Chapter 10
10.2 Plane curves and Parametric Equations
**Now we will start to study situations in which 3 variables are used to represent a curve in a
plane.
Ex. the path followed by an object that is propelled into the air at an angle of 45⁰
--If the initial velocity of the object is 48 feet per second, the object travels the parabolic path
y=
−𝑥 2
72
+ x (rectangular equation)
**this equation tells you: where the object has been but it doesn’t tell you when the object
was at a given point (x,y)
**to do this, you introduce a 3rd variable, t, called a PARAMETER
a) By writing both x and y as functions of t, you obtain PARAMETRIC EQUATIONS
b)
−𝑋 2
72
+x
 Y = -16y2 + 24√2 t
x 24√2 t
**From this set of equations you can determine that at a given time, (t=___) you know
the object is at a given point: (___, ____)
**this also shows that x and y are continuous functions of t, and the resulting path is a
Plane Curve
Definition of a Plane Curve: If f and g are continuous functions of t on the interval, I, then the
equations x = f(t) and y = g(t) are called parametric equations and t is called a parameter.
**The set of points (x,y) obtained as t varies over the interval, I, is called the graph of the
parameter equations Taken together, the parametric equations and the graph are called a
plane curve denoted by C.
Orientation of the Curve: when sketching a curve represented by a set of parametric
equations, you can plot points in the xy plane  By plotting the resulting points in order of
increasing values of t, the curve is traced out in a specific direction.
Ex. Sketch the curve described by x = t2 – 4
𝑡
y=2
-2≤ t ≤ 3
**Because graphs would not pass vertical line test  it does not define y as a function of x
#1. BENEFIT OF PARAMETRIC EQUATION = can be used to represent graphs that are more
general than graphs of functions
#2. Different parametric representations can be used to represent various speeds at which
objects travel along a given path.
ELIMINATING THE PARAMETER – when you find a rectangular equation that represents the
graph of a set of parametric equations.
Ex.
Parametric Equation Solve for t in 1 eq.--> Substitute into the 2nd eq.  Rect. Eq.
  
X = t2 – 4
𝑡

Y=2
  
x = (2y)2 – 4
x = 4y2 - 4
t = 2y
**The range of x and y implied by the parametric equations may be altered by the change to
rectangular form  if so the DOMAIN of rectangular equations must be adjusted so that its
graph matches the graph of the parametric equations.
Ex. Sketch the curve represented by the equations:
X=
1
and y =
√𝑡+1
𝑡
𝑡+1
t > -1
by eliminating the parameter and adjusting the domain.
Step 1: Solve 1 of the parametric equations for t:
1
(x)2 = (√𝑡+1)2  x2 =
1
𝑡+1
1
1
 flip so end up with t + 1 = 𝑥 2  t = 𝑥 2 - 1  t =
1−𝑥 2
𝑥2
Step 2: Now Substitute into parametric equation for y
Y=
1−𝑥2
𝑥2
1− 𝑥2
+1
𝑥2
= y = 1-x2 - which is defined for all values of x , but from the parametric
equation for x you can see the curve is defined only when t > -1
**So you restrict the domain of x to positive values
** The parameter – doesn’t have to represent time – next example if represents an angle
Ex. Sketch the curve represented by x = 3 cosθ y = 4sinθ 0≤ θ≤ 2π by eliminating the
parameter and finding rectangular equation
#1. Solve for sin and cos. In both:
𝑥
Cosθ = 3
𝑦
sinθ = 4
major =8
𝑥
𝑦

Use the identity: (3)2 + ( 4)2 = 1
𝑥2
9
𝑦2
+ 16 = 1 (ellipse centered at (0,0) (0,4)(0,-4) minor 6
Graph in Form of 1) x = h+ acosθ
counterclockwise
y= k + bsin θ
Graphs in form: 1) x = h+ asinθ
y = k + bcosθ
0≤ θ≤ 2π
0≤ θ≤ 2π
is an ellipse traced
is an ellipse traced clockwise
**PURPOSE IN ELIMINATING PARAMETER: aid to curve sketching
** Need parametric equations to: 1) tell you position 2) direction 3) speed at a given time
Finding Parametric Equations:
Ex. find a set of parametric equations to represent the graph of y = 1- x2, using each of the
parameters a. t = x
𝑑𝑦
b. the slope m= 𝑑𝑥 at the point (x,y)
a) Letting x = t produces:
y = 1 – t2
and x = t
𝑑𝑦
𝑚
b) To write x and y in terms of m: m = 𝑑𝑥 = −2𝑥  x = −2 is parametric eq. for x
𝑚
c) For y  y = 1 – (−2)2  y = 1 -
𝑚2
4
Parametric equations for a cycloid:
Ex. Determine the curve traced by a point P, on the circumference of a circle of radius a, rolling
along a straight line in a plane (cycloid)
#1. Let the parameter θ = the measure of the circle’s rotation. Let point P = (x,y) begin at
origin
When θ = 0 P = (0,0)
When θ = π P is at a maximum point (πa, 2a)
angle APC = 180 – θ
When θ = 2π P is back on x-axis at (2πa, 0) so sinθ = sin(180-θ) = sin APC =
𝐴𝐶
𝑎
=
𝐵𝐷
𝑎
𝐴𝐷
Cosθ = -cos(180 – θ) = -cos angle APC = −𝑎
Implies: AP = -acosθ and BD = asinθ
Parametric equations are: x = a(θ – sinθ)
y = a(1-cosθ)
X’(θ) = a – acosθ
y’(θ) = asinθ
x’(2π) = 0
y’(2π) = 0
Between these points the cycloid is called smooth.
