(1 – cos )
Area of one arch = 3πa2
Arc length of one arch = 8a
PROLATE CYCLOID
The path traced out by a fixed point at a radius b > a, where a is the radius of a rolling
circle, also sometimes called an extended cycloid. The prolate cycloid contains loops,
and has parametric equations
x = a  - b sin 
y = a – b cos 
HYPOCYCLOID WITH FOUR CUSPS
Equation in rectangular coordinates: x2/3 + y2/3 = a2/3
Equations in parametric form: x = a cos3 
y = a sin2 
Area bounded by curve = 3πa2/8
Arc length of entire curve = 6a
This is a curve described by a point P on a circle of radius a/4 as it rolls on the inside of a circle of radius a.
vast number of plane curves in rectangular, parametric and polar form:
http://www.math10.com/en/geometry/analytic-geometry/geometry5/special-plane-curves.html
3
PARAMETRIC FORM OF THE DERIVATIVE
If x and y are both differentiable functions of t and dx/dt ≠ 0, then
𝒅𝒚
𝒅𝒙
=
𝒅𝒚
𝒅𝒕
𝒅𝒙
𝒅𝒕
chain rule dy/dx =
𝑑𝑦 𝑑𝑡
𝑑𝑡 𝑑𝑥
=
𝑑𝑦 1
𝑑𝑡 𝑑𝑥
y’ =
𝒚̇
𝒙̇
y’’ =
𝒚′̇
𝒙̇
𝑑𝑡
If y’ = dy/dx is also differentiable function of t, then
𝒅𝟐 𝒚
𝒅𝟐 𝒙
=
𝒅
𝒅𝒙
𝒅𝒚
(𝒅𝒙)
=
𝒅
𝒅𝒕
𝒅𝒚 𝑑𝑡
(𝒅𝒙) 𝑑𝑥
=
𝒅 𝒅𝒚
( )
𝒅𝒕 𝒅𝒙
𝑑𝑥
𝑑𝑡
Tangent at point t1 : 1. find x1 = x(t1) and y1 = y(t1)
=
𝒅𝒚′/𝒅𝒕
𝒅𝒙/𝒅𝒕
2. find slope
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
at point t1
3. use y – y1 = m(x – x1)
Concavity at point t1: Find second derivate d2y/dx2 at t1. If second derivate is negative parametric curve will be concave down,
and for positive second derivate parametric curve will be concave up.
Horizontal tangents will occur where the derivative is zero and that means that we’ll get horizontal tangent at values of t for
which we have
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑡
=0 →
=0
provided
𝑑𝑥
𝑑𝑡
≠0
Vertical tangents will occur where the derivative is not defined and so we’ll get vertical tangents at values of t for which we
have,
𝑑𝑦
𝑑𝑥
𝑑𝑥
𝑑𝑡
=∞ →
=0
provided
𝑑𝑦
𝑑𝑡
≠0
Don’t be surprised if you get two tangents lines at a point. A quick graph of the parametric curve will explain what is going on
here.
If the parametric curve crosses itself there will
be more than one tangent line. There is one
tangent line for each instance that the curve
goes through the point where parametric curve
crosses itself
The arc length of functions in Cartesian plane:
𝑏
𝑏
𝑏
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √1 + (𝑦′)2 𝑑𝑥
The arc length of parametric form is :
𝑏
𝑏
𝑏
S = ∫𝑎 𝑑𝑠 = ∫𝑎 √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫𝑎 √(𝑥̇ )2 + (𝑦̇ )2 𝑑𝑡
S=