1
9.1—Intro to Parametric & Vector Calculus
Parametric Equations and Curves
In Algebra, equations are graphed in two variables, 𝑥 and 𝑦. There are many situations in which both, 𝑥 and 𝑦, depend
independently on a third variable, 𝑡 or 𝜃. There are a great many curves that we can’t even write down as a single equation in
terms of only 𝑥 and 𝑦. The third variable, 𝑡 or 𝜃 is called the parameter, and the separate equation
are called parametric equations.
In those cases instead of defining 𝑦 in terms of 𝑥 (𝑦 = 𝑦(𝑥)) we define both 𝑥 and 𝑦 in terms of 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.
If possible, Methods 1 and 2 could be used to get explicit relationship 𝑦 = 𝑦(𝑥) from parametric equations.
If it is not possible to get analytically explicit relationship = 𝑦(𝑥) , we use calculator (Method 3) to find point by point
(𝑥, 𝑦) = (𝑓(𝑡), 𝑔(𝑡)) for each value of 𝑡.
Method 1: 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
into x to give: x = ¼ y2 + y + ¾
t = ½ (y+1)
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
Method 2: Using trigonometry to eliminate a parameter
example 1: 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 process) we do this:
𝑥 2 𝑦2
+
= 𝑠𝑖𝑛2 𝑡 + 𝑐𝑜𝑠 2 𝑡 = 1
25
4
𝑒𝑙𝑖𝑝𝑠𝑒:
𝑥 2 𝑦2
+
=1
25
4
for direction of motion we have to go back to parameter 𝑡:
t
0
π/2
π
3π/2
2π
x
5
0
-5
0
5
y
0
2
0
–2
0
example 2: draw the parametric curve for the following set of parametric equations. Clearly indicate direction of motion.
2
x = 5 cos 3t
y = 2 sin 3t
0 ≤ t ≤ 2π
𝑥 2 𝑦2
+
= 𝑠𝑖𝑛2 (3𝑡) + 𝑐𝑜𝑠 2 (3𝑡) = 1
25
4
𝑒𝑙𝑖𝑝𝑠𝑒:
we replaced the t with 3t.
𝑥 2 𝑦2
+
=1
25
4
difference between example 1. and 2.:
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
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
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
parametric equations
x = a  - b sin 
y = a – b cos 
point
has
HYPOCYCLOID WITH FOUR CUSPS
Equation in rectangular coordinates: x2/3 + y2/3 = a2/3
Equations in parametric form: x = a cos3 
y = a sin3 
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/analyticgeometry/geometry5/special-plane-curves.html
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PARAMETRIC FORM OF THE DERIVATIVE
If 𝑥(𝑡) and 𝑦(𝑡) are differentiable functions of 𝑡, then the derivatives 𝑑𝑥/𝑑𝑡 and 𝑑𝑦/𝑑𝑡 measure the rates of change of 𝑥
and 𝑦 with respect to 𝑡: 𝑑𝑥/𝑑𝑡 and 𝑑𝑦/𝑑𝑡 tell how fast each variable is changing with respect to 𝑡.
The derivative 𝑑𝑦/𝑑𝑥 is the slope of the line tangent to the parametric graph (𝑥(𝑡), 𝑦(𝑡)).
To calculate 𝑑𝑦/𝑑𝑥 we need to use the Chain Rule:
𝑑𝑦
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
= 𝑑𝑡
𝑑𝑥
𝑑𝑡 𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑝𝑟𝑜𝑣𝑖𝑑𝑖𝑛𝑔 𝑡ℎ𝑎𝑡
𝑑𝑥
≠0
𝑑𝑡
If 𝑑𝑦/𝑑𝑥 is also differentiable function of 𝑡, then the second derivative is given by
𝑑 𝑑𝑦
[ ]
𝑑2𝑦
𝑑 𝑑𝑦
𝑑 𝑑𝑦 𝑑𝑡
𝑑𝑡 𝑑𝑥
=
[
]
=
[
]
=
𝑑𝑥
𝑑𝑥 2
𝑑𝑥 𝑑𝑥
𝑑𝑡 𝑑𝑥 𝑑𝑥
𝑑𝑡
keep in mind that you already found 𝑑𝑦/𝑑𝑥 as a
function of 𝑡 not 𝑥, so
𝑑2 𝑦
𝑑𝑥 2
is a function of 𝑡.