DEFINITION OF A SMOOTH CURVE: a curve c represented by x = f(t) and y = G(t) on an
interval, I, is called smooth if f’ and g’ are continuous on I and not simultaneously 0, except
possibly at the endpoints of I
Curve C is piecewise sm0oth = if it is smooth on each subinterval of some partition of I
10.3 Parametric Equations and Calculus
Theorem 10.7 Parametric Form of the Derivative: If a smooth curve
C is given by the equations x= f(t) and y = g(t). Then the slope of C at
(x,y) is
𝒅𝒚
𝒅𝒙
=
𝒅𝒚/𝒅𝒕
𝒅𝒚
𝒅𝒙/𝒅𝒕
𝒅𝒙
≠ 𝟎
Ex. Find dy/dx for the curve given by: x = sint and y=cost
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
=
− sin 𝑡
cos 𝑡
= - tant
For Higher Derivatives:
𝒅𝟐 𝒚
𝒅𝒙𝟐
=
𝒅
𝒅𝒚
[ ]=
𝒅𝒙 𝒅𝒙
𝒅 𝒅𝒚
[ ]
𝒅𝒕 𝒅𝒙
𝒅𝒙
𝒅𝒕
Ex. Finding Slope and Concavity
For the curve given by: x = √t and y = ¼(t2 – 4) t ≥ 0 find the slope
and concavity at (2,3)
Step #1
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
You could find the
=
−1
1
2
𝑡
2
1
𝑡
2
2nd
= t3/2
derivative to be
𝑑2 𝑦
𝑑𝑥 2
=
𝑑 𝑑𝑦
[ ]
𝑑𝑡 𝑑𝑥
𝑑𝑥
𝑑𝑡
=
3
𝑑
2]
[𝑡
𝑑𝑡
1 −1/2
𝑡
2
=
3 1/2
𝑡
2
1 −1/2
𝑡
2
=
3t
At (x,y) = (2,3) it follows that:
3 = ¼(t2-4)  12 = t2 – 4  16 = t2  t = 4
and slope is
𝑑𝑦
𝑑𝑥
= 43/2
=8
2nd Derivative: 3(4) = 12
12>0 so concave up at (2,3)
Ex. A Prolate Cycloid given by x = 2t- πsin t
and y = 2 –π cost t
crosses itself a the point (0,2). Find the equation of both tangent lines
at this point.
