Polar Coordinates (r, θ)
Polar Coordinates (r , θ) in the plane are described
by
Polar Coordinates (r, θ)
Polar Coordinates (r , θ) in the plane are described
by
r = distance from the origin
Polar Coordinates (r, θ)
Polar Coordinates (r , θ) in the plane are described
by
r = distance from the origin
and
θ ∈ [0, 2π) is the counter-clockwise
angle.
Polar Coordinates (r, θ)
Polar Coordinates (r , θ) in the plane are described
by
r = distance from the origin
and
θ ∈ [0, 2π) is the counter-clockwise
We make the convention
(−r, θ) = (r, θ + π).
angle.
Polar Coordinates (r, θ)
Polar Coordinates (r , θ) in the plane are described
by
r = distance from the origin
and
θ ∈ [0, 2π) is the counter-clockwise
We make the convention
(−r, θ) = (r, θ + π).
angle.
Plotting points
Example
Plot the points whose polar coordinates are given.
Plotting points
Example
Plot the points whose polar coordinates are given.
π
π
π
(a) 1, 5
(b) (2, 3π) (c) 2, −2
(d) −3, 3
4
3
4
Plotting points
Example
Plot the points whose polar coordinates are given.
π
π
π
(a) 1, 5
(b) (2, 3π) (c) 2, −2
(d) −3, 3
4
3
4
Solution The points are plotted in Figure 3.
Plotting points
Example
Plot the points whose polar coordinates are given.
π
π
π
(a) 1, 5
(b) (2, 3π) (c) 2, −2
(d) −3, 3
4
3
4
Solution The points are plotted in Figure 3. In part (d) the point
−3, 3 π4 is located three units from the pole in the fourth
quadrant because the angle 3 π4 is in the second quadrant and
r = −3 is negative.
Plotting points
Example
Plot the points whose polar coordinates are given.
π
π
π
(a) 1, 5
(b) (2, 3π) (c) 2, −2
(d) −3, 3
4
3
4
Solution The points are plotted in Figure 3. In part (d) the point
−3, 3 π4 is located three units from the pole in the fourth
quadrant because the angle 3 π4 is in the second quadrant and
r = −3 is negative.
Coordinate conversion - Polar/Cartesian
Coordinate conversion - Polar/Cartesian
x = r cos θ
y = r sin θ
Coordinate conversion - Polar/Cartesian
x = r cos θ
y = r sin θ
r 2 = x2 + y 2
tan θ =
y
x
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
π
3
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
π
1
=2·
3
2
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
π
1
=2· =1
3
2
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
y
π
1
=2· =1
3
2
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
y = r sin θ
π
1
=2· =1
3
2
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
x = r cos θ = 2 cos
y = r sin θ = 2 sin
π
3
π
1
=2· =1
3
2
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
π
1
=2· =1
3
2
√
π
3
y = r sin θ = 2 sin = 2 ·
3
2
x = r cos θ = 2 cos
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
π
1
=2· =1
3
2
√
π
3 √
y = r sin θ = 2 sin = 2 ·
= 3
3
2
x = r cos θ = 2 cos
Example Convert the point 2, π3 from polar to
Cartesian coordinates.
Solution Since r = 2 and θ = π3 ,
π
1
=2· =1
3
2
√
π
3 √
y = r sin θ = 2 sin = 2 ·
= 3
3
2
√
Therefore, the point is (1, 3) in Cartesian
coordinates.
x = r cos θ = 2 cos
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
12 + (−1)2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
12 + (−1)2 =
√
2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
tan θ
p
12 + (−1)2 =
√
2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
tan θ =
12 + (−1)2 =
y
x
√
2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
tan θ =
12 + (−1)2 =
y
= −1
x
√
2
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
tan θ =
12 + (−1)2 =
√
2
y
= −1
x
Since the point (1, −1) lies in the fourth quadrant,
we choose θ = − π4 or θ = 7 π4 .
Example Represent the point with Cartesian
coordinates (1, −1) in terms of polar coordinates.
Solution If we choose r to be positive, then
r=
p
x2 + y2 =
p
tan θ =
12 + (−1)2 =
√
2
y
= −1
x
Since the point (1, −1) lies in the fourth quadrant,
possible
we choose √
θ = − π4 or θ = 7 π4 . Thus,
√ one
π
π
answer is
2, − 4 ; another is
2, 7 4 .
Graph of a polar equation
Definition The graph of a polar equation
r = f (θ), or more generally F (r , θ) = 0, consists of
all points P that have at least one polar
representation (r , θ) whose coordinates satisfy the
equation.
Example What curve is represented by the polar
equation r = 2?
Example What curve is represented by the polar
equation r = 2?
Solution The curve consists of all points (r, θ) with
r = 2.
