5 parametric equations, tangents and curve lengths in polar coordinates

Parametric Equations
We may describe the motion of a point in the plane by
it's coordinate (x, y) at time t as (x(t), y(t)) where
x(t) and y(t) are functions in t.

The equations x(t), y(t) are called parametric equations
and the variable t is called the parameter.

Example A. a. Plot the path of the parametric equations
x(t) = √t , y(t) = t – 4 from t = 0 to t = 9.

x(t) = √t , y(t) = t – 4 from t = 0 to t = 9.
We make a table
by selecting
values for t and
plot the resulting
(x, y)’s.

x(t) = √t , y(t) = t – 4 from t = 0 to t = 9.
We make a table t x y
by selecting 0 0 -4
values for t and 1 1 -3
plot the resulting 4 2 0
(x, y)’s. 6 √6 2
9 3 5

x(t) = √t , y(t) = t – 4 from t = 0 to t = 9.
t = 9, (3,5)

We make a table t x y
by selecting 0 0 -4
t = 6, (√6,2)
values for t and 1 1 -3
plot the resulting 4 2 0 t = 4, (2,0)

(x, y)’s. 6 √6 2 t = 1, (1,-3)
9 3 5 t = 0, (0,-4)

It’s possible to find the x&y equation for the parametric
path if we can solve for t in terms of the x or the y.

b. Find the x&y equation given by the
parametric equations x(t) = √t , y(t) = t – 4

Since x(t) = √t,
we've t = x2.

Since x(t) = √t,
we've t = x2.
Hence y = x2 – 4 is the
x&y equation of the
curve.

t = 9, (3,5)
Since x(t) = √t,
we've t = x2.
Hence y = x2 – 4 is the t = 6, (√6,2)

x&y equation of the t = 4, (2,0)
curve.
t = 1, (1,-3)
In general, the parametric
t = 0, (0,-4)
equations do not generate x(t) = √t , y(t) = t – 4
the entire x&y graph.

The parameter t does not have to represent time but it
helps to view it as such.

Example B. a. Graph and interpret the parametric
equations x(θ) = –3c(θ), y(θ) = 3s(θ).

Make a table θ x y

and plot a few 0 –3 0
points. π/2 0 3
π 3 0

Make a table θ x y
(–3c(θ), 3sθ))
points. π/2 0 3
θ
π 3 0

Make a table θ x y
(–3c(θ), 3sθ))
points. π/2 0 3
θ
Note that x2 + y2 π 3 0
= (–3c(θ))2 + (3s(θ))2
= 9c2(θ) + 9s2(θ) = 9
or x2 + y2 = 9.

Make a table θ x y
(–3c(θ), 3sθ))
points. π/2 0 3
θ
Note that x2 + y2 π 3 0
= (–3c(θ))2 + (3s(θ))2
= 9c2(θ) + 9s2(θ) = 9
or x2 + y2 = 9. So the path is the
circle with r = 3. The point starts at the initial position
(θ = 0) at (–3, 0) circling once every 2π clockwisely.

Remark: We also use the word
“parameterization” to describe curves
that change according to some
specific coefficient such as r = k*c(θ)
which are circles parametrized by k.

“parameterization” to describe curves k
which are circles parametrized by k. Circles r = k*c(θ)
parametrized by k

parametrized by k
Parametrization of x&y Curves

parametrized by k
Given a function y = f(x) and its graph, we may put it
into the "standard“ parametric form as x = t, y = f(t).

parametrized by k
Hence for the equation y = x2 x(t) = t
the standard parametric equations is y(t) = t2

parametrized by k
Hence for the equation y = x2 x(t) = t
Another set of parametric
equations that has the path as y = x is
2 x(t) = t3
y(t) = t6

parametrized by k
Hence for the equation y = x2. x(t) = t
Another set of parametric
equations that has the path as y = x is
2 x(t) = t3
y(t) = t6
The two different parameterizations
represent two different motions on the same parabola.

