3 - Arc Length and Arc Length Parametrization - Class
3 - Arc Length and Arc Length Parametrization - Class
3 - Arc Length and Arc Length Parametrization - Class
Consider an object moving at constant (uniform) speed over time. The object travels for
a distance and the distance (or arc length) can be modelled as multiplication of speed
and time spent.
If the object is travelling with varying speed, one may compute the distance by means of
integration:
You may find similarity between the result above and the way we find area of a
rectangle by integration.
1
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
Formally, we have
Theorem 3.1
Given a parametric curve 𝒓𝒓(𝐷𝐷). The arc length of the given curve from the point 𝐷𝐷 = 𝐷𝐷
to the point 𝐷𝐷 = 𝑏𝑏 is given by
𝑏𝑏
� |𝒓𝒓′ (𝐷𝐷)|𝑠𝑠𝐷𝐷
𝑎𝑎
In the previous section, we discussed 𝒓𝒓(𝐷𝐷) = 〈𝑥𝑥(𝐷𝐷), 𝑦𝑦(𝐷𝐷), 𝑧𝑧(𝐷𝐷)〉 over some intervals,
and let’s recall they are parametric equation of a curve. We will also have the following
and
2 2 2
|𝒓𝒓′ (𝐷𝐷)| = ��𝑥𝑥 ′ (𝐷𝐷)� + �𝑦𝑦 ′ (𝐷𝐷)� + �𝑧𝑧 ′ (𝐷𝐷)�
If 2-D case is concerned, we simply regard the last component 𝑧𝑧(𝐷𝐷) to be zero, the result
still holds.
2
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
Example 3.1
Given a helix 𝒓𝒓(𝐷𝐷) = 〈cos 𝐷𝐷 , sin 𝐷𝐷 , 𝐷𝐷〉. Find the arc length from 𝐷𝐷 = 0 to 𝐷𝐷 = 𝜋𝜋.
𝜋𝜋 𝜋𝜋
= � |𝒓𝒓(𝐷𝐷)|𝑠𝑠𝐷𝐷 = � √2 𝑠𝑠𝐷𝐷 = √2𝜋𝜋
0 0
Example 3.2
Find the arc length of the curve 𝒓𝒓(𝐷𝐷) = 𝐷𝐷𝒊𝒊 + 3 sin 𝐷𝐷 𝒋𝒋 + 3 cos 𝐷𝐷 𝒌𝒌 from (0, 0, 3) to
(2𝜋𝜋, 0, 3).
Firstly, we need to check the starting and ending in terms of 𝐷𝐷 . The starting points
corresponds to 𝐷𝐷 = 0 for the point (0, 0, 3) and the ending point corresponds to 𝐷𝐷 = 2𝜋𝜋
for the point (2𝜋𝜋, 0, 3). (Verify it.)
Then, we need
= √1 + 9 = √10
3
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
= 2√10𝜋𝜋
Note that although the last two examples return |𝒓𝒓′ (𝐷𝐷)| with a number, it, in general,
may not be the case and the computation will become somewhat complicated.
Consider that if we wish to find out the general expression of arc length in terms of , we
term this function arc length function 𝐷𝐷(𝐷𝐷) see the next example.
Example 3.3
Find the arc length function for the curve in Example 3.2. From previous example, we
have
𝑡𝑡
𝐷𝐷(𝐷𝐷) = � √10 𝑠𝑠𝑥𝑥
0
= √10𝐷𝐷
y so as to avoid confusion of 𝐷𝐷 and 𝑠𝑠𝐷𝐷 since 𝑠𝑠𝑥𝑥 and 𝑠𝑠𝐷𝐷 carry the same physical meaning
and result in the computation.
4
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
𝐷𝐷(𝐷𝐷)
𝐷𝐷 = .
√10
We hope to plug in the just solved 𝐷𝐷 back to the original curve 𝒓𝒓(𝐷𝐷).
By so doing, we can simply tell where the object is located on the curve if we know the
distance of 𝐷𝐷 along the curve.