Example What curve is represented by the polar
equation r = 2?
Solution The curve consists of all points (r, θ) with
r = 2. Since r represents the distance from the
point to the pole, the curve r = 2 represents the
circle with center O and radius 2.
Example What curve is represented by the polar
equation r = 2?
Solution The curve consists of all points (r, θ) with
r = 2. Since r represents the distance from the
point to the pole, the curve r = 2 represents the
circle with center O and radius 2. In general, the
equation r = a represents a circle with center O and
radius |a|.
Example What curve is represented by the polar
equation r = 2?
Solution The curve consists of all points (r, θ) with
r = 2. Since r represents the distance from the
point to the pole, the curve r = 2 represents the
circle with center O and radius 2. In general, the
equation r = a represents a circle with center O and
radius |a|.
Sketch the curve with polar equation r = 2 cos θ.
Sketch the curve with polar equation r = 2 cos θ.
Solution Plotting points we find what seems to be
a circle:
Sketch the curve with polar equation r = 2 cos θ.
Solution Plotting points we find what seems to be
a circle:
Example Find the Cartesian coordinates for
r = 2 cos θ.
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x = r 2
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x = r 2 = x 2 + y 2
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x = r 2 = x 2 + y 2
or
x 2 − 2x + y 2 = 0
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x = r 2 = x 2 + y 2
or
x 2 − 2x + y 2 = 0
or
(x − 1)2 + y 2 = 1.
Example Find the Cartesian coordinates for
r = 2 cos θ.
Solution Since x = r cos θ, the equation r = 2 cos θ
becomes r = 2xr or
2x = r 2 = x 2 + y 2
or
x 2 − 2x + y 2 = 0
or
(x − 1)2 + y 2 = 1.
This is the equation of a circle of radius 1 centered
at (1, 0).
Cardioid
Example Sketch the curve r = 1 + sin θ.
Cardioid
Example Sketch the curve r = 1 + sin θ. This
curve is called a cardioid.
Cardioid
Example Sketch the curve r = 1 + sin θ. This
curve is called a cardioid. Solution
Cardioid
Example Sketch the curve r = 1 + sin θ. This
curve is called a cardioid. Solution
Four-leaved rose
Example Sketch the curve r = cos 2θ.
Four-leaved rose
Example Sketch the curve r = cos 2θ. This curve
is called a four-leaved rose.
Four-leaved rose
Example Sketch the curve r = cos 2θ. This curve
is called a four-leaved rose. Solution
Four-leaved rose
Example Sketch the curve r = cos 2θ. This curve
is called a four-leaved rose. Solution
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
dy
dx
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
dy
=
dx
dy
dθ
dx
dθ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
dy
=
dx
dy
dθ
dx
dθ
=
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
dy
=
dx
dy
dθ
dx
dθ
=
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
We locate horizontal tangents by finding the points where
dy
dθ
= 0 (provided that
dx
dθ
6= 0.)
Tangents to Polar Curves
To find a tangent line to a polar curve r = f(θ) we regard θ as a
parameter and write its parametric equations as
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
Then, using the method for finding slopes of parametric curves and
the Product Rule, we have
dy
=
dx
dy
dθ
dx
dθ
=
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
We locate horizontal tangents by finding the points where
dy
dθ
= 0 (provided that
dx
dθ
6= 0.) Likewise, we locate vertical
tangents at the points where
dx
dθ
= 0 (provided that
dy
dθ
6= 0).
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 .
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
dx
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
=
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ(1 + 2 sin θ)
1 − 2 sin2 θ − sin θ
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
=
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
cos π3 (1 + 2 sin π3 )
dy =
dx θ= π
(1 + sin π3 )(1 − 2 sin π3 )
3
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
cos π3 (1 + 2 sin π3 )
dy =
dx θ= π
(1 + sin π3 )(1 − 2 sin π3 )
3
√
=
(1
1
+
2 (1
√
3
+ 2 )(1
3)
√
− 3)
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
cos π3 (1 + 2 sin π3 )
dy =
dx θ= π
(1 + sin π3 )(1 − 2 sin π3 )
3
√
=
(1
1
+
2 (1
√
3
+ 2 )(1
√
3)
1+ 3
√
√
=
√
(2 + 3)(1 − 3)
− 3)
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
cos π3 (1 + 2 sin π3 )
dy =
dx θ= π
(1 + sin π3 )(1 − 2 sin π3 )
3
√
=
(1
1
+
2 (1
√
3
+ 2 )(1
√
√
3)
1+ 3
1+ 3
√
√
√
=
=
√
(2 + 3)(1 − 3)
−1 − 3
− 3)
Example For the cardioid r = 1 + sin θ find the
slope of the tangent line when θ = π3 . Solution
dy
=
dx
dr
dθ
dr
dθ
sin θ + r cos θ
cos θ − r sin θ
=
cos θ sin θ + (1 + sin θ) cos θ
cos θ cos θ − (1 + sin θ) sin θ
cos θ(1 + 2 sin θ)
cos θ(1 + 2 sin θ)
=
2
(1 + sin θ)(1 − 2 sin θ)
1 − 2 sin θ − sin θ
The slope of the tangent at the point where
θ = π3 is
=
cos π3 (1 + 2 sin π3 )
dy =
dx θ= π
(1 + sin π3 )(1 − 2 sin π3 )
3
√
=
(1
1
+
2 (1
√
3
+ 2 )(1
√
√
3)
1+ 3
1+ 3
√
√
√ = −1
=
=
√
(2 + 3)(1 − 3)
−1 − 3
− 3)
Area under a polar graph r = f(θ)
The area of a region ”under” a polar function
r = f(θ) is described by either of the following
formulas.