Parametrization of Polar Curves.
Given the polar function r = f(θ),

y r = f(θ)

r
θ
x

Given the polar function r = f(θ), a point on the curve
with polar coordinate (r = f(θ), θ) corresponds to the
rectangular coordinate (x, y) where
x = r*c (θ)
y = r*s (θ)
y r = f(θ)

r
θ
x

x = r*c (θ) x(θ) = f(θ)c(θ)
y = r*s (θ) or y(θ) = f(θ)s(θ)
(r=f(θ), θ) or y r = f(θ)
(x=r*c(θ), y=r*s(θ))

r
θ
x

x = r*c (θ) x(θ) = f(θ)c(θ)
This is the standard (r=f(θ), θ) or y r = f(θ)
parametrization of the polar (x=r*c(θ), y=r*s(θ))

curve r = f(θ) with θ as the parameter.
r
θ
x

x = r*c (θ) x(θ) = f(θ)c(θ)

Example C. Parametrize the polar r
θ
function r = 1 – sin(θ).
x

x = r*c (θ) x(θ) = f(θ)c(θ)

Example C. Parametrize the polar r
θ
function r = 1 – sin(θ).
x
The standard parametrization is
x = r*c (θ) x(θ) = (1 – s(θ))c(θ)
or
y(θ) = (1 – s(θ))s(θ)

Tangent Lines for Parametric Curves

Given the differentiable parametric equations
x = x(t) dy dy/dt
y = y(t) then the derivative at t0 is dx = dx/dt
t=t
0 t = t0

x = x(t) dy dy/dt
t=t
0 t = t0

Example D. Given x(t) = t3, y(t) = t2. Find the slope
of the tangent at the (–8, 4).

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt
dx = dx/dt

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt 2t 2
dx = dx/dt = 3t2 = 3t

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt 2t 2
dx = dx/dt = 3t2 = 3t
The corresponding t for the
point (–8, 4) satisfies
x(t) = t3 = –8 so t = –2.

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt 2t 2
dx = dx/dt = 3t2 = 3t
x(t) = t3 = –8 so t = –2.
Hence the slope at (–8, 4) is dy =
dx
t = –2

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt 2t 2
dx = dx/dt = 3t2 = 3t
x(t) = t3 = –8 so t = –2.
Hence the slope at (–8, 4) is dy = 2 = –1
dx 3
3t
t = –2

x = x(t) dy dy/dt
t=t
0 t = t0

dy dy/dt 2t 2
dx = dx/dt = 3t2 = 3t (–8, 4)

point (–8, 4) satisfies –1
slope =
x(t) = t = –8 so t = –2.
3 3

Hence the slope at (–8, 4) is dy = 2 = –1
dx 3
3t
t = –2

Arc Length for Parametric Curves

Given the parametric equations x = x(t), y = y(t)
from t = a to t = b where x'(t) and y'(t) are continuous,
t =b

then the arc length is ∫
t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

Example E. Given x(t) = t3/3, y = t2/2 find the arc length
from t = 0 to t =1. Find an x&y equation and draw the
segment of the curve in question.

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

We have x'(t) = t2, y'(t) = t,

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

We have x'(t) = t2, y'(t) = t, hence the arc length is
t=1 t=1 t=1

∫
t =0
(t 2 ) 2 + (t ) 2 dt = ∫
t =0
t 4 + t 2 dt = ∫c
t t 2 +1dt
t =0

1 2 1 23 / 2 1
= (t +1) 3 / 2 | = −
3 0c 3 3

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

t=1 t=1 t=1

∫
t =0
(t 2 ) 2 + (t ) 2 dt = ∫
t =0
t 4 + t 2 dt = ∫ t t 2 +1dt
t =0

1 2 1 23 / 2 1
= (t +1) 3 / 2 | =c −
3 0 3 3

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

t=1 t=1 t=1

∫
t =0
(t 2 ) 2 + (t ) 2 dt = ∫
t =0
t 4 + t 2 dt = ∫ t t 2 +1dt
t =0

1 2 1 23 / 2 1
= (t +1) 3 / 2 | =c − Substitution Method
3 0 3 3 Set u = t2+1

t =b

t =a
( x ' (t )) 2 +( y ' (t )) 2 dt

t=1 t=1 t=1

∫
t =0
(t 2 ) 2 + (t ) 2 dt = ∫
t =0
t 4 + t 2 dt = ∫ t t 2 +1dt
t =0

1 2 1 23 / 2 1
= (t +1) 3 / 2 | = − Substitution Method
3 0 3 3 Set u = t2+1

To find an x&y equation, solve for t in terms of x(t)
then plug in the results into y(t).

3
x = t /3
3
3x = t or t = √3x
3

3
x = t /3
3
3x = t or t = √3x
3
3
y = t2/2 y = √9x2
2

3
x = t /3
3
3x = t or t = √3x
3
3
y = t /2
2
y = √9x2 3
√9x2
y= 2
2
As t goes from 0 to 1, (1/3,1/2) , t = 1
we obtain the curve
segment from (0,0), t = 0
(0,0) to (1, 1).

Tangents and Arc Lengths in Polar
Equations
Given the polar function r = f(θ), it may be realized as
the parametric equations:
x(θ) = f(θ)c(θ)
y(θ) = f(θ)s(θ).

