arc length and curvature by dr. julia arnold. objectives: 1.find the arc length of a space curve....
TRANSCRIPT
Arc Length and Arc Length and CurvatureCurvature
Arc Length and Arc Length and CurvatureCurvature
By Dr. Julia ArnoldBy Dr. Julia Arnold
Objectives:
1. Find the arc length of a space curve.
2. Use the arc length parameter to describe a plane curve or space curve.
3. Find the curvature of a curve at a point on the curve.
4. Use a vector-valued function to find frictional force.
Objective 1Objective 1Objective 1Objective 1
1.1. Find the arc length of a Find the arc length of a space curve.space curve.
Given a smooth plane curve C that has parametric equations
x = x(t) and y = y(t) where , the arc length s is given bya t b
2 2( ) ( )
b
a
s x t y t dt
In vector form, where C is given by r(t)=x(t)i + y(t)j, the above equation can be written as
( )b
a
s t dtrWe can extend this formula to space quite naturally as follows:
(See Section 10.3)
2 2 2( ) ( ) ( ) ( )
b b
a a
s x t y t z t dt t dt r
If C is a smooth curve given by r(t)= x(t)i + y(t)j +z(t)k on an interval [a,b], then the arc length C on the interval is
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
To help you visualize what is taking place look at the curve and imagine taking steps from point to point.
Now let the points get closer and closer together and sum them up.
Let’s look a a few examples:
Example 1: Find the length of the space curve over the given interval.( ) 2sin ,5 ,2cos 0,r t t t t
Set up the integral
Find the derivative of the vector.
Substitute into the formula.
Simplify
Integrate and evaluate.
2 2
0
2 2
0
0 0
( ) 2sin ,5 ,2cos 0,
( ) 2cos ,5, 2sin
4cos 25 4sin
4(sin cos ) 25
4 25 29 29 290
t t t t
t t t
t t dt
t t dt
dt dt t
r
r
Objective 2Objective 2Objective 2Objective 2
2. Use the arc length parameter 2. Use the arc length parameter to describe a plane curve or to describe a plane curve or space curve.space curve.
Curves can be represented by vector-valued functions in different ways depending on the choice of parameter.
For example the following two representations are equivalent.
For motion along a curve the most convenient parameter is time t. However, for studying the geometric properties of a curve, the convenient parameter is often arc length s.
2 3 2 3
2 2 3 3
2 3
( ) , , 1 2 ( ) , , 0 ln 2
ln , , ,
ln1 ln ln 2 0 ln 2
( ) , , 0 ln 2
u u u
u u u
u u u
t t t t on t and u e e e on u
Let u t then e t e t e t
and t u
thus
u e e e on u
r r
r
Proof
2 2 2( ) ( ) ( ) ( ) ( )
t t
a a
s t u du x u y u z u du r
If C is a smooth curve given by r(t)= x(t)i + y(t)j +z(t)k on an interval [a,b], then the arc length of C on the interval [a,b] , with a<t<b is C = s(t) which is
The arc length s is called the arc length parameter.
Example 2: Consider the curve represented by the vector-valued function
23( ) 4 sin cos ,4 cos sin ,
2r t t t t t t t t
A. Write the length of the arc s as a function of t by evaluating the integral:
2 2 2
0
( ) ( ) ( )t
s x u y u z u du Solution:
2 2 2 2 2
0
2 2
0
22 2
0 0
( ) 4 cos sin cos ,4 sin cos sin ,3
( ) 4 sin ,4 cos ,3
16 sin 16 cos 9
16 9
525 5 5
02 2
t
t
t t
r t t t t t t t t t t
r t t t t t t
s u u u u u du
s u u du
tus u du s udu t
Example 2: Consider the curve represented by the vector-valued function
23( ) 4 sin cos ,4 cos sin ,
2r t t t t t t t t
B. Solve for t in part A and substitute the result into the original set of parametric equations. This yields a parameterization of the curve in terms of the arc length parameter s.