𝑻𝒂𝒏𝒈𝒆𝒏𝒕 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 𝒕𝟏 ∶ 1. 𝑓𝑖𝑛𝑑 𝑥1 = 𝑥(𝑡1 ) 𝑎𝑛𝑑 𝑦1 = 𝑦(𝑡1 )
𝑑𝑦
𝑑𝑦
2. 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒
= 𝑑𝑡
𝑑𝑥 𝑑𝑥
𝑑𝑡
𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑡1
3. 𝑢𝑠𝑒 𝑦 – 𝑦1 = 𝑦’(𝑥 – 𝑥1 )
𝑪𝒐𝒏𝒄𝒂𝒗𝒊𝒕𝒚 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 𝒕𝟏 : 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑒
𝐼𝑓
𝑑2𝑦
𝑎𝑡 𝑡1 .
𝑑𝑥 2
𝑑2 𝑦
𝑑2𝑦
< 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑑𝑜𝑤𝑛. 𝐼𝑓
> 0 → 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑒 𝑢𝑝
2
𝑑𝑥
𝑑𝑥 2
𝑯𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 𝒍𝒊𝒏𝒆 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜:
𝑑𝑦
𝑑𝑦
=0 →
=0
𝑑𝑥
𝑑𝑡
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝑑𝑥
≠0
𝑑𝑡
𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒂𝒔𝒚𝒎𝒑𝒕𝒐𝒕𝒆𝒔 𝑤𝑖𝑙𝑙 𝑜𝑐𝑐𝑢𝑟 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑:
𝑑𝑦
𝑑𝑥
=∞ →
=0
𝑑𝑥
𝑑𝑡
𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑
𝑑𝑦
≠0
𝑑𝑡
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
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𝑇ℎ𝑒 𝒂𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 𝑜𝑓 𝑦 = 𝑦(𝑥) 𝑖𝑛 𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑝𝑙𝑎𝑛𝑒 𝑖𝑛 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒐𝒓𝒎:
2
2
𝑏
𝑏
𝑡2
𝑑𝑥
𝑑𝑦
𝑆 = ∫ 𝑑𝑠 = ∫ √(𝑑𝑥)2 + (𝑑𝑦)2 = ∫ √( 𝑑𝑡) + ( 𝑑𝑡)
𝑑𝑡
𝑑𝑡
𝑎
𝑎
𝑡1
𝑏
𝑡2
𝑆 = ∫ 𝑑𝑠 = ∫ √(
𝑎
𝑡1
𝑑𝑥 2
𝑑𝑦 2
) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑡1 𝑎𝑛𝑑 𝑡2 𝑎𝑟𝑒 𝑡𝑖𝑚𝑒𝑠 𝑐𝑜𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛𝑠 𝑎 𝑎𝑛𝑑 𝑏
𝑇ℎ𝑒 𝒂𝒓𝒆𝒂 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝑏𝑦 𝑦 = 𝑦(𝑥) 𝑎𝑛𝑑 𝑥 𝑎𝑥𝑖𝑠 𝑖𝑛 𝐶𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 𝑝𝑙𝑎𝑛𝑒 𝑖𝑛 𝒑𝒂𝒓𝒂𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒐𝒓𝒎:
The area enclosed by function y = y(x) and x-axis in Cartesian plane in parametric form:
𝑏
𝑏
𝐴 = ∫ 𝑑𝐴 = ∫ 𝑦(𝑥)𝑑𝑥
𝑎
𝑎 ≤ x ≤ b
𝑎
𝛽
𝐴 = ∫ 𝑦(𝑡)
𝛼
𝑑𝑥
𝑑𝑡
𝑑𝑡
α ≤ t ≤ β
Example 1: The location of an object is given by the parametric equations x(t) = t3 + 1 feet and y(t) = t2 + t feet at time t seconds.
(a) Evaluate x(t) and y(t) at t = –2, –1, 0, 1, and 2, and then graph the path of the object for –2 ≤ t ≤ 2.
(b) Evaluate dy/dx for t = –2, –1, 0, 1, and 2. Do your calculated values for dy/dx agree with the shape of your graph in part (a)?
