Lecture 36: Polar Coordinates Π Π Π 2 2 5

Home , Polar coordinate system

Polar Co-ordinates Polar to Cartesian coordinates Cartesian to Polar coordinates Example 3 Graphing Equations in Polar Coordinates Example 5 Example 5 Example 5 Example 6 Example 6 Using Symmetry Using Symmetry Using Symmetry Example (Symmetry) Circles Tangents to Polar Curves Tangents to Polar Curves Example 9 Polar Co-ordinates

A polar coordinate system, gives the co-ordinates of a point with reference to a point O and a half line or ray starting at the point O. We will look at polar coordinates for points in the xy-plane, using the origin (0, 0) and the positive x-axis for reference.

A point P in the plane, has polar coordinates (r, θ), where r is the distance of the point from the origin and θ is the angle that the ray |OP| makes with the positive x-axis.

Annette Pilkington Lecture 36: Polar Coordinates Π Π Π 2 2 5. 2 3. 3 Π Π 3 Π Π 3 Π Π 4. 2.5 4 4 4 4 4 4 2. 3. 2. 1.5 2. 1. 1. 0.5 Π 0. 0 Π 0. 0 Π 0

5 Π 7 Π 5 Π 7 Π 5 Π 7 Π 4 4 4 4 4 4 3 Π 3 Π 3 Π 2 2 2

Example 1 Plot the points whose polar coordinates are given by π π 7π 5π (2, ) (3, − ) (3, ) (2, ) 4 4 4 2

Annette Pilkington Lecture 36: Polar Coordinates Polar Co-ordinates Polar to Cartesian coordinates Cartesian to Polar coordinates Example 3 Graphing Equations in Polar Coordinates Example 5 Example 5 Example 5 Example 6 Example 6 Using Symmetry Using Symmetry Using Symmetry Example (Symmetry) Circles Tangents to Polar Curves Tangents to Polar Curves Example 9 Example 1

Example 1 Plot the points whose polar coordinates are given by π π 7π 5π (2, ) (3, − ) (3, ) (2, ) 4 4 4 2

Π Π Π 2 2 5. 2 3. 3 Π Π 3 Π Π 3 Π Π 4. 2.5 4 4 4 4 4 4 2. 3. 2. 1.5 2. 1. 1. 0.5 Π 0. 0 Π 0. 0 Π 0

5 Π 7 Π 5 Π 7 Π 5 Π 7 Π 4 4 4 4 4 4 3 Π 3 Π 3 Π 2 2 2

Annette Pilkington Lecture 36: Polar Coordinates Π

2 5. 3 Π Π 4. 4 4

Π 0. 0

5 Π 7 Π

4 4

3 Π

Note the representation of a point in polar coordinates is not unique. For instance for any θ the point (0, θ) represents the pole O. We extend the meaning of polar coordinate to the case when r is negative by agreeing that the two points (r, θ) and (−r, θ) are in the same line through O and at the same distance |r| but on opposite side of O. Thus

(−r, θ) = (r, θ + π)

3π Example 2 Plot the point (−3, 4 )

(−r, θ) = (r, θ + π)

3π Example 2 Plot the point (−3, 4 )

2 5. 3 Π Π 4. 4 4

Π 0. 0

5 Π 7 Π

4 4

3 Π

Annette Pilkington Lecture 36: Polar Coordinates For (2, π ), we have r = 2, θ = π . In Cartesian co-ordinates, we get IIIII 4 4 √ x = r cos θ = 2 cos(π/4) = 2 √1 = 2 2 √ we get y = r sin θ = 2 sin(π/4) = 2 √1 = 2 2 π π I For (3, − 3 ), we have r = 3, θ = − 3 . In Cartesian co-ordinates, we get x = r cos θ = 3 cos(− π ) = 3 1 = 3 3 2 2 √ √ π − 3 −3 3 we get y = r sin θ = 3 sin(− 3 ) = 3 2 = 2