2
2 2 2 2 2 2 3 2( ( )) 4 sin cos ,4 cos sin ,
5 5 5 5 5 5 2 5
s s s s s s sr t s
Solution:
2 25 2 2
2 5 5
s ss t t t
Example 2: Consider the curve represented by the vector-valued function
23( ) 4 sin cos ,4 cos sin ,
2r t t t t t t t t
C. Find the coordinates of the point on the curve for arc lengths
2
2 5 2 5 2 5 2 5 2 5 2 5 3 2 5( ( 5)) 4 sin cos ,4 cos sin ,
5 5 5 5 5 5 2 5
( ( 5)) 1.030,5.408,1.342
r t
r t
Solution:
2 55
5s so t
Example 2: Consider the curve represented by the vector-valued function
23( ) 4 sin cos ,4 cos sin ,
2r t t t t t t t t
C. Find the coordinates of the point on the curve for arc lengths
2
8 8 8 8 8 8 3 8( (4)) 4 sin cos ,4 cos sin ,
5 5 5 5 5 5 2 5
8 8 8 8 8 8( (4)) 4 sin cos ,4 cos sin ,2.4
5 5 5 5 5 5
( (4)) 2.291,6.029,2.4
r t
r t
r t
Solution:
2(4) 84
5 5s thus t
Example 2: Consider the curve represented by the vector-valued function
23( ) 4 sin cos ,4 cos sin ,
2r t t t t t t t t
D. Verify that
Solution:
( ) 1r s
2
2 2 2 2 2 2 3 2( ( )) 4 sin cos ,4 cos sin ,
5 5 5 5 5 5 2 5
2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 1 2( ) 4 cos sin cos ,4 sin cos
5 2 5 5 5 2 5 2 5 5 5 2 5 5 5 2 5
s s s s s s sr t s
s s s s s s sr s
s s s s s
2 2
1 2 2 3sin ,
2 5 5 5
4 2 4 2 3( ) sin , cos ,
5 5 5 5 5
16 2 16 2 9 16 9( ) sin cos 1
25 5 25 5 25 25 25
s
s
s sr s
s sr s
This brings us to a Theorem about the arc length parameter, namely
If C is a smooth curve given by ( ) ( ) ( ) ( ) ( ) ( )s x s y s s x s y s z(s) r i j or r i j + k
Where s is the arc length parameter, then( ) 1s r
Moreover, if t is any parameter for the vector-valued function r such that
Then t must be the arc length parameter.
( ) 1t r
This theorem is stated without proof.
Objective 3Objective 3Objective 3Objective 3
3. Find the curvature of a curve 3. Find the curvature of a curve at a point on the curve.at a point on the curve.
Curvature
An important use of the arc length parameter is to find curvature.Curvature is the measure of how sharply the curve bends.
For example, in this helix we get more bend
here
Than here.
We can calculate curvature by calculating the magnitude of the rate of change of the unit tangent vector T with respect to the arc length s.
T1
T2 T3
Definition of Curvature
Let C be a smooth curve ( in the plane or in space) given by r(s), where s is the arc length parameter. The curvature K at s is given by
dK (s)
ds
TT
Example 3: Find the curvature using s is the arc length parameter, for
2( ) 4(sin cos ) 4(cos sin )3
t t t t t t t t2
r i j + k
dK (s)
ds
TT
Solution: This was the problem we did earlier and found the arc length parameter to be: and the function to be
Using the formula for curvature K in terms of arc length s, namelyand knowing that
( )( ) ( ) 1
( )
r sT s and r s
r s
we get:
2
2 2 2 2 2 2 3 2( ( )) 4 sin cos ,4 cos sin ,
5 5 5 5 5 5 2 5
s s s s s s sr t s
2
5
st
4 2 4 2 3( ) ( ) sin , cos ,
5 5 5 5 5
s sT s r s Since curvature K is
Example 3: Find the curvature using s is the arc length parameter, for
2( ) 4(sin cos ) 4(cos sin )3
t t t t t t t t2
r i j + k
dK (s)
ds
TT
Solution:
2 2
2 2
4 2 4 2 3( ) ( ) sin , cos ,
5 5 5 5 5
1 4 2 2 1 4 2 2( ) ( ) cos , sin ,0
2 5 5 5 2 5 5 5
2 2 2 2 2 2( ) ( ) cos , sin ,0
5 5 5 5 5 5
4 2 2 4 2 2( ) cos sin 0
25 5 5 25 5 5
8 2 2( ) cos sin
125 5 5
8 2 2( )
125
s sT s r s
s sT s r s
s s
s sT s r s
s s
s sT s
s s
s sT s
s
T ss
5 2 10
255 5 5
s s
ss s
xy
Using winplot, this is the curve in question.
Since s = 25
2t
In terms of t the curvature would be
22 2
2 22 2
52 10
2 10 2 5 5 4 25 20 425 12525 125 125 25252 2
ts t t t
Ks t t tt t
c
We can see that the curvature of a circle is the same everywhere and reason it to be a constant which turns out to be 1/r where r is the radius of the circle.
See example 4 in your text.
Other formulas for curvature.
Since the previous definition depends on the arc length parameter, it might be good to have some alternative definitions which depend on an arbitrary parameter t.
Two formulas for curvature
Theorem 12.8
If C is a smooth curve given by r(t), then the curvature K of C at t is given by
3
( ) ( ) ( )
( ) ( )
t t tK
t t
T r r
r r
( ) 4 sin ,4 cos ,3
( ) 5
r t t t t t t
r t t
Example 4: using the alternative curvature formula on the same vector-valued function
2( ) 4(sin cos ) 4(cos sin )3
t t t t t t t t2
r i j + k
We can compare our answers. From Example 2A we already know that
Next we need to find T(t).