Area under a polar graph r = f(θ)
The area of a region ”under” a polar function
r = f(θ) is described by either of the following
formulas. These formulas arise from the fact that
the area of a θ1 ≤ θ ≤ θ2 portion of a circle of
radius r is given by 12 (θ2 − θ1 )r 2 .
Area under a polar graph r = f(θ)
The area of a region ”under” a polar function
r = f(θ) is described by either of the following
formulas. These formulas arise from the fact that
the area of a θ1 ≤ θ ≤ θ2 portion of a circle of
radius r is given by 12 (θ2 − θ1 )r 2 .
A=
Rb 1
2
a 2 [f(θ)]
dθ,
Area under a polar graph r = f(θ)
The area of a region ”under” a polar function
r = f(θ) is described by either of the following
formulas. These formulas arise from the fact that
the area of a θ1 ≤ θ ≤ θ2 portion of a circle of
radius r is given by 12 (θ2 − θ1 )r 2 .
A=
Rb 1
2
a 2 [f(θ)]
A=
Rb 1
a 2r
2
dθ,
dθ,
Area under a polar graph r = f(θ)
The area of a region ”under” a polar function
r = f(θ) is described by either of the following
formulas. These formulas arise from the fact that
the area of a θ1 ≤ θ ≤ θ2 portion of a circle of
radius r is given by 12 (θ2 − θ1 )r 2 .
A=
Rb 1
2
a 2 [f(θ)]
A=
Rb 1
a 2r
2
Also see the two figures below.
dθ,
dθ,
Area under a polar graph r = f(θ)
Area under a polar graph r = f(θ)
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ.
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
Area
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
Z
π
4
Area =
− π4
1 2
r dθ
2
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
Z
π
4
Area =
− π4
1 2
1
r dθ =
2
2
Z
π
4
− π4
cos2 2θ dθ
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
Z
π
4
Area =
− π4
1 2
1
r dθ =
2
2
Z
π
4
− π4
1
cos 2θ dθ =
2
2
Z
π
4
− π4
cos2 2θ dθ
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
π
4
Z
Area =
− π4
1
=
2
Z
π
4
− π4
1 2
1
r dθ =
2
2
Z
π
4
− π4
1
(1 + cos 4θ) dθ
2
1
cos 2θ dθ =
2
2
Z
π
4
− π4
cos2 2θ dθ
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
π
4
Z
Area =
− π4
1
=
2
Z
π
4
− π4
1 2
1
r dθ =
2
2
Z
π
4
− π4
1
cos 2θ dθ =
2
2
Z
π
4
cos2 2θ dθ
− π4
π
4
1
1
1
(1 + cos 4θ) dθ = (θ + sin 4θ)
2
4
4
−π
4
Example Find the area enclosed by one loop of
the four-leaved rose r = cos 2θ. Solution First
recall the picture of this curve:
By our area formulas,
π
4
Z
Area =
− π4
1
=
2
Z
π
4
− π4
1 2
1
r dθ =
2
2
Z
π
4
− π4
1
cos 2θ dθ =
2
2
Z
π
4
cos2 2θ dθ
− π4
π
4
1
1
1
π
(1 + cos 4θ) dθ = (θ + sin 4θ)
= .
2
4
4
8
−π
4
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)).
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is
s(t) = |C0 (t)|
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is p
s(t) = |C0 (t)| = (x0 (t))2 + (y0 (t))2 .
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is p
s(t) = |C0 (t)| = (x0 (t))2 + (y0 (t))2 .
Since the integral of the speed is the distance
traveled or length for C : [a, b] → R2 ,
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is p
s(t) = |C0 (t)| = (x0 (t))2 + (y0 (t))2 .
Since the integral of the speed is the distance
traveled or length for C : [a, b] → R2 ,
Length of a curve(C)
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is p
s(t) = |C0 (t)| = (x0 (t))2 + (y0 (t))2 .