Solution: Because x = 0 and y = 2 when t = ± π/2 and
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
=
𝜋𝑠𝑖𝑛𝑡
2− 𝜋𝑐𝑜𝑠𝑡
( 2 = 2-πcost  0 = -πcost  0
When x =0 and y=2 when t = ±π/2
= cos t  ±𝜋/2
You have
𝑑𝑦
𝑑𝑥
=
−𝜋
2
when t =
−𝜋
2
and
𝑑𝑦
𝑑𝑥
=
𝜋
2
when t =
𝜋
2
𝜋
𝜋
2
2
** So the 2 tangent lines at (0,2) are y – 2 = - - (x) and y -2 = (x)
--**If dy/dt = 0 and dx/dt ≠ 0 when t = to  the curve represented by
x = f(t) and y = g(t) has a horizontal tangent at (f(to), g(to)) ---- in
above example there is a horizontal tan. At (0,2-π) when t = 0
--** If dx/dt = 0 and dy/dt≠ 0 when t = to, the curve represented by x
= f(t) and y = g(t) has a vertical tangent at (f(to), g(to))
ARC LENGTH - parametric equations can be used to describe the path
of a particle moving in the plane.  is also used to determine THE
DISTANCE traveled by the particle along its path
Theorem 10.8 Arc Length in Parametric Form:
If a smooth curve C is given by x = f(t) and y = g(t) such that C does not
intersect itself on the interval a≤ t ≤ b (except possibly at endpoints)
then the arc length of C over the interval is given by:
𝒃
𝒅𝒙
𝒅𝒚
𝒃
S = ∫𝒂 √( )𝟐 + ( )𝟐 dt = ∫𝒂 √[𝒇′ (𝒕)]𝟐 + [𝒈′ (𝒕)]𝟐 𝒅𝒕
𝒅𝒕
𝒅𝒕
(When applying formula  make sure that the curve is traced out only
once on the interval of integration)
EPICYCLOID = if a circles’ path rolls around the circumference of
another circle
Ex. Finding Arc Length
A circle of radius 1 rolls around the circumference of a larger circle or
radius 4 as shown below. The epicycloid traced by a point on the
circumference of the smaller circle is given by:
X = 5cost – cos5t
and y = 5sint – sin5t
Find the distance traveled by the point in 1 complete trip about the
larger circle
Notice: curve has sharp points when t – 0 and t =
 Between these 2 points
𝑑𝑥
𝑑𝑡
and
𝑑𝑥
𝑑𝑡
𝜋
2
are not simultaneously
0
𝜋
 So the position of the curve generated from t = 0 to t = is
2
smooth
Now to find total distance traveled by the point  you find the
arc length of that position in the 1st Quadrant and multiply by 4
𝜋/2
S = 4 ∫0
𝜋/2
4∫0
𝑑𝑥
𝑑𝑦
𝑑𝑡
𝑑𝑡
√( )2 + ( )2 dt
=
√(−5𝑠𝑖𝑛𝑡 + 5𝑠𝑖𝑛5𝑡)2 + (5𝑐𝑜𝑠𝑡 − 5 cos 5𝑡)2 𝑑𝑡
𝜋/2
= 20 ∫0
√2 − 2𝑠𝑖𝑛𝑡𝑠𝑖𝑛5𝑡 − 2𝑐𝑜𝑠𝑡𝑐𝑜𝑠5𝑡 𝑑𝑡
𝜋/2
= 20 ∫0
𝜋/2
= 20 ∫0
𝜋/2
= 40 ∫0
√2 − 2𝑐𝑜𝑠4𝑡 𝑑𝑡
√4𝑠𝑖𝑛3 2𝑡 𝑑𝑡
𝑠𝑖𝑛2𝑡 𝑑𝑡
= -20[cos 2t] evaluated at
𝜋
2
𝑎𝑛𝑑 0 = 40
Do Length of Recording Tape Ex.
Area of a Surface of Revolution - If a smooth curve C given by x = f(t)
and y = g(t) does not cross itself on an interval a≤ t ≤ b, then the area S
of the surface of revolution formed by revolving C about the coordinate
axes is given by the following:
𝒃
𝒅𝒙
𝒅𝒚
𝒅𝒕
𝒅𝒕
1. S = 2π∫𝒂 𝒈(𝒕)√( )𝟐 + ( )𝟐 dt
REVOLUTION ABOUT THE
X-AXIS g(t) ≥ 0
𝒃
𝒅𝒙
𝒅𝒚
𝒅𝒕
𝒅𝒕