4 sin ,4 cos ,3( ) 4sin 4cos 3( ) , ,
( ) 5 5 5 5
t t t t tt t tT t
t t
r
r
Now we need T’(t)4sin 4cos 3
( ) , ,5 5 5
4cos 4sin( ) , ,0
5 5
t tT t
t tT t
3
( ) ( ) ( )
( ) ( )
t t tK
t t
T r r
r rUsing the formula
Now we find ( )T t
2 2
4cos 4sin( ) , ,0
5 5
16 16 4( ) cos sin
25 25 5
t tT t
T t t t
4( ) 45( ) 5 25
tK
t t t
T
r
Which is what we got back on slide 37
Click on the purple crayon to get back to this slide.
3
( ) ( )
( )
t tK
t
r r
r
Using the other formula
We have
( ) 4 sin ,4 cos ,3
( ) 5
( ) 4 cos 4sin , 4 sin 4cos ,3
( ) ( ) 4 sin 4 cos 3
4 cos 4sin 4 sin 4cos 3
12 cos 3 ( 4 cos 4sin ) ( 4 sin )( 4 sin 4cos )
3 ( 4 sin 4co
r t t t t t t
r t t
we need r t t t t t t t
i j k
r t r t t t t t t
t t t t t t
t t i t t t t j t t t t t k
t t t
2 2
2 2 2 2
2 2 2 2 2 2
s ) ( 12 sin ) 4 cos ( 4 cos 4sin )
(12 cos 12 sin 12 cos ) ( 12 cos 12 sin 12 sin )
(16 sin 16 sin cos 16 cos 16 cos sin )
(12 sin ) ( 12 cos ) (16 sin 16 cos ) (12
t i t t j t t t t t k
t t t t t t i t t t t t t j
t t t t t t t t t t k
t t i t t j t t t t k t
2 2 2
4 2 4 2 4 4 4 4 2
sin ) ( 12 cos ) (16 )
( ) ( )
144 sin 144 cos 256 144 256 400 20
t i t t j t k
we need r t r t
t t t t t t t t t
2
3 3
( ) ( ) 20 4
125 25( )
t t tK
t tt
r r
r
4( ) 45( ) 5 25
tK
t t t
T
r
Example 5
At (4,0) the curvature would be:
3 3 52 2 2 2
1 , 4, 1 2 12
1
21 1
12 2 .177
[1 ( 1) ] [2] 2
xSince y then at x y
y
K
Thus 25/2 would be the radius of the circle which would be approximately 5.66
x
y
Using winplot I found the normal to the tangent line at x = 4 (blue line) and then found the center to be approximately at (0,-4).
Figure 12.37
Solution to question on previous slide.
Objective 4Objective 4Objective 4Objective 4
Use a vector-valued function to find frictional force.
Example 6
2 3
2
2 4 2 2 2
2
2
2
22
2 2 2 2 2
4 22 2
22
22
1( ) 2
3
2 2
4 4 ( 2) 2
2 2( )
2
4 4 2 4( )
2
16 4 2 16 16 16( )
2 2
2 2)2 4 4( )
2
2 4
24
t t t t
(t)= t t
(t) t t t t
(t) t tt
(t) t
t t tt
t
t t t t 16 - 16t +4t tt
t t
(tt tt
t
2
r i j k
r i j k
r
r i j kT
r
i + j kT
+ +T
T 2 22
2
22 tt
Solution:
Continued
Example 6
2
2
2
22 2
2
2( )
22
22( ) 2 2
ds(t) t which is also
dt
tt
(t) tKt t t
r
T
T
r
Solution continued
2
2
2
22
22
2
2
2
22 2
2
dst
dt
d sa t
dt
dsa K t
dt t
T
N
Figure 12.38
Hint for HWConfusion of Mass and Weight
A few further comments should be added about the single force which is a source of much confusion to many students of physics - the force of gravity. The force of gravity acting upon an object is sometimes referred to as the weight of the object. Many students of physics confuse weight with mass. The mass of an object refers to the amount of matter that is contained by the object; the weight of an object is the force of gravity acting upon that object. Mass is related to how much stuff is there and weight is related to the pull of the Earth (or any other planet) upon that stuff. The mass of an object (measured in kg) will be the same no matter where in the universe that object is located. Mass is never altered by location, the pull of gravity, speed or even the existence of other forces. For example, a 2-kg object will have a mass of 2 kg whether it is located on Earth, the moon, or Jupiter; its mass will be 2 kg whether it is moving or not (at least for purposes of our study); and its mass will be 2 kg whether it is being pushed upon or not.
On the other hand, the weight of an object (measured in Newtons) will vary according to where in the universe the object is. Weight depends upon which planet is exerting the force and the distance the object is from the planet. Weight, being equivalent to the force of gravity, is dependent upon the value of g.
On earth's surface g is 9.8 m/s2 (often approximated as 10 m/s2) or 32 feet/s2. On the moon's surface, g is 1.7 m/s2. Go to another planet, and there will be another g value. Furthermore, the g value is inversely proportional to the distance from the center of the planet. So if we were to measure g at a distance of 400 km above the earth's surface, then we would find the g value to be less than 9.8 m/s2.