To convert from Polar to Cartesian coordinates, we use the identities:

x = r cos θ, y = r sin θ

Example 3 Convert the following (given in polar co-ordinates) to Cartesian π π coordinates (2, 4 ) and (3, − 3 )

Annette Pilkington Lecture 36: Polar Coordinates π π I For (3, − 3 ), we have r = 3, θ = − 3 . In Cartesian co-ordinates, we get x = r cos θ = 3 cos(− π ) = 3 1 = 3 3 2 2 √ √ π − 3 −3 3 we get y = r sin θ = 3 sin(− 3 ) = 3 2 = 2

To convert from Polar to Cartesian coordinates, we use the identities:

x = r cos θ, y = r sin θ

To convert from Polar to Cartesian coordinates, we use the identities:

x = r cos θ, y = r sin θ

Example 3 Convert the following (given in polar co-ordinates) to Cartesian π π coordinates (2, 4 ) and (3, − 3 ) For (2, π ), we have r = 2, θ = π . In Cartesian co-ordinates, we get I 4 4 √ x = r cos θ = 2 cos(π/4) = 2 √1 = 2 2 √ we get y = r sin θ = 2 sin(π/4) = 2 √1 = 2 2 π π I For (3, − 3 ), we have r = 3, θ = − 3 . In Cartesian co-ordinates, we get x = r cos θ = 3 cos(− π ) = 3 1 = 3 3 2 2 √ √ π − 3 −3 3 we get y = r sin θ = 3 sin(− 3 ) = 3 2 = 2

Annette Pilkington Lecture 36: Polar Coordinates I For (2, 2), we have x = 2, y = 2. Therefore√ r 2 = x 2 + y 2 = 4 + 4 = 8, and r = 8. y We have tan θ = x = 2/2 = 1. Since this point is in the ﬁrst quadrant, we have θ = π . 4 √ Therefore the polar co-ordinates for the point (2, 2) are ( 8, π ). √ √ 4 I For (1, − 3), we have x = 1, y = − 3. Therefore r 2 = x 2 + y 2 = 1 + 3 = 4, and r = 2. y √ We have tan θ = x = − 3. Since this point is in the fourth quadrant, we have θ = −π . √ 3 −π Therefore the polar co-ordinates for the point (1, − 3) are (2, 3 ).

Annette Pilkington Lecture 36: Polar Coordinates √ √ I For (1, − 3), we have x = 1, y = − 3. Therefore r 2 = x 2 + y 2 = 1 + 3 = 4, and r = 2. y √ We have tan θ = x = − 3. Since this point is in the fourth quadrant, we have θ = −π . √ 3 −π Therefore the polar co-ordinates for the point (1, − 3) are (2, 3 ).

To convert from Cartesian to polar coordinates, we use the following identities y r 2 = x 2 + y 2, tan θ = x When choosing the value of θ, we must be careful to consider which quadrant the point is in, since for any given number a, there are two angles with tan θ = a, in the interval 0 ≤ θ ≤ 2π. Example 3 Give polar coordinates√ for the√ points (given in Cartesian co-ordinates) (2, 2), (1, − 3), and (−1, 3) I For (2, 2), we have x = 2, y = 2. Therefore√ r 2 = x 2 + y 2 = 4 + 4 = 8, and r = 8. y We have tan θ = x = 2/2 = 1. Since this point is in the ﬁrst quadrant, we have θ = π . 4 √ π Therefore the polar co-ordinates for the point (2, 2) are ( 8, 4 ).

