Tangent Lines and Arc Length for Parametric
and Polar Curves.
In earlier lectures we looked at parametric curves, that is
those that can be written as
x = f (t); y = g (t );a ≤ t ≤b
Then it can be shown that
dy = dy / dx
dx dt dt
This allows us to find the slope of tangent lines without
eliminating the parameter.
Example. Find the tangent line to the curve
x = t − 2sin t; y = 2 − cost; t ≥ 0
at the point where t = 9π/4.
We have
Thus
dx =1− 2cost =1− 2 =
dt
dy = sint = 1
dt
2
(
)
dy = dy / dx = 1 / 1− 2 =
1
= 1 ≈−1.7
dx dt dt
2
2(1− 2) 2 − 2
This situation is shown below.
Problem. Find dy/dx and d2y/dx2 at the point on the curve
x = 1 t 2; y = 1t3
2
3
where t = 2.
dy = t 2 Therefore dy = dy / dx = t 2 / t = t
Solution. dx = t
dt
dt
dx dt dt
At the point where t = 2, dy/dx = 2.
Now if we let y′ = dy/dx = t, then
d 2 y = dy′ = dy′ / dx =1/ t
dx2 dx dt dt
At the point where t = 2, d2y/dx2 = 1/2.
Problem. Find dy/dx and d2y/dx2 at the point on the curve
x = cosϕ; y = 3sinϕ
where ϕ = 5π/6
Solution.
Therefore
dy = 3cosϕ
dx =−sinϕ
dϕ
dϕ
dy = dy / dx = 3cosϕ / −sinϕ =−3cotϕ
dx dϕ dϕ
At the point where ϕ = 5π/6, dy/dx = 3√3.
Now if we let y′ = dy/dx = −3cot ϕ , then
d 2 y = dy′ = dy′ / dx = =3csc2 ϕ =−3csc3ϕ
dx2 dx d ϕ d ϕ − sin ϕ
At the point where ϕ = 5π/6, d2y/dx2 = −24.
Problem. Find all values of t at which the parametric curve
x = 2t3 −15t 2 + 24t +7; y = t 2 +t +1
has (a) a horizontal tangent line and (b) a vertical tangent line.
Solution.
dx = 6t 2 − 30t + 24 = 6(t 2 −5t +4) = 6(t − 4)(t −1)
dt
dy = 2t +1 Thus dy = 2t +1
dx 6(t − 4)(t −1)
dt
We see that the tangent line is horizontal when 2t + 1 = 0, or
t = −1/2. The tangent line is vertical when t = 4 or t = 1.
We can apply this procedure to find the tangent line to a polar
curve. A polar curve r = f(θ) can be written parametrically by
noting that x = rcos(θ) and y = rsin(θ). Thus
x = f (θ )cosθ
y = f (θ )sinθ
This means that
dx =− f (θ )sinθ + f ′(θ )cosθ =−r sinθ + dr cosθ
dθ
dθ
dy = f (θ )cosθ + f ′(θ )sinθ = r cosθ + dr sinθ
dθ
dθ
Therefore at any point we have
dr
dy = r cosθ + sin θ dθ
dx −r sin θ + cosθ dr
dθ
Problem. Find the slope of the tangent line to the polar curve
r =1+sinθ
for θ = π/4.
Solution. dr/dθ = cosθ . Thus
dr
dy = r cosθ + sin θ dθ = (1+sin θ )cos θ +sin θ cosθ
dx −r sin θ + cosθ dr −(1+ sin θ )sin θ + cosθ cosθ
dθ
1 1
1 1
1
(1+ )
+
1+
2 2
2 2 =
2 =− 2 −1
=
1 1
1 1
1
−(1+ ) +
−
2 2
2 2
2
Problem. Find the slope of the tangent line to the polar curve
r = asec2θ
for θ = π/6.
Solution. dr/dθ = 2asec2θtan2θ . Thus at θ = π/6, we have
dr/dθ = 2asec2θ tan2θ = 4a√3, r = 2a, sinθ = 1/2, cosθ =
√3/2.
 3 1
Thus
dr
2a 
 + 4a 3
r cosθ + sin θ
dy =
 2  2
dθ =
dx −r sin θ + cosθ dr
1  3
dθ −2a  2  + 2  4a 3


