areas and approximations - math.berkeley.edu
TRANSCRIPT
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Areas and Approximations
Jeff Hicks
Apr. 17, 2019
UC Berkeley
Jeff Hicks (UC Berkeley) Areas and Approximations Apr. 17, 2019 1 / 28
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Outline
• Computing Areas of Triangles
• Computing Areas of other polygons
• Approximations of sin(θ) and cos(θ) for small θ.
• Calculating distances
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Area of a Triangle
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Area of a Triangle
Computing the Area of a Triangle
Formula
A triangle has area b·h2 where b is the length of one of the sides (the
“base”), and h is the height of the triangle as measured by an altitude.
b
h
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Area of a Triangle
How can we use trig to help
Let’s take a closer look at this triangle:
b
hc
θ
We see that the height h = c · sin(θ). This makes the area
1
2b · h =
1
2b · c · sin(θ)
Notice that the area formula does not contain the height of the triangle.
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Area of a Triangle
How can we use trig to help
Let’s take a closer look at this triangle:
b
hc
θ
We see that the height h = c · sin(θ). This makes the area
1
2b · h =
1
2b · c · sin(θ)
Notice that the area formula does not contain the height of the triangle.
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Area of a Triangle
Formula
The area of a triangle with sides b and c meeting at an angle θ is:
A =1
2b · c · sin(θ)
Example
Find the area of an equalateral triangle which has a side length of 5.
5
5
π3
1
2b · c sin(θ) =
1
2· 5 · 5 · sin(π/3)
=1
2c 2̇5 ·
√3
2
=25√
3
4.
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Area of a Triangle
Formula
The area of a triangle with sides b and c meeting at an angle θ is:
A =1
2b · c · sin(θ)
Example
Find the area of an equalateral triangle which has a side length of 5.
5
5
π3
1
2b · c sin(θ) =
1
2· 5 · 5 · sin(π/3)
=1
2c 2̇5 ·
√3
2
=25√
3
4.
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Area of a Triangle
Some other observations about the rule
When θ is close to 0, our triangle looks like this:
b
c
θ
Notice that even if b and c are large, the area of this triangle will be small.
This is because when θ ∼ 0, then sin(θ) ∼ 0.
1
2b · c · sin(θ) ∼ 1
2b · c · 0 ∼ 0
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Area of a Triangle
Some other observations about the rule
When θ is close to 0, our triangle looks like this:
b
c
θ
Notice that even if b and c are large, the area of this triangle will be small.
This is because when θ ∼ 0, then sin(θ) ∼ 0.
1
2b · c · sin(θ) ∼ 1
2b · c · 0 ∼ 0
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Area of a Triangle
Some other observations about the rule
When θ is close to π/2, our triangle looks like this:
b
c
θ
When θ ∼ 0, then sin(θ) ∼ 1.
1
2b · c · sin(θ) ∼ 1
2b · c
which looks a lot like the formula for the area of a right triangle.
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Area of a Triangle
Some other observations about the rule
When θ is close to π/2, our triangle looks like this:
b
c
θ
When θ ∼ 0, then sin(θ) ∼ 1.
1
2b · c · sin(θ) ∼ 1
2b · c
which looks a lot like the formula for the area of a right triangle.
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Area of a Triangle
Some Other observations about the rule
The formula1
2b · c sin(θ)
treats the variables b and c on equal footing.
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Area of a Triangle
What about other shapes?
b
hc
θ
The area of a parallelogram is
A = b · h
Formula
The area of a parallelogram with sides b and c meeting at an angle θ is:
A = b · c · sin(θ)
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Area of a Triangle
Example
What is the area of the following parallelogram
5
3√
2
π/4
A =b · c · sin(θ)
=5 · 3√
2 · sin(π/4)
=5 · 3√
2 ·√
2
2
=15.
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Area of a Triangle
Example
What is the area of the following parallelogram
5
3√
2
π/4
A =b · c · sin(θ)
=5 · 3√
2 · sin(π/4)
=5 · 3√
2 ·√
2
2
=15.
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Area of a Triangle
Relation between Parallelograms and Triangles
We could have also could have gotten this formula from cutting the
parallelogram into two .
5π/4
π/4
This is a trick we can use to find the area of many different shapes.
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Area of a Triangle
Area of a regular Pentagon
What is the area of a regular pentagon when the distance from the center
to the vertex is 3?
3θ
The area of each triangle is
1
23 · 3 sin(θ)
The angle θ = 2π5 . This gives us that
the area of the whole pentagon is
5·(
1
23 · 3 sin(2π/5)
)=
75
3((√
10 + (2√
5))/4)
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Area of a Triangle
Area of a regular Pentagon
What is the area of a regular pentagon when the distance from the center
to the vertex is 3?
3θ
The area of each triangle is
1
23 · 3 sin(θ)
The angle θ = 2π5 . This gives us that
the area of the whole pentagon is
5·(
1
23 · 3 sin(2π/5)
)=
75
3((√
10 + (2√
5))/4)
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Area of a Triangle
Area of a n-gon
Repeating on a shape with n equal sides and radius r ,
r
Area is
n·(
1
2r · r · sin
(2π
n
))=
n
2
(sin
(2π
n
))r2
This looks suspiciously like the formula
for the area of the circle. Notice that
as n becomes very large, this better and
better approximates the area of a circle.
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Estimates for Trig Functions
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Estimates for Trig Functions
Comparing circles
Our estimate from the last slide says that when n is very large
n
2
(sin
(2π
n
))r2 ≈ π · r2
where the right hand side is the area of a circle.