To convert from Cartesian to polar coordinates, we use the following identities y r 2 = x 2 + y 2, tan θ = x When choosing the value of θ, we must be careful to consider which quadrant the point is in, since for any given number a, there are two angles with tan θ = a, in the interval 0 ≤ θ ≤ 2π. Example 3 Give polar coordinates√ for the√ points (given in Cartesian co-ordinates) (2, 2), (1, − 3), and (−1, 3) I For (2, 2), we have x = 2, y = 2. Therefore√ r 2 = x 2 + y 2 = 4 + 4 = 8, and r = 8. y We have tan θ = x = 2/2 = 1. Since this point is in the ﬁrst quadrant, we have θ = π . 4 √ Therefore the polar co-ordinates for the point (2, 2) are ( 8, π ). √ √ 4 I For (1, − 3), we have x = 1, y = − 3. Therefore r 2 = x 2 + y 2 = 1 + 3 = 4, and r = 2. y √ We have tan θ = x = − 3. Since this point is in the fourth quadrant, we have θ = −π . √ 3 −π Therefore the polar co-ordinates for the point (1, − 3) are (2, 3 ).

Annette Pilkington Lecture 36: Polar Coordinates √ √ I For (−1, 3), we have x = −1, y = 3. Therefore r 2 = x 2 + y 2 = 1 + 3 = 4, and r = 2. y √ We have tan θ = x = − 3. Since this point is in the second quadrant, we have θ = 2π . √ 3 2π Therefore the polar co-ordinates for the point (−1, 3) are (2, 3 ).

Example 3 Give polar√ coordinates for the point (given in Cartesian co-ordinates) (−1, 3)

Example 3 Give polar√ coordinates for the point (given in Cartesian co-ordinates) (−1, 3) √ √ I For (−1, 3), we have x = −1, y = 3. Therefore r 2 = x 2 + y 2 = 1 + 3 = 4, and r = 2. y √ We have tan θ = x = − 3. Since this point is in the second quadrant, we have θ = 2π . √ 3 2π Therefore the polar co-ordinates for the point (−1, 3) are (2, 3 ).

Annette Pilkington Lecture 36: Polar Coordinates I Lines: A line through the origin (0, 0) has equation θ = θ0 I Circle centered at the origin: A circle of radius r0 centered at the origin has equation r = r0 in polar coordinates.

I The equation r = 5 describes a circle of radius 5 centered at the origin. π The equation θ = 4 describes a line through the origin making an angle π of 4 with the positive x axis.

3 Π Π

4 4

Π 0

5 Π 7 Π

4 4

3 Π

The graph of an equation in polar coordinates r = f (θ) or F (r, θ) = 0 consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.

π Example 4 Graph the following equations r = 5, θ = 4

Annette Pilkington Lecture 36: Polar Coordinates I Circle centered at the origin: A circle of radius r0 centered at the origin has equation r = r0 in polar coordinates.

I The equation r = 5 describes a circle of radius 5 centered at the origin. π The equation θ = 4 describes a line through the origin making an angle π of 4 with the positive x axis.

3 Π Π

4 4

Π 0

5 Π 7 Π

4 4

3 Π

The graph of an equation in polar coordinates r = f (θ) or F (r, θ) = 0 consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.

IILines: A line through the origin (0, 0) has equation θ = θ0

π Example 4 Graph the following equations r = 5, θ = 4

Annette Pilkington Lecture 36: Polar Coordinates I The equation r = 5 describes a circle of radius 5 centered at the origin. π The equation θ = 4 describes a line through the origin making an angle π of 4 with the positive x axis.

3 Π Π

4 4

Π 0

5 Π 7 Π

4 4

3 Π

The graph of an equation in polar coordinates r = f (θ) or F (r, θ) = 0 consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.

Annette Pilkington Lecture 36: Polar Coordinates Π

3 Π Π

4 4

Π 0

5 Π 7 Π

4 4

3 Π

The graph of an equation in polar coordinates r = f (θ) or F (r, θ) = 0 consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.

IILines: A line through the origin (0, 0) has equation θ = θ0 I Circle centered at the origin: A circle of radius r0 centered at the origin has equation r = r0 in polar coordinates. π Example 4 Graph the following equations r = 5, θ = 4 I The equation r = 5 describes a circle of radius 5 centered at the origin. π The equation θ = 4 describes a line through the origin making an angle π of 4 with the positive x axis.