2. S= 2π∫𝒂 𝒇(𝒕)√( )𝟐 + ( )𝟐 dt
REVOLUTION ABOUT
THE Y-AXIS f(t) ≥ 0
3 3√3
Ex. Let C be the arc of the circle x2 + y2 = 9 from (3,0) to ( ,
2
2
). Find
the are of the surface formed by revolving C about the x-axis.  You
can represent C parametrically by the equations x = 3cost y = 3sint 0
≤ t ≤ π/3
𝜋/3
S = 2π∫0
3𝑠𝑖𝑛𝑡√(−3𝑠𝑖𝑛𝑡)2 + (3𝑐𝑜𝑠𝑡)2 𝑑𝑡
𝜋/3
= 6𝜋 ∫0
𝜋/3
= 6𝜋 ∫0
𝑠𝑖𝑛𝑡√9(𝑠𝑖𝑛2 𝑡 + 𝑐𝑜𝑠 2 𝑡)𝑑𝑡
3𝑠𝑖𝑛𝑡 𝑑𝑡
= -18π[cost] evaluated at π/3 to 0
= -18π(1/2 – 1) = 9𝜋
10.4 Polar Coordinates and Polar Graphs
To form polar coordinate system:
1. Fix a point O  called POLE (or origin)
2. Construct from O an initial ray called  POLAR AXIS
3. Each point P, in plane can be assigned polar coordinates (r,θ)
r = directed distance from 0 to P
θ = directed angle, counterclockwise from polar axis to OP
*with polar (r,θ) and (r, 2π+θ) represent the same point
* since r is directed distance, then (r,θ) and (-r,θ+π) represent same
point
***So (r,θ) can be written as: = (r, θ + 2nπ) or (-r, θ + (2n+1)π)
Coordinate Conversion:
Theorem 10.10 The polar coordinates (r, θ) of a point are related to the
rectangular coordinates (x,y) of the point as follows:
X = rcosθ
y = rsinθ
tanθ =
𝒙
𝒚
r2 = x2 + y2 r = √𝒙𝟐 + 𝒚𝟐
Ex. From Polar to Rectangular (2,π)
X = 2cosπ
y = 2sinπ
2(0) = 0  (-2,0)
2(-1) = -2
Ex. Rectangular to Polar (-1,1)
R = √(−1)2 + (1)2 = √2
So (√2 ,
3𝜋
4
tan =
1
−1
= -1  tan-1(-1) =
3𝜋
4
)
Polar Graphs
Ex. Describe the graph of each polar equation. Confirm each
description by converting to a rectangular equation
a. r = 2
b. θ=
𝜋
3
c. r= secθ
a) R = 2 consists of all points that are 2 units from the pole (rect.
Equation = 22 = x2 + y2
b) θ=
𝜋
3
consists of all the points on the line that makes an angle of
𝜋
3
with a positive x – axis
c) r = secθ must be converted to rectangular form: rcosθ = 1 x = 1
Ex. Sketch the graph of r=2cos3θ
**write polar equation in parameter mode:
X = 2cos3θcosθ
y = 2cos3θsinθ
or make a table of values:
Θ
r
0
2
π/6
0
π/3
-2
π/2
0
3π/2
2
Letting θ vary from 0 to π gives you a rose curve
ROSE CURVE – of form r = acosnθ and r = asinnθ
Slope and Tangent Lines  find by using parametric equations
X = rcosθ = f(θ)cosθ
section
Theorem 10.11
and y = rsinθ = f(θ)sinθ then use dy/dx from last
𝒅𝒚
𝒅𝒙
=
𝒅𝒚/𝒅𝜽
𝒅𝒙/𝒅𝜽
=
𝒇(𝜽)𝒄𝒐𝒔𝜽 +𝒇′ (𝜽)𝒔𝒊𝒏𝜽
** if f is
−𝒇(𝜽)𝒔𝒊𝒏𝜽+𝒇′ (𝜽)𝒄𝒐𝒔𝜽
differentiable function of θ then the slope of the tangent line to the
graph of r=f(θ) at point (r,θ) is given by above formula
** 1. Solutions to
** 2. Solutions to
** If
𝑑𝑦
𝑑𝜃
and
𝑑𝑥
𝑑𝜃
𝒅𝒚
𝒅𝜽
𝒅𝒙
𝒅𝜽
= 0 yield Horizontal tangents provided that
= 0 yields Vertical tangents provided that
𝑑𝑦
𝑑𝜃
𝑑𝑥
𝑑𝜃
≠0
≠0
are simultaneously 0, no conclusion can be drawn about
tangent lines.
