Chapter 4

Parametric Equations and Polar

Coordinates
Chapter Outline
 Plane Curves and Parametric Equations.
 Calculus and Parametric Equations.
 Polar Coordinates.
 Calculus and Polar Coordinates.
Plane Curves
Plane curve is described by giving
y as a function of x [y=f(x)] or
x as a function of y [x=g(y)] or
a relation between x & y that defines y
implicitly as a function of x [f(x,y)=0].
Describing Curves: There are 2 methods for
describing curves.
(i) Parametric Equations.
(ii) Polar Coordinate system.
 Curves defined by Parametric Equations:
When the path of a particle moving in the plane is not
the graph of a function, we cannot describe it using a
formula that express y directly in terms of x.
Imagine that a particle moves along the curve C as
shown below

It is impossible to describe C by an equation of the

form y=f(x) because C fails the vertical line test.
Vertical Line Test: It says that a curve in the xy-plane is
the graph of a function of x iff no vertical line
intersects the curve more than once.
So, the above curve C is a set of ordered pairs
(x(t),y(t)) where x & y are now thought of a functions
of parameter t with a≤t≤ b
The equations
x=f(t), y=g(t) are called parametric equations.
Parametric Curve: As t varies, the point (x,y)=(f(t),g(t))
varies and traces out a curve C, which is Parametric
Curve.
The equation of the Parametric curve can be
represented in x and y by eliminating the Parameter
t.
The initial point of Parametric curve is (f(a), g(a))
The terminal point of Parametric curve is (f(b), g(b))
 Example:
Sketch and identify the curve defined by the parametric
equations x=t2-2t, y=t+1.
Solution:

A particle whose position is given by the parametric equations

moves along the curve in the direction of the arrows as t
increases.
Eliminate the parameter t as follows
t=y-1 (from y=t+1)
Substitute t in equation of x, x=t2-2t= (y-1)2-2(y-1)
=y2-2y+1-2y-2= y2-4y+3
So, the curve is x= y2-4y+3
Calculus and Parametric Equations:
Apply the methods of Calculus like tangent, area, surface area
to the Parametric curves.
Tangents with Parametric Equations:
We know that the equation of tangent line at coordinate point
(x1,y1) is given by

So, for Parametric equations x=f(t) & y=g(t), which can also be
expressed, by eliminating the parameter, as y=F(x)--------(1)
Substitute x=f(t) & y=g(t) in (1) g(t)=F(f(t))
Differentiate above, g’(t)=F’(f(t)).f’(t) (By chain Rule)
=F’(x)f’(t)
If f’(t)=0, then
So, Slope of tangent to the curve y=F(x) at (x,F(x))is
if dx/dt =0.
Horizontal Tangent: The Curve will have Horizontal Tangent if
and only if
Vertical Tangent: The Curve will have Vertical Tangent if and
only if
Example:
Find equation of Tangent line to the Parametric Equations
x=1+4t-t2, y=2-t3 at t=1.
Solution: At t=1, x=1+4(1)-(1)2= 4
y= 2-(1)3=1
The coordinate point on the line is (4,1).
Then find

So, the equation of line is

(y-1)=(-3/2)(x-4)
y= (-3/2)x+7.
Polar Coordinates:
Introduced by Newton.
Choose a point in the plane that is called the pole( or origin) & is
labeled O.
Then draw a ray starting at O called the Polar-axis. This axis is
usually drawn horizontally to the right & corresponds to the
positive x-axis in the Cartesian Coordinates.

If P is any point in the plane, let r be the distance

From O to P & let θ be the angle between the Polar axis & the
line OP.
Then point P is represented by the ordered pair (r, θ) & r, θ are
called Polar coordinates of P.
Angle is positive if measured in the counter clockwise direction
from the polar axis & negative in the clockwise direction.
If P=0 then r=0 & (0, θ) represents the pole for any θ.
Polar Coordinates:
In Figure the point (-r,θ) & (r,θ) lie on
the same line through O & at the
same distance IrI from O, but on
opposite sides of O.
If r>0, the point (r,θ) lies in the same quadrant as θ. If r<0, it lies
in the quadrant on the opposite side of Pole.
Example: Plot the points whose Polar coordinates are given
(a) (1, 5π/4) (b) (2, 3π)
Solution:
The points are plotted as follows
Connection between Polar & Cartesian Coordinates:
We know that Pole Corresponds to the origin & Polar axis
coincides with the positive x-axis.
If the point P has Cartesian coordinates (x,y) & polar coordinates
(r, θ), then we have
y-axis
cosθ= x/r P(x,y)
sinθ=y/r r
y
θ
O x x-axis
Converting Polar coordinates to Cartesian coordinates:
If r and θ are known is given by
x= rcosθ , y=r sinθ
Converting Cartesian coordinates to Polar coordinates:
If x & y are given then
r2=x2+y2, tanθ=y/x.
Example:Convert the point (2,π/3) from Polar to Cartesian
coordinates.
Solution: Since r=2 & θ= π/3
We know,
x=rcosθ= 2cos(π/3)= 2(1/2)=1
y= rsinθ=2sin(π/3)= 2(√3/2)=√3
Therefore the point in Cartesian coordinates is (1, √3).
Example: Represent the point with Cartesian coordinates (1,-1)
in Polar coordinates.
Solution: If we choose to be positive, then
r2= x2 + y2 = 12 +(-1)2 =2, r= √2
tanθ=y/x=-1
Since the point (1,-1) lies 4th quadrant, θ= -π/4 or θ= 7π/4.
Thus Polar coordinates are (√2, -π/4) or (√2, 7π/4)
Calculus & Polar Coordinates:
Apply the methods of Calculus like tangent, area, surface area
to the Polar curves .
Tangents to the Polar Curves:
We know that the equation of tangent line at coordinate point
(x1,y1) is given by

For Polar Curve r=f(θ),we can get

x=f(θ)cosθ, y=f(θ)sinθ &
Example: What is the slope of the Polar curve r=2cos4θ at any
angle θ?
Solution: r’= -8sin4θ

Example: What is the equation of the tangent line to the Polar

curve r=cos2 θ, when θ =π/4?
Solution: r’=-2sin2θ, so

At θ =π/4, x=rcosθ =cos2θcosθ=cos2π/4cosπ/4=0

y=rsinθ =cos2θsinθ=cos2π/4sinπ/4=0
Equation of Tangent is (y-0)=1(x-0), y=x.