The graph of an equation in polar coordinates r = f (θ) or F (r, θ) = 0 consists of all points P that have at least one polar representation (r, θ) whose coordinates satisfy the equation.

3 Π Π

4 4

Π 0

5 Π 7 Π

4 4

3 Π

2 Annette Pilkington Lecture 36: Polar Coordinates IIWe ﬁnd the value of r for speciﬁc values of θ in order to plot some points on the curve. We note that it is enough to sketch the graph on the interval 0 ≤ θ ≤ 2π, since (r(θ), θ) = (r(θ + 2π), θ + 2π).

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

θ r 0

π 4

π 2

3π 4

5π 4

2π

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

θ r IIWe ﬁnd the value of r for 0 speciﬁc values of θ in order to plot some points on the curve. π 4 We note that it is enough to sketch the graph on the interval π 2 0 ≤ θ ≤ 2π, since (r(θ), θ) = (r(θ + 2π), θ + 2π). 3π 4

5π 4

2π

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

5π 4

2π

Annette Pilkington Lecture 36: Polar Coordinates that they lie on a circle of radius 3 centered at (0, 3) and that the curve traces out this circle twice in an anticlockwise direction as θ increases from 0 to 2π. Π

2 8. 3 Π Π

4 6. 4

Π 0. 0

5 Π 7 Π

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I The equation in Cart. co-ords is 2 2 I When we plot the points, we see x + (y − 3) = 9.

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

θ r 0 √0 π 4 6/ 2 π 6 2 √ 3π 4 6/ 2 π 0√ 5π 4 −6/ 2 3π 2 −6

IIWe ﬁnd the value of r for speciﬁc values of θ in order to plot some points on the curve. We note that it is enough to sketch the graph on the interval 0 ≤ θ ≤ 2π, since (r(θ), θ) = (r(θ + 2π), θ + 2π).

Annette Pilkington Lecture 36: Polar Coordinates I The equation in Cart. co-ords is x 2 + (y − 3)2 = 9.

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

θ r that they lie on a circle of radius 0 √0 3 centered at (0, 3) and that the π 4 6/ 2 curve traces out this circle twice π 6 in an anticlockwise direction as θ 2 √ 3π 6/ 2 increases from 0 to 2π. 4 Π π 0 2 √ 8. 5π 3 Π Π 4 −6/ 2 4 6. 4 3π 2 −6 4. 2. We ﬁnd the value of r for I Π 0. 0 speciﬁc values of θ in order to plot some points on the curve.

We note that it is enough to 5 Π 7 Π 4 4

sketch the graph on the interval 3 Π 0 ≤ θ ≤ 2π, since 2 (r(θ), θ) = (r(θ + 2π), θ + 2π).

I When we plot the points, we see

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

We note that it is enough to 5 Π 7 Π 4 4

sketch the graph on the interval 3 Π 0 ≤ θ ≤ 2π, since 2 (r(θ), θ) = (r(θ + 2π), θ + 2π). I The equation in Cart. co-ords is 2 2 I When we plot the points, we see x + (y − 3) = 9.

Example 5 Graph the equation r = 6 sin θ and convert the equation to an equation in Cartesian coordinates.

θ r Cartesian Co-ords (x, y) Π 0 0 (0, 0) 2 √ 8. π 6/ 2 (3, 3) 3 Π Π 4 4 6. 4 π 6 (0, 6) 2 √ 4. 3π 4 6/ 2 (−3, 3) 2. π 0√ (0, 0) 5π Π 0. 0 4 −6/ 2 (3, 3) 3π 2 −6 (0, 6)

5 Π 7 Π

4 4

3 Π

It may help to calculate the cartesian co-ordinates in order to sketch the curve.