= 3a 3 = 3 3
5a
5
We can find out some useful information about polar curves
that pass through the pole.
Theorem. Suppose that the curve r = f(θ) passes through the
pole, or origin, at θ = θ0. If dr/dθ is not 0 at θ = θ0, then the
line θ = θ0 is the tangent line to the curve at θ = θ0 at the
origin.
Proof.
dr
dr
dy = r cosθ + sin θ dθ = sin θ dθ = tanθ
dx −r sin θ + cosθ dr cosθ dr
dθ
dθ
at the origin, under the conditions given.
Problem. Sketch the polar curve
r = 4cosθ
and find the polar equations of the tangent line to the curve at
the pole.
Solution. The curve can be written as r 2 = 4r cosθ
or x2 + y2 = 4x. On completing the square, this becomes
x − 2 2 + y2 = 4. The graph is:
(
)
Since at θ = π/2, r = 0 and dr/dθ = −4sinθ = −4, the previous
theorem tells us that the tangent line at the pole is θ = π/2.
Problem. Sketch the polar curve
r = sin2θ
and find the polar equations of the tangent line to the curve at
the pole.
Solution. The curve is a 4 leaf rose whose graph is:
We have r = 0 when θ = 0, and θ = π/2. At these points, the
derivative is 2cos 2θ = 1 or −1. Thus the tangent lines at the
pole have equations θ = 0 and θ = π/2, by the previous
theorem.
Arc Length
Recall from earlier lectures that the differential element of
arc length, ds can be represented by the diagram
ds
dy
dx
From which we see that ds = dx2 + dy 2
Then arc length can be found by integration,
s = ∫ ds = ∫ dx2 + dy 2
We previously investigated various forms of this equation,
corresponding to y = f(x), x = g(y), and the parametric curve
x = x(t), y = y(t). The parametric form follows from the
differential substitutions
dx = dx dt
dy = dy dt
dt
dt
so that
2 dy 2
dx
   
s = ∫   +  dt
 dt   dt 
We apply this formula to find a formula for arc length of a
polar curve. We know that a parametric form of the polar curve
r = f(θ) is given by
x = f (θ )cos(θ ); y = f (θ )sin(θ )
Then
dx =− f (θ )sin(θ ) + f ′(θ )cos(θ ) =−r sin(θ ) + dr cos(θ )
dθ
dθ
dy = f (θ )cos(θ ) + f ′(θ )sin(θ ) = r cos(θ ) + dr sin(θ )
dθ
dθ
By squaring these expressions, we get
2
 dx 
2 sin 2(θ ) − 2r  dr
=
r
 dθ 
 dθ



2

 dr 
2(θ )
sin(
θ
)cos(
θ
)
+
cos

 dθ 



2
dy


 dr
2
2
 dθ  = r cos (θ ) + 2r  dθ



2
dr



2
cos(
θ
)sin(
θ
)
+

 dθ  sin (θ )



We add these equations to get:
 dx
 dθ

2
2
  dy   2  dr
 + dθ  = r +  dθ
 
 
2
   2
2(θ ) 
cos
(
θ
)
+
sin
 

 

2
dr


= r2 +

d
θ


Thus we finally arrive at the formula for polar arc length.
θ 2  dx 2  dy
s= ∫ 
+

 dθ   dθ
θ1
2
θ 2 2  dr

 dθ = ∫ r +  dθ


θ1
2

 dθ

Problem. Find the length of the spiral r = eθ between θ = 0 and
θ = π/2.
This curve is as shown below.
Using the arc length formula, we have
π
π
2
2
2
 dr 
2
s = ∫ r +
dθ = ∫

 dθ 
0
0
( ) ( )
π
2 θ 2
2 θ
θ
e
+ e
dθ = 2 ∫ e dθ
0
π
= 2(e 2 −1) ≈ 31.3
Problem. Find the length of the entire circle r = 2acos(θ).
2
dr

dr =−2asin(θ ) so r2 +
2 cos2 (θ )+ sin 2(θ )  = 4a2
=
4
a
 dθ 


dθ


π
π
2
2
2
 dr 
2
s = ∫ r +
dθ = ∫ 2adθ =2π a.

dθ 

π
−
−π
2
2
Problem. Find the length of the polar curve r = sin 2(θ /2)
from θ = 0 to θ = π .
dr = sin(θ )cos(θ ) so
2
2
dθ
2
dr


4(θ ) +sin2(θ )cos2(θ ) = sin 2(θ )
r2 +
=
sin

2
2
2
2
d
θ


π
s = ∫ sin(θ )dθ =−2cos(θ )  = 2
2
2  0
0
π
Problem. Find the length of the polar curve r = sin3(θ /3)
from θ = 0 to θ = π .
dr = sin2(θ )cos(θ ) so
3
3
dθ
2
dr


6(θ ) + sin 4(θ )cos2(θ ) = sin 4(θ )
r2 +
=
sin

3
3
3
3
d
θ


π /2
π / 2
π /2 2
1
1
3


s = ∫ sin (θ 3)dθ = ∫ 1−cos(2θ 3) dθ = θ − sin(2θ 3)  


2
2 2
 0
0
0

π 3  π  π 3 3
1
=  − sin    = −
2 2 2  3  4 8