(c) Find the equation of the line tangent to the graph of the parametric equations when t = 3.
a)
b)
t
–2
–1
0
1
2
𝑑𝑦
𝑑𝑡
⟹
y
2
0
0
2
6
= 2𝑡 + 1 and
t
–2
–1
0
1
2
c)
x
–7
0
1
2
9
𝑑𝑥
𝑑𝑡
𝑑𝑦
= 3𝑡 2 so
𝑑𝑥
=
2𝑡+1
3𝑡 2
.
dy/dx
–1/4
–1/3
∞
1
5/12
x(t) = t3 + 1
when t = 3
tangent:
y(t) = t2 + t
𝑥 = 28
y – y1 = y’(x – x1)
𝑑𝑦
𝑑𝑡
𝑑𝑥
= 2𝑡 + 1
𝑑𝑦
𝑦 = 12
y – 12 =
𝑑𝑡
7
27
𝑑𝑡
= 7
(x-28)
𝑑𝑦
= 3𝑡 2
𝑑𝑥
𝑑𝑡
𝑑𝑥
=
= 27
⟹
7
296
⟹ y =
12
x+
27
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
7
27
5
Particle Motion
Velocity and Acceleration Component VECTORS, Speed and Direction of Motion
Suppose a particle moves along a smooth curve in the plane so that its position at any time t is ⟨𝑥(𝑡), 𝑦(𝑡)⟩ ,
where x and y are differentiable functions of t.
•
•
𝒅𝒙
is velocity of a particle in the horizontal direction.
𝒅𝒕
𝒅𝒚
𝒅𝒕
is velocity of a particle in the vertical direction.
⃗ = ⟨𝒙(𝒕), 𝒚(𝒕)⟩ ≡ (𝒙(𝒕), 𝒚(𝒕)) is the particle’s position vector at time 𝑡
•𝒔
𝒅𝒙
⟨
⃗ =
•𝒗
𝒅𝒕
,
𝒅𝒚
𝒅𝒕
⟩
≡ (𝒙′(𝒕), 𝒚′(𝒕)) is the velocity vector at time 𝑡
⃗ = ⟨𝒙′′(𝒕), 𝒚′′(𝒕)⟩ ≡ (𝒙′′(𝒕), 𝒚′′(𝒕)) is the acceleration vector at time 𝑡
•𝒂
𝟐
𝟐
⃗ | ≡ ‖𝒗
⃗‖=
• |𝒗
√(𝒅𝒙) + (𝒅𝒚)
⃗ | ≡ ‖𝒂
⃗ ‖=
• |𝒂
√(𝒅 𝟐𝒙) + (𝒅 𝟐𝒚)
𝒅𝒕
𝒅𝒕
𝒅𝒕
𝟐
•
⃗
𝒗
𝒅𝒕
𝟐
𝟐
is the speed of a particle or magnitude/ length/ norm of the velocity vector.
𝟐
is the acceleration of a particle or magnitude/ length/ norm of acceleration.
is the direction vector representing the particle’s direction of motion
|𝑣
⃗|
Displacement & Distance Traveled
⃗ = ⟨𝒙′ (𝒕), 𝒚′ (𝒕)⟩
Suppose a particle moves along a path in the plane so that its velocity at any time 𝑡 is 𝒗
𝒕𝟐
•
𝒕𝟐
∫ 𝒙′(𝒕) 𝒅𝒕, ∫ 𝒚′(𝒕) 𝒅𝒕
𝒕𝟏
vector is the displacement from t1 to t2
𝒕𝟏
𝒕𝟐
𝑛𝑒𝑤 𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏:
⟨𝑥, 𝑦⟩ = ⟨𝑥0 , 𝑦0 ⟩ +
∫ 𝒙′(𝒕) 𝒅𝒕, ∫ 𝒚′(𝒕) 𝒅𝒕
𝒕𝟏
𝑡2
𝒕𝟐
𝒕𝟏
𝑡2
𝑡2
𝑑𝑥 2
𝑑𝑦 2
• ∫ |𝑣 (𝑡)|𝑑𝑡 = ∫ √(𝑣𝑥 )2 + (𝑣𝑦 )2 𝑑𝑡 = ∫ √( ) + ( ) 𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑡1
𝑡1
𝑡1
is the distance traveled from t1 to t2