Example 6 Graph the equation r = 1 + cos θ . Check the variations shown at end of lecture notes.

θ r Cartesian Co-ords (x, y) 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π

Annette Pilkington Lecture 36: Polar Coordinates I When we plot the points, we see that they lie on a heart shaped curve.

2 2.5 3 Π Π 2. 4 4

1.5

0.5

Π 0. 0

5 Π 7 Π

4 4

3 Π

Example 6 Graph the equation r = 1 + cos θ . Check the variations shown at end of lecture notes.

θ r Cartesian Co-ords (x, y) 0 ≤ θ ≤ 2π, since 0 2 (2, 0) √ √ √ (r(θ), θ) = (r(θ + 2π), θ + 2π). π √2+1 ( 2+1 , 2+1 ) 4 2 2 2 π 1 (0, 1) 2 √ √ √ −( 2−1) 3π √2−1 ( , 2−1 ) 4 2 2 2 π 0 (0, 0) √ √ √ −( 2−1) −( 2−1) 5π √2−1 ( , ) 4 2 2 2 3π 1 (0, −1) 2 √ √ √ ( 2+1) −( 2+1) 7π √2+1 ( , ) 4 2 2 2 2π 2 (2, 0)

I We ﬁnd the value of r for speciﬁc values of θ in order to plot some points on the curve. We note that it is enough to sketch the graph on the interval

Example 6 Graph the equation r = 1 + cos θ . Check the variations shown at end of lecture notes.

θ r Cartesian Co-ords (x, y) 0 ≤ θ ≤ 2π, since 0 2 (2, 0) √ √ √ (r(θ), θ) = (r(θ + 2π), θ + 2π). π √2+1 ( 2+1 , 2+1 ) 4 2 2 2 When we plot the points, we see π 1 (0, 1) I 2 √ √ √ −( 2−1) that they lie on a heart shaped 3π √2−1 ( , 2−1 ) 4 2 2 2 curve. π 0 (0, 0) √ √ √ Π 5π 2−1 −( 2−1) −( 2−1) 2 √ ( , ) 2.5 4 2 2 2 3 Π Π 2. 3π 1 (0, −1) 4 4 2 √ √ √ 1.5 7π 2+1 ( 2+1) −( 2+1) √ 1. 4 ( 2 , 2 ) 2 0.5

2π 2 (2, 0) Π 0. 0

I We ﬁnd the value of r for speciﬁc values of θ in order to 5 Π 7 Π plot some points on the curve. 4 4 We note that it is enough to 3 Π 2 sketch the graph on the interval

We have 3 common symmetries in curves which often shorten our work in graphing a curve of the form r = f (θ):

I If f (−θ) = f (θ) the curve is symmetric about the horizontal line θ = 0. In this case it is enough to draw either the upper or lower half of the curve, drawing the other half by reﬂecting in the line θ = 0(the x-axis). [Example r = 1 + cos θ].

2 2.5 3 Π Π 2. 4 4

1.5

0.5

Π 0. 0

5 Π 7 Π

4 4

3 Π

I If f (θ) = f (θ + π), the curve has central symmetry about the pole or origin. Here it is enough to draw the graph in either the upper or right half plane and then rotate by 180 degree to get the other half. [Example r = sin 2θ.]

2 3 Π Π

4 4

Π 0

5 Π 7 Π

4 4 3 Π

π I If f (θ) = f (π − θ), the curve is symmetric about the vertical line θ = 2 . It is enough to draw either the right half or the left half of the curve in this case. [Example r = 6 sin θ.]