Ex. Find the Horizontal and Vertical tangent lines of r=sinθ 0≤θ≤π
Step 1: write in parametric form x = rcosθ = sinθcosθ
And y = rsinθ = sinθsinθ = sin²θ
Step 2: differentiate x and y with respect to θ and set each derivative=0
𝑑𝑥
𝑑𝜃
𝑑𝑦
𝑑𝜃
𝜋 3𝜋
= cos2θ – sin2θ = cos2θ (double ange identity) = 0 θ = ,
4
= 2sinθcosθ = sin2θ (double angle identity) = 0
√2 𝜋
2 4
θ = 0,
4
𝜋
2
√2 3𝜋
)
2 4
So graph has vertical tangents at ( , ) and ( ,
𝜋
And horizontal tangents at (0,0) and (1, )
2
Ex. find the horizontal and vertical tangents to the graph of r=2(1-cosθ)
y = rsinθ = 2(1-cosθ)sinθ
= 2[(1-cosθ)(cosθ+sinθ(sinθ))] = -2(2cosθ+1)(cosθ -1) = 0
Cosθ = ½ cosθ = 1
So
𝑑𝑦
𝑑𝜃
= 0 when θ =
2𝜋 4𝜋
3
3
and 0
X = rcosθ = 2cosθ – 2cos²θ
𝑑𝑥
𝑑𝜃
= -2sinθ + 4cosθsinθ = 2sinθ(2cosθ – 1) = 0
Sinθ = 0 or cos θ = ½
So
𝑑𝑥
𝑑𝜃
= 0 when θ =
𝜋 5𝜋
3
3
and 0, π
Symmetry Test for Polar Graphs:
1. Symmetry about x-axis: if point (r,θ) lies on the graph, the point
(r,-θ) or (-r,π-θ) lies on the graph
2. Symmetry about y-axis: if point (r,θ) lies on the graph, the point
(r,π-θ) or (-r,-θ) lies on the graph
3. Symmetry about the origin: If point (r,θ) lies on the graph, the
point (-r,θ) or (r, θ+π) lies on the graph.
Ex. identify the symmetries of the curves:
a) R = 1+cosθ
b) R = 2 + sinθ
Slopes of Polar Curves
#17 ex. cardiod r = -1 + cosθ
θ=±
𝜋
2
𝜋
𝜋
2
2
so (-1, ) or (-1, - )
10.5 Area and Arc Length in Polar Coordinates:
**using sectors of a circle instead of rectangles
Area of a Sector: = ½θr2
**graph is bound by radial lines θ=α and θ=β
1) To find area of the region, partition the interval [α,β] into n equal
subintervals
2) Then approximate the area of the region by the sum of the areas
of the n sectors
**all of this leads to the proof of the following theorem:
Theorem 10.13 Area in Polar Coordinates:
If f is continuous and nonnegative on the interval [α,β] and
0< β-α≤2π then the area of the region bounded by the graph of
R=f(θ) between the radial lines θ=α and θ=β is given by:
𝛽
𝛽
A = ½∫𝛼 [𝑓(𝜃0]2 𝑑𝜃  ½ ∫𝛼 𝑟 2 𝑑𝜃
Ex. Find the area of 1 petal of the rose curve given by r = 3cos3θ
 Notice right petal is traced as θ increases from –π/6 to π/6 so area
is:
𝜋/6
=½∫−𝜋/6(3𝑐𝑜𝑠3𝜃)2 𝑑𝜃
Ex. Find the area of the region lying between the inner and outer loops
of the limoncon r = 1- 2sinθ
5𝜋/6
=½∫𝜋/6 (1 − 2𝑠𝑖𝑛𝜃)2 dθ
Points of Intersection of Polar Graphs – sketch the graphs b/c may
miss a point. To find, just set equations equal to each other or
substitute one equation in for another.
Ex. Find the area of the region common to the 2 regions bounded by
the following:
r=6cosθ
r=2-2cosθ
Arc Length in Polar Form:
Theorem 10.14 Arc Length of Polar Curve
Let f be a function whose derivative is continuous on an interval α≤θ≤β.
The length of the graph of r = f(θ) from θ=α to θ=β is
𝜷
S = ∫∝ √[𝒇(𝜽)]𝟐 + [𝒇′ (𝜽)]𝟐
𝜷
𝒅𝒓
S = ∫∝ √𝒓𝟐 + [ ]𝟐 𝒅𝜽
𝒅𝜽
Ex. Find the length of the arc from θ = 0 to θ = 2π for the cardiod
r=f(θ) = 2-2cosθ  f’(θ) = 2sinθ
Area of a Surface of Revolution – use parametric equation x=rcosθ
y = rsinθ
Theorem 10.15 Area of a Surface of Revolution: Let f be a function
whose derivative is continuous on an interval α≤θ≤β. The area of the
surface formed by revolving the graph of r=f(θ) from θ=α to θ=β is
1.
𝜷
S = 𝟐𝝅 ∫∝ 𝒇(𝜽)𝒔𝒊𝒏𝜽√[𝒇(𝜽)]𝟐 + [𝒇′ (𝜽)]𝟐 dθ about the polar
axis
𝜷
2. S = 𝟐𝝅 ∫∝ 𝒇(𝜽)𝒄𝒐𝒔𝜽√[𝒇(𝜽)]𝟐 + [𝒇′ (𝜽)]𝟐 dθ about the line θ=
Ex. Find the area of the surface formed by revolving the circle
r=f(θ)=cosθ about the line
𝜋
2
𝝅
𝟐