We conclude that when n is very large,
n
2
(sin
(2π
n
))≈π
sin
(2π
n
)≈2π
n
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Estimates for Trig Functions
Comparing circles
Our estimate from the last slide says that when n is very large
n
2
(sin
(2π
n
))r2 ≈ π · r2
where the right hand side is the area of a circle.
We conclude that when n is very large,
n
2
(sin
(2π
n
))≈π
sin
(2π
n
)≈2π
n
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Estimates for Trig Functions
Letting θ = 2πn when n is very big...
Identity
When θ is very small, sin(θ) ≈ θ.
Remark
This only works if θ is measured in radians! Fun Fact: This is the real
reason why we use radians!
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Estimates for Trig Functions
Identity
When θ is very small, sin(θ) ≈ θ.
sin(θ) θ
A large difference be-
tween sin(θ) and θ
sin(θ) θ
A large difference be-
tween sin(θ) and θ
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Estimates for Trig Functions
In red, we have the graph of y = sin(θ), while in blue we have the graph of
y = θ.
−1 1
−1
0
1
θ
y
0θ
y
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Estimates for Trig Functions
Example
Apprxomate the value of 5 · sin(.01).
5 · sin(0.1) ≈ 5 · (0.1) = .5
Example
Given that sin(θ) = .02, approximate θ.
θ is approximately .02.
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Estimates for Trig Functions
Example
Apprxomate the value of 5 · sin(.01).
5 · sin(0.1) ≈ 5 · (0.1) = .5
Example
Given that sin(θ) = .02, approximate θ.
θ is approximately .02.
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Estimates for Trig Functions
Example
Apprxomate the value of 5 · sin(.01).
5 · sin(0.1) ≈ 5 · (0.1) = .5
Example
Given that sin(θ) = .02, approximate θ.
θ is approximately .02.
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Estimates for Trig Functions
Example
Apprxomate the value of 5 · sin(.01).
5 · sin(0.1) ≈ 5 · (0.1) = .5
Example
Given that sin(θ) = .02, approximate θ.
θ is approximately .02.
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Estimates for Trig Functions
In fact, there are better and better approximations.
Fact
sin(θ) = θ − θ3
3! + θ5
5! −θ7
7! + · · ·
−3−2−1 1 2 3
−2
−1
0
1
2
θ
y
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Estimates for Trig Functions
A similar statement is true of cos(θ).
Identity
When θ is very small, cos(θ) ≈ 1− θ2
2 .
In red, we have the graph of y = cos(θ), while in blue we have the graph
of y = 1− θ2/2.
−1 10
1
θ
y
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Estimates for Trig Functions
Fact
cos(θ) = 1− θ2
2! + θ4
4! −θ6
6! + · · ·
−3−2−1 1 2 3
−2
−1
0
1
2
θ
y
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Estimates for Trig Functions
For Beyond this Course
In almost every field of quantitative study, one desires to study some
function in terms of simpler functions. In this class so far we’ve seen:
ex = 1 + x +x2
2!+
x3
3!+
x4
4!+ · · ·
cos(θ) = 1− θ2
2!+θ4
4!− θ6
6!+ · · ·
sin(θ) = θ − θ3
3!+θ5
5!− θ7
7!+ · · ·
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Estimates for Trig Functions
The approximations which have been important to us in this course come
from considering just the first few terms of these sums:
ex ≈ 1 + x
cos(θ) ≈ 1− θ2
2‘
sin(θ) ≈ θ
We call this a “leading order approximation”. One portion of calculus is
understanding why we can make these approximations in general
situations.
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Estimates for Trig Functions
Measing distance to a star
r
θ
H
The length H is given by H = rsin(θ) .
Since θ is very small, we can approximate sin(θ) ≈ θ
H ≈ r
θ
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Estimates for Trig Functions
Measing distance to a star
r
θ
H
The length H is given by H = rsin(θ) .
Since θ is very small, we can approximate sin(θ) ≈ θ
H ≈ r
θ
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Estimates for Trig Functions
Measing distance to a star
rθ
H
The length H is given by H = rsin(θ) .
Since θ is very small, we can approximate sin(θ) ≈ θ
H ≈ r
θ
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Estimates for Trig Functions
Rule of Thumb for far away things
The distance to something far away is about
H ≈ r
θ
Where θ is the number of radians it moves in your view, and r is the
distance between the two points that you make an observation from.
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Estimates for Trig Functions
Where else does this occur? In humans, when judging the distance to
far away objects! Our depth position is controlled by our abilities to
resolve angles.
• Without additional information, humans can tell the distance of a
point to about 10 meters.
• Human eyes are about 66mm apart.
The angle which a point moves between the left and right eye when at 10
meters is roughly
10 meters =33
sin(θ)≈ 33mm
θ
θ ≈ 0.0033 Radians.
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Estimates for Trig Functions
Where else does this occur? In humans, when judging the distance to
far away objects! Our depth position is controlled by our abilities to
resolve angles.
• Without additional information, humans can tell the distance of a
point to about 10 meters.
• Human eyes are about 66mm apart.
The angle which a point moves between the left and right eye when at 10
meters is roughly
10 meters =33
sin(θ)≈ 33mm
θ
θ ≈ 0.0033 Radians.
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Estimates for Trig Functions
θ ≈ 0.0033 Radians
That angle roughly corresponds to the size of a small beachball at 1
football field away. In fact, humans can resolve angles as small as
0.0003radians.
That’s about a small beachball at a kilometer away. This means that there
are other difficulties involved in measuring distance (for instance, the
relative angle of your eyes to eachother.)
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Estimates for Trig Functions
Short Survey
https://forms.gle/3jQ1h2HAJVab1W5H7
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