Since the integral of the speed is the distance
traveled or length for C : [a, b] → R2 ,
Z b
Length of a curve(C) =
s(t) dt
a
Speed and length
Definition The velocity vector of a curve
C(t) = (x(t), y(t)) is C0 (t) = (x0 (t), y0 (t)). The
speed of C(t) is p
s(t) = |C0 (t)| = (x0 (t))2 + (y0 (t))2 .
Since the integral of the speed is the distance
traveled or length for C : [a, b] → R2 ,
Z b
Length of a curve(C) =
s(t) dt
a
Z
=
a
bq
(x0 (t))2 + (y0 (t))2 dt.
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞.
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)).
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t)
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos t).
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos t). Recall that the speed s(t) of c(t)
is |c0 (t)|
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2 and the length is
the integral of the speed:
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2 and the length is
the integral of the speed:
Z ∞q
e−2t (cos2 t + sin2 t + 2 cos t sin t + sin2 t + cos2 t − 2 cos t sin t)dt
0
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2 and the length is
the integral of the speed:
Z ∞q
e−2t (cos2 t + sin2 t + 2 cos t sin t + sin2 t + cos2 t − 2 cos t sin t)dt
0
Z ∞
√
=
e−t 2 dt
0
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2 and the length is
the integral of the speed:
Z ∞q
e−2t (cos2 t + sin2 t + 2 cos t sin t + sin2 t + cos2 t − 2 cos t sin t)dt
0
Z ∞
t
√
√
=
e−t 2 dt = lı́m − 2e−t 0
t→∞
0
Problem 30
As the parameter t increases forever, starting at
t= 0, the curve with parametric equations
x = e −t cos t,
y = e −t sin t
spirals inward toward the origin, getting ever closer to the origin (but
never actually reaching) as t → ∞. Find the length of this spiral curve.
Solution The tangent vector to the curve
c(t) = (x(t), y(t)) = (e−t cos t, e−t sin t| is c0 (t) and it is found by
taking the derivative of the coordinate functions:
c0 (t) = (x0 (t), y0 (t)). So,
c0 (t) = (−e−t cos t − e−t sin t, −e−t sin t + e−t cos t) =
−e−t (cos t + sin t, sin t − cos
p t). Recall that the speed s(t) of c(t)
is |c0 (t)| which is equal to (x0 (t))2 + (y0 (t))2 and the length is
the integral of the speed:
Z ∞q
e−2t (cos2 t + sin2 t + 2 cos t sin t + sin2 t + cos2 t − 2 cos t sin t)dt
0
Z ∞
t
√
√
√
=
e−t 2 dt = lı́m − 2e−t = 2.
0
t→∞
0
Length formula
In polar coordinates x = r cos θ, y = r sin θ.
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dr
=
cos θ − r sin θ
dθ
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
=
cos θ − r sin θ
dθ
dθ
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
2 2
dx
dy
+
dθ
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
2
2 2
dr
dx
dy
dr
=
+
cos2 θ − 2r cos θ sin θ + r2 sin2 θ
dθ
dθ
dθ
dθ
2
dr
dr
sin2 θ + 2r sin θ cos θ + r2 sin2 θ
+
dθ
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
2
2 2
dr
dx
dy
dr
=
+
cos2 θ − 2r cos θ sin θ + r2 sin2 θ
dθ
dθ
dθ
dθ
2
dr
dr
sin2 θ + 2r sin θ cos θ + r2 sin2 θ
+
dθ
dθ
2
dr
=
+ r2 .
dθ
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
2
2 2
dr
dx
dy
dr
=
+
cos2 θ − 2r cos θ sin θ + r2 sin2 θ
dθ
dθ
dθ
dθ
2
dr
dr
sin2 θ + 2r sin θ cos θ + r2 sin2 θ
+
dθ
dθ
2
dr
=
+ r2 .
dθ
Thus the length L of a polar curve r = f(θ), a ≤ θ ≤ b, is:
Length formula
In polar coordinates x = r cos θ, y = r sin θ. Then
dx
dy
dr
dr
=
cos θ − r sin θ
=
sin θ + r cos θ.
dθ
dθ
dθ
dθ
Using cos2 θ + sin2 θ = 1, we get
2
2 2
dr
dx
dy
dr
=
+
cos2 θ − 2r cos θ sin θ + r2 sin2 θ
dθ
dθ
dθ
dθ
2
dr
dr
sin2 θ + 2r sin θ cos θ + r2 sin2 θ
+
dθ
dθ
2
dr
=
+ r2 .
dθ
Thus the length L of a polar curve r = f(θ), a ≤ θ ≤ b, is:
Z
b
s
r2
L=
a
+
dr
dθ
2
dθ.