Multivariable Calculus
Summary 5: Curves in space-curvature
Definition. Unit tangent vector T



v (t )


T (t ) 
 v  vT where v  v
v
Arc-length parameterization of a position vector.

Let r (t )  x(t ), y(t ), z (t ) be a position vector .
The position point (x, y, z) on a curve C is a function of the parameter t.
b
Arc length s   v(t )dt is a function of the parameter t, therefore s=s(t)
a
The position point (x, y, z) on a curve C is a function of the parameter s because it
is uniquely determined by its distance s(t) along the curve.
Thus, the curve C can also be described by a function
of s.


r (t )  x(t ), y (t ), z (t )  x( s), y ( s), z ( s)  R( s)

dR 
Theorem 1.
 T (t ) , the unit tangent vector to C.
ds

d
R





dR dR ds dr
dR dt v (t ) 


 



 T (t ), where v  v (t )
dt ds dt dt
ds ds v(t )
dt

dR 
Conclusion:
 T (t ) , the unit tangent vector to C.
ds


Theorem 2: If u (t ) is a unit vector then u (t ) is orthogonal (perpendicular) to

u (t ) .
Proof:
 2 


Step 1. u (t )  u (t )  u (t )  1 by the fact that u (t )  1
2  
Recall the property v  v  v






Step 2. Differentiating u (t )  u (t )  1 u (t )  u (t )  u (t )  u (t )  0






 2u (t )  u (t )  0  u (t )  u (t )  0  u (t ) and u (t ) are perpendicular
-1-


dT
Theorem 3: T is perpendicular to
ds
Proof:


dT
Step 1. By the previous theorem T and
are perpendicular because
dt

T is a unit vector





dT dT dt dT 1 dT
dT
1
Step 2.

 
 
is parallel to
because is a scalar
ds dt ds dt v
ds
dt
v


dT
Ste 3. Therefore, T is perpendicular to
because if a vector is perpendicular to
ds
one of two parallel vectors, it is also perpendicular to the other.
Definition of curvature:

dT
The curvature  of a curve is defined by  
, magnitude of the change of the
ds
unit tangent vector with respect to arc length s.
Note: curvature has to do with the rate at which the velocity vector turns. A
straight line has zero curvature. The curvature of a circle decreases as the radius
increases.
Definition: the radius of curvature of a curve at a given point, is defined as the
1
reciprocal of the curvature at that point, that is   . They are inversely

proportional, that is, as one increase the other decreases.
We say a line has an infinite radius of curvature because its curvature is zero.
Example: The larger the circle, the smaller the curvature
A circle of radius 10 has curvature of 0.1
A circle of radius 100 has curvature of 0.01
A straight line has 0 curvature, therefore it has infinite radius.
-2-
Theorem 4.

dT
d

which is equivalent to  
in the plane.  is the
ds
ds
angle between T and the positive x  axis.
Proof:

Let  be the angle of inclination of the unit tangent vector T  i cos  j sin 


dT dT d
d


  i sin   j cos 
ds d ds
ds

dT
d
d

  i sin   j cos 
  i sin   j cos 
ds
ds
ds
but  i sin   j cos   1 because  i sin   j cos  is a unit vector


dT
d
dT
d
Therefore

or  

ds
ds
ds
ds

1 dT
Theorem 5.  
v dt
Proof:





dT
dT dt
dT dt
dT 1 1 dT





 
ds
dt ds
dt ds
dt v v dt
Definition of Principal unit Normal vector N

dT




1
d
T
d
T
N  ds 

 N
 ds
ds
dT
ds
  


Definition: The binormal vectoris defined by B  T  N . Since T and N are
perpendicular
unit vectors then B is also a unit vector. The vector triple
  
T , N , B forms a moving trihedral along the curve.
-3-
Tangential and Normal components of the acceleration:
Theorem 6:

dv
a  aT T  a N N where aT 
and a N  v 2
dt
Proof:




dv (t ) d vT
dT dv  dv 
dT ds

a (t ) 

v
 T  T v

dt
dt
dt dt
dt
ds dt


dv 
dT
dv 

2 dT
 a (t )  T  v
v  T  v
dt
ds
dt
ds


dT
but
 kN by definition of principal unit normal
ds
 


dv 

Therefore, a (t )  T  kv 2 N  a (t )  aT  kv 2 N
dt

dv
Note : the component of acceleration on the T is aT   a
dt
 
and the normal component is a n  kv 2
dv
 a is called the scalar acceleration.
dt
 
v a
Theorem 7: k  3
v




 
 