Note that we begin the measurement of distance travelled from the location/point at 𝐷𝐷
equal to zero. i.e. 𝐷𝐷 = 0.
This technique is called reparametrization. The function 𝒓𝒓(𝐷𝐷) is reparametrized into the
form
𝒓𝒓�𝐷𝐷(𝐷𝐷)�.
𝐷𝐷 𝐷𝐷 𝐷𝐷
𝒓𝒓�𝐷𝐷(𝐷𝐷)� = 〈 , 3 sin , 3 cos 〉
√10 √10 √10
|𝒓𝒓′ (𝐷𝐷)| = 1
5
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
Consider
Also, we have
𝐷𝐷
𝐷𝐷 = � �𝒓𝒓′ (𝑢𝑢)� 𝑠𝑠𝑢𝑢.
𝐷𝐷
𝑠𝑠𝐷𝐷
= |𝒓𝒓′ (𝐷𝐷)|.
𝑠𝑠𝐷𝐷
Therefore,
𝑠𝑠𝐷𝐷
|𝒓𝒓′ (𝐷𝐷)| = |𝒓𝒓′ (𝐷𝐷)| � �
𝑠𝑠𝐷𝐷
1
= |𝒓𝒓′ (𝐷𝐷)| ∙ =1
|𝒓𝒓′ (𝐷𝐷)|
for any 𝐷𝐷, we say that the curve is parametrized by arc length and the curve 𝒓𝒓(𝐷𝐷) moves
with unit speed.
One should be aware that it may be complicated to find an arc length parametrization
when the given curve 𝒓𝒓(𝐷𝐷) is not nice to work with…
6
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
𝐷𝐷
𝐷𝐷 = � �𝒓𝒓′ (𝑢𝑢)� 𝑠𝑠𝑢𝑢
𝐷𝐷
2) From 1), we note that the upper limit of the integral is 𝐷𝐷, the function 𝐷𝐷 should be a
function of 𝐷𝐷 after integration. When the integration gets done, we express 𝐷𝐷 in terms of
𝐷𝐷 whenever possible, then 𝐷𝐷 becomes a function of 𝐷𝐷, i.e.
𝐷𝐷 = 𝐷𝐷(𝐷𝐷).
3) Substitute all quantities related to 𝐷𝐷 with 𝐷𝐷 = 𝐷𝐷(𝐷𝐷) in the given curve 𝒓𝒓(𝐷𝐷).
Example 3.4
Recall Example 3.1 find the arc length parametrization of the helix,
from the point (1,0,0) with the same orientation as the given helix.
7
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ENGG1130
2020-2021 Term 2 Topic 3 – Arc length and Arc length Parametrization
|𝒓𝒓′ (𝐷𝐷)| = √2
(Refer to Example 3.1 for details)
𝑡𝑡 𝑡𝑡
′ (𝑢𝑢)
∴ 𝐷𝐷 = � �𝒓𝒓 �𝑠𝑠𝑢𝑢 = � √2 𝑠𝑠𝑢𝑢 = √2𝐷𝐷
0 0
𝑠𝑠
Therefore, 𝐷𝐷 = and the reparametrization of the helix in terms of 𝐷𝐷 become
√2
𝐷𝐷 𝐷𝐷 𝐷𝐷
𝒓𝒓(𝐷𝐷) = 〈cos , sin , 〉.
√2 √2 √2
Practice:
3
𝒓𝒓(𝐷𝐷) = 3𝐷𝐷 2 𝒊𝒊 + √32𝐷𝐷 2 𝒋𝒋 + 6𝐷𝐷𝒌𝒌
ii) Given (𝐷𝐷) = 〈4𝐷𝐷 , −2𝐷𝐷 , √5𝐷𝐷 2 〉 , show the arc length function
Suggested Answer:
i)
3
2
2
𝒓𝒓(𝐷𝐷(𝐷𝐷)) = 〈3�√9 + 3𝐷𝐷 − 3� , √32 �√9 + 3𝐷𝐷 − 3� , 6(�√9 + 3𝐷𝐷 − 3�〉
8
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