Equations
x(θ) = f(θ)c(θ)
y(θ) = f(θ)s(θ).
and the derivative dy for the parametric equations is
dx
dy dy/dθ
=
dx dx/dθ

Equations
x(θ) = f(θ)c(θ)
y(θ) = f(θ)s(θ).
dx
dy dy/dθ
=
dx dx/dθ
Hence for r = f(θ), the slope at the point (r =f(θ), θ) is
dy dy/dθ d(f(θ)s(θ))/dθ
dx = dx/dθ = d(f(θ)c(θ))/dθ

Equations
x(θ) = f(θ)c(θ)
y(θ) = f(θ)s(θ).
dx
dy dy/dθ
=
dx dx/dθ
Hence for r = f(θ), the slope at the point (r =f(θ), θ) is
dy dy/dθ d(f(θ)s(θ))/dθ
dx = dx/dθ = d(f(θ)c(θ))/dθ
Expand using the product rule, we have that
dy f '(θ)s(θ) + f(θ)c(θ)
dx = f '(θ)c(θ) – f(θ)s(θ)

Equations
Example F. Find the slope at r = cos(2θ) at θ = π/6.

Equations
In the parametric form, we have
x(θ) = c(2θ)c(θ) and y(θ) = c(2θ)s(θ),

Equations
x(θ) = c(2θ)c(θ) and y(θ) = c(2θ)s(θ), hence
dy dy/dθ
dx = dx/dθ =

Equations
dy dy/dθ –2s(2θ)s(θ) + c(2θ)c(θ)
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)

Equations
dy dy/dθ –2s(2θ)s(θ) + c(2θ)c(θ) at θ = π/6,
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)
dy –2s(π/3)s(π/6) + c(π/3)c(π/6)
dx = –2s(π/3)c(π/6) – c(π/3)s(π/6)

Equations
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)
dy –2s(π/3)s(π/6) + c(π/3)c(π/6)
dx = –2s(π/3)c(π/6) – c(π/3)s(π/6)
–√3*½ + ½*√3/2
= –√3*√3/2 – ½*½

Equations
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)
dy –2s(π/3)s(π/6) + c(π/3)c(π/6)
dx = –2s(π/3)c(π/6) – c(π/3)s(π/6)
–√3*½ + ½*√3/2
= –√3*√3/2 – ½*½
–2√3 + √3
= –6–1

= √3
7

Equations
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)
dy –2s(π/3)s(π/6) + c(π/3)c(π/6)
dx = –2s(π/3)c(π/6) – c(π/3)s(π/6)
–√3*½ + ½*√3/2
= –√3*√3/2 – ½*½
(1/2,π/6)
P

–2√3 + √3
= –6–1

= √3
7

Equations
dx = dx/dθ = –2s(2θ)c(θ) – c(2θ)s(θ)
dy –2s(π/3)s(π/6) + c(π/3)c(π/6)
dx = –2s(π/3)c(π/6) – c(π/3)s(π/6)
–√3*½ + ½*√3/2 slope = √3
7
= –√3*√3/2 – ½*½
(1/2,π/6)P

–2√3 + √3
= –6–1

= √3
7

Equations
Example G. Find where the slope is 0 for r = 1+ cos(θ).

Equations
We have x(θ) = (1+c(θ))c(θ) and y(θ) = (1+c(θ))s(θ).

Equations
dy dy/dθ dy
We want dx = dx/dθ = 0, or dθ = 0

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0
–s2 + c + c2 = 0

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0
–s2 + c + c2 = 0
–(1 – c2) + c + c2 = 0
2c2 + c – 1 = 0
(2c – 1)(c + 1) = 0

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0
–s2 + c + c2 = 0
–(1 – c2) + c + c2 = 0
2c2 + c – 1 = 0
(2c – 1)(c + 1) = 0
So c(θ) = ½  θ = ±π/3, c(θ) = –1 θ = –π

Equations
dy dy/dθ dy
dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0
–s2 + c + c2 = 0
–(1 – c2) + c + c2 = 0
2c2 + c – 1 = 0
(2c – 1)(c + 1) = 0
So c(θ) = ½  θ = ±π/3, c(θ) = –1 θ = –π
But dx/dθ = 0 (check this!) so that r = 1 + c(θ) is not
differentiable at –π. Hence the slope is 0 at θ = ±π/3.