2 8. 3 Π Π

4 6. 4

Π 0. 0

5 Π 7 Π

4 4

3 Π

Annette Pilkington Lecture 36: Polar Coordinates I Since cos(4θ) = cos(4(−θ)), we have symmetry with respect to the x-axis. I Since cos(4θ) = cos(4(π − θ)) = cos(4π − 4θ), we have symmetry with respect to the x-axis. π I Therefore it is enough to draw part of the graph for 0 ≤ θ ≤ 2 and ﬁnd the rest of the graph (for 0 ≤ θ ≤ 2π or −π ≤ θ ≤ π) by symmetry. θ r = cos(4θ) 0 1 π 8 0 I π 4 −1 3π 8 0 π 2 1

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

Annette Pilkington Lecture 36: Polar Coordinates I Since cos(4θ) = cos(4(π − θ)) = cos(4π − 4θ), we have symmetry with respect to the x-axis. π I Therefore it is enough to draw part of the graph for 0 ≤ θ ≤ 2 and ﬁnd the rest of the graph (for 0 ≤ θ ≤ 2π or −π ≤ θ ≤ π) by symmetry. θ r = cos(4θ) 0 1 π 8 0 I π 4 −1 3π 8 0 π 2 1

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

IISince cos(4θ) = cos(4(−θ)), we have symmetry with respect to the x-axis.

Annette Pilkington Lecture 36: Polar Coordinates π I Therefore it is enough to draw part of the graph for 0 ≤ θ ≤ 2 and ﬁnd the rest of the graph (for 0 ≤ θ ≤ 2π or −π ≤ θ ≤ π) by symmetry. θ r = cos(4θ) 0 1 π 8 0 I π 4 −1 3π 8 0 π 2 1

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

IISince cos(4θ) = cos(4(−θ)), we have symmetry with respect to the x-axis. I Since cos(4θ) = cos(4(π − θ)) = cos(4π − 4θ), we have symmetry with respect to the x-axis.

Annette Pilkington Lecture 36: Polar Coordinates θ r = cos(4θ) 0 1 π 8 0 I π 4 −1 3π 8 0 π 2 1

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

IISince cos(4θ) = cos(4(−θ)), we have symmetry with respect to the x-axis. I Since cos(4θ) = cos(4(π − θ)) = cos(4π − 4θ), we have symmetry with respect to the x-axis. π I Therefore it is enough to draw part of the graph for 0 ≤ θ ≤ 2 and ﬁnd the rest of the graph (for 0 ≤ θ ≤ 2π or −π ≤ θ ≤ π) by symmetry.

Annette Pilkington Lecture 36: Polar Coordinates Π

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

3 Π Π

4 4 1.

Π 0

5 Π 7 Π

4 4

3 Π

Example 7 Sketch the rose r = cos(4θ)

Π Π Π

2 2 2

3 Π Π 3 Π Π 3 Π Π

4 4 4 4 4 4 1. 1. 1.

Π 0 Π 0 Π 0

5 Π 7 Π 5 Π 7 Π 5 Π 7 Π

4 4 4 4 4 4

3 Π 3 Π 3 Π

2 2 2

I I We reﬂect in the y-axis to get the graph for 0 ≤ θ ≤ π and subsequently in the x axis to get the graph for −π ≤ θ ≤ π.

We have many equations of circles with polar coordinates: r = a is the circle centered at the origin of radius a, r = 2a sin θ is the circle of radius a centered π at (a, 2 ) (on the y-axis), and r = 2a cos θ is the circle of radius a centered at (a, 0) (on the x-axis). Below, we show the graphs of r = 2, r = 4 sin θ and r = 4 cos θ.

Π Π Π 2 2 2 5. 5. 3 Π Π 3 Π Π 3 Π Π 4. 4. 4 4 4 4 4 4 3. 3.

2. 2. 1. 1. 1.

