Lengths in Polar Coordinatesapilking/Math10560/Lectures/Lecture 37.pdfLengths in Polar...
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Lengths in Polar Coordinates
Given a polar curve r = f (θ), we can use the relationship between Cartesiancoordinates and Polar coordinates to write parametric equations whichdescribe the curve using the parameter θ
x(θ) = f (θ) cos θ y(θ) = f (θ) sin θ
To compute the arc length of such a curve between θ = a and θ = b, we needto compute the integral
L =
Z b
a
r(dx
dθ)2 + (
dy
dθ)2dθ
We can simplify this formula because( dx
dθ)2 + ( dy
dθ)2 = (f ′(θ) cos θ − f (θ) sin θ)2 + (f ′(θ) sin θ + f (θ) cos θ)2
= [f ′(θ)]2 + [f (θ)]2 = r 2 + ( drdθ
)2 Thus
L =
Z b
a
rr 2 + (
dr
dθ)2dθ
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 1
Compute the length of the polar curve r = 6 sin θ for 0 ≤ θ ≤ π
I Last day, we saw that the graph of this equation is a circle of radius3 and as θ increases from 0 to π, the curve traces out the circle exactly once.
0
Π
4
Π
2
3 Π
4
Π
5 Π
4
3 Π
2
7 Π
4
1.
2.
3.
4.
5.
6.
I We have L =R π
0
qr 2 + ( dr
dθ)2dθ
I r = 6 sin θ and drdθ
= 6 cos θ.
I L =R π
0
p36 sin2 θ + 36 cos2 θdθ =
R π0
6p
sin2 θ + cos2 θdθ = 6π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 1
Compute the length of the polar curve r = 6 sin θ for 0 ≤ θ ≤ πI Last day, we saw that the graph of this equation is a circle of radius
3 and as θ increases from 0 to π, the curve traces out the circle exactly once.
0
Π
4
Π
2
3 Π
4
Π
5 Π
4
3 Π
2
7 Π
4
1.
2.
3.
4.
5.
6.
I We have L =R π
0
qr 2 + ( dr
dθ)2dθ
I r = 6 sin θ and drdθ
= 6 cos θ.
I L =R π
0
p36 sin2 θ + 36 cos2 θdθ =
R π0
6p
sin2 θ + cos2 θdθ = 6π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 1
Compute the length of the polar curve r = 6 sin θ for 0 ≤ θ ≤ πI Last day, we saw that the graph of this equation is a circle of radius
3 and as θ increases from 0 to π, the curve traces out the circle exactly once.
0
Π
4
Π
2
3 Π
4
Π
5 Π
4
3 Π
2
7 Π
4
1.
2.
3.
4.
5.
6.
I We have L =R π
0
qr 2 + ( dr
dθ)2dθ
I r = 6 sin θ and drdθ
= 6 cos θ.
I L =R π
0
p36 sin2 θ + 36 cos2 θdθ =
R π0
6p
sin2 θ + cos2 θdθ = 6π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 1
Compute the length of the polar curve r = 6 sin θ for 0 ≤ θ ≤ πI Last day, we saw that the graph of this equation is a circle of radius
3 and as θ increases from 0 to π, the curve traces out the circle exactly once.
0
Π
4
Π
2
3 Π
4
Π
5 Π
4
3 Π
2
7 Π
4
1.
2.
3.
4.
5.
6.
I We have L =R π
0
qr 2 + ( dr
dθ)2dθ
I r = 6 sin θ and drdθ
= 6 cos θ.
I L =R π
0
p36 sin2 θ + 36 cos2 θdθ =
R π0
6p
sin2 θ + cos2 θdθ = 6π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 1
Compute the length of the polar curve r = 6 sin θ for 0 ≤ θ ≤ πI Last day, we saw that the graph of this equation is a circle of radius
3 and as θ increases from 0 to π, the curve traces out the circle exactly once.
0
Π
4
Π
2
3 Π
4
Π
5 Π
4
3 Π
2
7 Π
4
1.
2.
3.
4.
5.
6.
I We have L =R π
0
qr 2 + ( dr
dθ)2dθ
I r = 6 sin θ and drdθ
= 6 cos θ.
I L =R π
0
p36 sin2 θ + 36 cos2 θdθ =
R π0
6p
sin2 θ + cos2 θdθ = 6π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Areas in Polar Coordinates
Suppose we are given a polar curve r = f (θ) and wish to calculate the areaswept out by this polar curve between two given angles θ = a and θ = b. Thisis the region R in the picture on the left below:
Dividing this shape into smaller pieces (on right) and estimating the areas ofthe small pieces with pie-like shapes (center picture), we get a Riemann sumapproximating A = the area of R of the form
A ≈nX
i=1
f (θ∗i )2
2∆θ
Letting n→∞ yields the following integral expression of the area
Z b
a
f (θ)2
2dθ =
Z b
a
r 2
2dθ
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 2
Compute the area bounded by the curve r = sin 2θ for 0 ≤ θ ≤ π2
.