Equations
dy dy/dθ dy
We want dx = dx/dθ = 0, or dθ = 0 (1+√3/2, π/3)

dy
Set dθ = f '(θ)s(θ) + f(θ)c(θ)= 0
–s(θ)s(θ) +(1+ c(θ))c(θ) = 0
–s2 + c + c2 = 0
–(1 – c2) + c + c2 = 0
2c2 + c – 1 = 0
(1+√3/2, –π/3)
(2c – 1)(c + 1) = 0
So c(θ) = ½  θ = ±π/3, c(θ) = –1 θ = –π
But dx/dθ = 0 (check this!) so that r = 1 + c(θ) is not
differentiable at –π. Hence the slope is 0 at θ = ±π/3.

Equations
Given the polar equation r = f(θ) in parametric form
x = r*cos(θ), y = r*sin(θ), we may easily verify that
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
Hence the arc-length from θ = a to θ = b is:
θ =b θ =b

∫ ( dx / dθ ) + (dy / dθ )) dθ = ∫ r 2 + ( r ' (θ )) 2 dθ
2 2

θ=a θ=a

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
θ =b θ =b

2 2

θ=a θ=a

Example H. Let r = 2sin(θ). Find the arc-length for
θ from 0 to 2π.

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
θ =b θ =b

2 2

θ=a θ=a

θ from 0 to 2π.
r2 + (r')2 = (2s(θ))2 + (2c(θ))2

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
θ =b θ =b

2 2

θ=a θ=a

θ from 0 to 2π.
r2 + (r')2 = (2s(θ))2 + (2c(θ))2 = 4s2(θ)+4c2(θ) = 4.

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
θ =b θ =b

2 2

θ=a θ=a

θ from 0 to 2π.
r2 + (r')2 = (2s(θ))2 + (2c(θ))2 = 4s2(θ)+4c2(θ) = 4.
Hence the arc-length is
θ =2π θ =2π θ =2π

∫ r + ( r ' (θ )) dθ = ∫ 4dθ = ∫ 2dθ = 4π
2 2

θ =0 θ =0 θ =0

Equations
(dx/dθ)2 + (dy/dθ)2 = r2 + (dr/dθ)2.
θ =b θ =b

2 2

θ=a θ=a

θ from 0 to 2π.
r2 + (r')2 = (2s(θ))2 + (2c(θ))2 = 4s2(θ)+4c2(θ) = 4.
Hence the arc-length is r = 2sin(θ)
θ =2π θ =2π θ =2π
r =1
∫ r + ( r ' (θ )) dθ = ∫ 4dθ = ∫ 2dθ = 4π
2 2

θ =0 θ =0 θ =0

We get the circumference of two circles since the
graphs traced out two circles as θ goes from 0 to 2π.

Equations
Example I. Find the arc-length of r = 1 – cos(θ).

Equations
The graph is a cardioid

r = 1 – cos(θ)

Equations
The graph is a cardioid and
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2

r = 1 – cos(θ)

Equations
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
= 2 – 2c.
r = 1 – cos(θ)

Equations
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
= 2 – 2c. We can't integrate
√2 – 2c = √2 √1 – c(θ) as is.
r = 1 – cos(θ)

Equations
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
√2 – 2c = √2 √1 – c(θ) as is. Using the
r = 1 – cos(θ)
double angle formula c(2A) = 1 – 2s2(A),
we write c(θ) = 1 – 2s2(θ/2).

Equations
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
r = 1 – cos(θ)
double angle formula c(2A) = 1 – 2s (A),
2

we write c(θ) = 1 – 2s2(θ/2). By symmetry, we may
double the integral from 0 to π to find answer which is:

Equations
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
r = 1 – cos(θ)
2

θ =b θ =π θ =π
2
θ
∫=a
r 2 + (r ' (θ )) 2 dθ = 2
θ
∫
=0
2(1 − cos(θ ) )dθ = 2
θ
∫
=0
2 * 2 sin 2 ( θ ) dθ
2

θ =π θ =π
=4
θ
∫ sin(
=0
θ
2 )dθ = −8 cos(6’ ) | = 0 − (−8) = 8
θ
2
θ =0

Equations
The graph is a cardioid and.
r2 + (r')2 = (1 – c(θ))2 + (s(θ))2
= 1 – 2c + c2 + s2
r = 1 – cos(θ)
2

θ =b θ =π θ =π
2
θ
∫=a
r 2 + (r ' (θ )) 2 dθ = 2
θ
∫
=0
2(1 − cos(θ ) )dθ = 2
θ
∫
=0
2 * 2 sin 2 ( θ ) dθ
2

θ =π θ =π
=4
θ
∫ sin(
=0
θ
2 )dθ = −8 cos( 2 ) | = 0 − (−8) = 8
θ
θ =0

5 parametric equations, tangents and curve lengths in polar coordinates

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