Π 0 Π 0 Π 0

5 Π 7 Π 5 Π 7 Π 5 Π 7 Π 4 4 4 4 4 4 3 Π 3 Π 3 Π 2 2 2

If we want to ﬁnd the equation of a tangent line to a curve of the form r = f (θ), we write the equation of the curve in parametric form, using the parameter θ.

x = r cos θ = f (θ) cos θ, y = r sin θ = f (θ) sin θ

From the calculus of parametric equations, we know that if f is diﬀerentiable and continuous we have the formula:

dy dy f 0(θ) sin θ + f (θ) cos θ dr sin θ + r cos θ = dθ = = dθ dx dx f 0(θ) cos θ − f (θ) sin θ dr dθ dθ cos θ − r sin θ Note As usual, we locate horizontal tangents by identifying the points where dy/dx = 0 and we locate vertical tangents by identifying the points where dy/dx = ∞

If we want to ﬁnd the equation of a tangent line to a curve of the form r = f (θ), we write the equation of the curve in parametric form, using the parameter θ.

x = r cos θ = f (θ) cos θ, y = r sin θ = f (θ) sin θ

From the calculus of parametric equations, we know that if f is diﬀerentiable and continuous we have the formula:

Annette Pilkington Lecture 36: Polar Coordinates I We have parametric equations x = θ cos θ and y = θ sin θ. dy dy/dθ sin θ+θ cos θ I dx = dx/dθ = cos θ−θ sin θ π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ).

π Example 8 Find the equation of the tangent to the curve r = θ when θ = 2

Annette Pilkington Lecture 36: Polar Coordinates dy dy/dθ sin θ+θ cos θ I dx = dx/dθ = cos θ−θ sin θ π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ).

π Example 8 Find the equation of the tangent to the curve r = θ when θ = 2

I We have parametric equations x = θ cos θ and y = θ sin θ.

Annette Pilkington Lecture 36: Polar Coordinates π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ).

π Example 8 Find the equation of the tangent to the curve r = θ when θ = 2

I We have parametric equations x = θ cos θ and y = θ sin θ. dy dy/dθ sin θ+θ cos θ I dx = dx/dθ = cos θ−θ sin θ

Annette Pilkington Lecture 36: Polar Coordinates π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ).

π Example 8 Find the equation of the tangent to the curve r = θ when θ = 2

I We have parametric equations x = θ cos θ and y = θ sin θ. dy dy/dθ sin θ+θ cos θ I dx = dx/dθ = cos θ−θ sin θ π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π

π Example 8 Find the equation of the tangent to the curve r = θ when θ = 2

I We have parametric equations x = θ cos θ and y = θ sin θ. dy dy/dθ sin θ+θ cos θ I dx = dx/dθ = cos θ−θ sin θ π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ).

π Find the equation of the tangent to the curve r = θ when θ = 2

-2 2 4 6

-1

-2

-3

-4

I We have parametric equations x = θ cos θ and y = θ sin θ. and dy dy/dθ sin θ+θ cos θ dx = dx/dθ = cos θ−θ sin θ π dy 1+0 2 I When θ = , = π = − . 2 dx 0− 2 π π I When θ = 2 , the corresponding point on the curve in polar co-ordinates is π π π given by ( 2 , 2 ) and in Cartesian co-ordinates by (0, 2 ). π I Therefore the equation of the tangent line, when θ = 2 is given by π 2 (y − 2 ) = − π x.

Annette Pilkington Lecture 36: Polar Coordinates I In parametric form, we have x = r cos θ = 6 sin θ cos θ and y = r sin θ = 6 sin2 θ. dy dy/dθ 12 sin θ cos θ 2 sin θ cos θ I We have dx = dx/dθ = 6 cos2 θ−6 sin2 θ = cos2 θ−sin2 θ √ √ π dy 2 3/4 I When θ = 3 , we have dx = (1/4)−(3/4) = − 3. √ π 3 3 9 I When θ = , x = and y = . Therefore the equation of the tangent 3 2 √ 2 √ 9 3 3 at this point is y − 2 = − 3(x − 2 ).