I Using the formula A =R b
af (θ)2
2dθ =
R π2
0(sin 2θ)2
2dθ
I Using the half-angle formula, we get A = 14
R π2
0 1− cos(4θ)dθ
= 14
ˆθ − sin(4θ)
4
˜π/20
= π8
.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 2
Compute the area bounded by the curve r = sin 2θ for 0 ≤ θ ≤ π2
.
I Using the formula A =R b
af (θ)2
2dθ =
R π2
0(sin 2θ)2
2dθ
I Using the half-angle formula, we get A = 14
R π2
0 1− cos(4θ)dθ
= 14
ˆθ − sin(4θ)
4
˜π/20
= π8
.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 2
Compute the area bounded by the curve r = sin 2θ for 0 ≤ θ ≤ π2
.
I Using the formula A =R b
af (θ)2
2dθ =
R π2
0(sin 2θ)2
2dθ
I Using the half-angle formula, we get A = 14
R π2
0 1− cos(4θ)dθ
= 14
ˆθ − sin(4θ)
4
˜π/20
= π8
.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Areas of Region between two curves
If instead we consider a region bounded between two polar curves r = f (θ) andr = g(θ) then the equations becomes
1
2
Z b
a
f (θ)2 − g(θ)2dθ
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Example 3
Example 3 Find the area of the region that lies inside the circle r = 3 sin θ andoutside the cardioid r = 1 + sin θ.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
I We use the formula A = 12
R b
af (θ)2 − g(θ)2dθ
I To find the angles where the curves intersect, we find the values of θwhere 3 sin θ = 1 + sin θ or 2 sin θ = 1. That is sin θ = 1/2
I Therefore θ = π6, 5π
6.
I 12
R b
af (θ)2 − g(θ)2dθ = 1
2
R 5π6
π6
9 sin2 θ − (1 + sin θ)2dθ
I = 12
R 5π6
π6
8 sin2 θ − 1 − 2 sin θdθ = 4R 5π
6π6
sin2 θdθ − 12
R 5π6
π6
1dθ −R 5π
6π6
sin θdθ
= 2R 5π
6π6
1− cos(2θ)dθ − π3
+ cos θ˜5π/6π/6
= 4π3− sin(2θ)
˜5π/6π/6− π
3+√
3= π.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Word of Warning
NOTE The fact that a single point has many representation in polarcoordinates makes it very difficult to find all the points of intersections of twopolar curves. It is important to draw the two curves!!!Example 4 Find all possible intersection points of the curves r = cos 2θ andr = 1
2.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
1.
I Here we must solve two equations 12
= cos(2θ) and − 12
= cos(2θ) sincethe curve r = 1
2is the same as the curve r = − 1
2.
I From the first equation, we get 2θ = ±π3, ± 5π
3and from the second we
get 2θ = ± 2π3, ± 4π
3.
I Therefore we have 8 points of intersection θ = ±π6, ±π
3, ± 2π
3, ± 5π
6.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Word of Warning
NOTE The fact that a single point has many representation in polarcoordinates makes it very difficult to find all the points of intersections of twopolar curves. It is important to draw the two curves!!!Example 4 Find all possible intersection points of the curves r = cos 2θ andr = 1
2.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
1.
I Here we must solve two equations 12
= cos(2θ) and − 12
= cos(2θ) sincethe curve r = 1
2is the same as the curve r = − 1
2.
I From the first equation, we get 2θ = ±π3, ± 5π
3and from the second we
get 2θ = ± 2π3, ± 4π
3.
I Therefore we have 8 points of intersection θ = ±π6, ±π
3, ± 2π
3, ± 5π
6.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Word of Warning
NOTE The fact that a single point has many representation in polarcoordinates makes it very difficult to find all the points of intersections of twopolar curves. It is important to draw the two curves!!!Example 4 Find all possible intersection points of the curves r = cos 2θ andr = 1
2.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
1.
I Here we must solve two equations 12
= cos(2θ) and − 12
= cos(2θ) sincethe curve r = 1
2is the same as the curve r = − 1
2.
I From the first equation, we get 2θ = ±π3, ± 5π
3and from the second we
get 2θ = ± 2π3, ± 4π
3.
I Therefore we have 8 points of intersection θ = ±π6, ±π
3, ± 2π
3, ± 5π
6.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates
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Lengths in Polar Coordinates Areas in Polar Coordinates Areas of Region between two curves Warning
Word of Warning
NOTE The fact that a single point has many representation in polarcoordinates makes it very difficult to find all the points of intersections of twopolar curves. It is important to draw the two curves!!!Example 4 Find all possible intersection points of the curves r = cos 2θ andr = 1
2.
0
Π
6
Π
3
Π
22 Π
3
5 Π
6
Π
7 Π
6
4 Π
3 3 Π
2
5 Π
3
11 Π
6
1.
I Here we must solve two equations 12
= cos(2θ) and − 12
= cos(2θ) sincethe curve r = 1
2is the same as the curve r = − 1
2.
I From the first equation, we get 2θ = ±π3, ± 5π
3and from the second we
get 2θ = ± 2π3, ± 4π
3.
I Therefore we have 8 points of intersection θ = ±π6, ±π
3, ± 2π
3, ± 5π
6.
Annette Pilkington Lecture 37: Areas and Lengths in Polar Coordinates