I The tangent is horizontal when dy/dθ = 0 (as long as dx/dθ 6= 0). I dy/dθ = 0 if 12 sin θ cos θ = 0, this happens if either sin θ = 0 or cos θ = 0. That is if θ = nπ/2 for any integer n. (Note dx/dt 6= 0 at any of these poinrs).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

π (a) Find the equation of the tangent to the circle r = 6 sin θ when θ = 3

(b) At which points do we have a horizontal tangent?

Annette Pilkington Lecture 36: Polar Coordinates dy dy/dθ 12 sin θ cos θ 2 sin θ cos θ I We have dx = dx/dθ = 6 cos2 θ−6 sin2 θ = cos2 θ−sin2 θ √ √ π dy 2 3/4 I When θ = 3 , we have dx = (1/4)−(3/4) = − 3. √ π 3 3 9 I When θ = , x = and y = . Therefore the equation of the tangent 3 2 √ 2 √ 9 3 3 at this point is y − 2 = − 3(x − 2 ).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

π (a) Find the equation of the tangent to the circle r = 6 sin θ when θ = 3 I In parametric form, we have x = r cos θ = 6 sin θ cos θ and y = r sin θ = 6 sin2 θ.

(b) At which points do we have a horizontal tangent?

Annette Pilkington Lecture 36: Polar Coordinates √ √ π dy 2 3/4 I When θ = 3 , we have dx = (1/4)−(3/4) = − 3. √ π 3 3 9 I When θ = , x = and y = . Therefore the equation of the tangent 3 2 √ 2 √ 9 3 3 at this point is y − 2 = − 3(x − 2 ).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

(b) At which points do we have a horizontal tangent?

Annette Pilkington Lecture 36: Polar Coordinates √ π 3 3 9 I When θ = , x = and y = . Therefore the equation of the tangent 3 2 √ 2 √ 9 3 3 at this point is y − 2 = − 3(x − 2 ).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

(b) At which points do we have a horizontal tangent?

Annette Pilkington Lecture 36: Polar Coordinates I The tangent is horizontal when dy/dθ = 0 (as long as dx/dθ 6= 0). I dy/dθ = 0 if 12 sin θ cos θ = 0, this happens if either sin θ = 0 or cos θ = 0. That is if θ = nπ/2 for any integer n. (Note dx/dt 6= 0 at any of these poinrs).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

π (a) Find the equation of the tangent to the circle r = 6 sin θ when θ = 3 I In parametric form, we have x = r cos θ = 6 sin θ cos θ and y = r sin θ = 6 sin2 θ. dy dy/dθ 12 sin θ cos θ 2 sin θ cos θ I We have dx = dx/dθ = 6 cos2 θ−6 sin2 θ = cos2 θ−sin2 θ √ √ π dy 2 3/4 I When θ = 3 , we have dx = (1/4)−(3/4) = − 3. √ π 3 3 9 I When θ = , x = and y = . Therefore the equation of the tangent 3 2 √ 2 √ 9 3 3 at this point is y − 2 = − 3(x − 2 ). (b) At which points do we have a horizontal tangent?

Annette Pilkington Lecture 36: Polar Coordinates I dy/dθ = 0 if 12 sin θ cos θ = 0, this happens if either sin θ = 0 or cos θ = 0. That is if θ = nπ/2 for any integer n. (Note dx/dt 6= 0 at any of these poinrs).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

I The tangent is horizontal when dy/dθ = 0 (as long as dx/dθ 6= 0).

Annette Pilkington Lecture 36: Polar Coordinates I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

Annette Pilkington Lecture 36: Polar Coordinates 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0).

I The tangent is vertical when dx/dθ = 0 (as long as dy/dθ 6= 0). 2 2 I This happens if cos θ = sin θ, which is true if cos θ = ± sin θ. True if π θ = (2n + 1) 4 .

3 Π Π

4 6. 4 6 5. 4. 3. 2. 4 1. Π 0

5 Π 7 Π

4 4

-2 2 4 3 Π

Annette Pilkington Lecture 36: Polar Coordinates