  
 
 
v  a  v  aT  v 2 N  av  T  v 2 v  N  avT  T  v 3T  N



 
 0  v 3 B  v 3 B where B is the unit vector T  N
 
Therefore v  a  v 3
 
va
r   r 
and   3 
3
v
r
-4-
Theorem 8:
 
 


v
a
v

a

a  aT T  a N N where aT 
and a N 
v
v
 
dv v  a

dt
v



  dv  


 
 dv
v  a  vT   T  kv 2 N   vT   T   vT  kv 2 N
 dt

 dt 
 
 
dv  
dv
dv
v  a dv
 
3
3
v
T  T  v T  N  v
(1)  v  (0)  v  a  v


 aT Fro
dT
dT
dT
v
dT
 
v a
Therefore aT 
v
Show that aT 


 
v a
m theorem 8, k  3
v
 
va
 
2
Therefore, v  v  a  v 
 a N
v
Therefore,
3

a  aT T  a N N where
 
 
v
 a r (t )  r (t )
dv v  a r (t )  r (t )
aT 


and a N  v 2 

dt
v
r (t )
v
r (t )

Theorem 9: The principal unit normal vector N can be computed from the formula

 a  aT T
N
a N




Proof: the proof follows from a  aT T  aN N by solving for N
-5-
Theorem 10. For vectors in the plane xy,  
x y   x y 
x
2
  y  2

3/ 2
Proof:

 
 
Let r  x(t ), y (t ), 0 , v  r   x (t ), y (t ), 0 and a  r   x (t ), y (t ), 0
i
 
v  a  x
x 
j k
 
y  0  oi  oj  ( x y   x y )k  v  a  x y   x y 
y  0
 
va
x y   x y 
x y   x y 
Therefore,   3 

3
3/ 2
v
  x 2   y 2 
 x 2   y 2




Theorem 11: if y = f(x) show that the curvature formula is given by
y 

3/ 2
1   y 2
Proof: if y = f(x), use the parametric representation, x(t)=x and y=f(x)
x y   x y 
1y   0 y 
y 



3/ 2
3/ 2
3/ 2
 x  2   y  2
12   y 2
1   y  2








Definition. Osculating circle: the circle that best fits at a point of a curve is called
1
k
the osculating circle. The radius r== . The osculating circle must be tangent to
the curve at the given point.
Theorem 12: The position vector




 r N

 for the center of the osculating circle is given
where r is the position vector of the point in the curve at which the osculating
circle is constructed.
Proof:

1. Let (h, k) be the center of the osculating circle. Therefore   h, k . Let P=(x, y) be the
point on the curve that is used to construct the osculating circle. The vector connecting the given
point P on the curve and the center C of the osculating circle must be perpendicular to the unit


vector T , therefore it must be a scalar multiple of the vector N with length  .Therefore the


vector PC  N . If the point O represents the origin of the coordinate system then by vector






addition, OC  OP  PC    r  N
-6-
Practice problems:
1. Find the curvature, tangential and normal components of the acceleration of the vector

function r ( t )  t , t 2 , t 3  at t=2.
Solution:



r (t )  t , t 2 , t 3  v (t )  1, 2t , 3t 2  a (t )  0, 2, 6t



 at t  2, r (2)  2, 4, 8 , v (2)  1, 4, 12 , a (2)  0, 2, 12 , v  161
i j k




v (2)  a (2)  0  8  144  152, v (2)  a (2)  1 4 12  24,  12, 2
0 2 12


v (2)  a (2)
576  144  4
776
 


 0.013636
3
161
161
v3
161


v (2)  a (2) 152
aT 

 11.979
v
161


v (2)  a (2)
776
a N 

 2.1954
v
161


Curvature: 0.013636_______
Tangential component: _11.979____
Normal component: __2.1954_____
2. Find the curvature of y  x 3at ( 1,1)
answer: __
Method 1.



r (t )  t , t 3 , 0 , v (t )  1, 3t 2 ,0 , a (t )  0, 6t , 0


 at(1,  1), t  1, v (1)  1, 3, 0 , a (t )  0,  6, 0
i


, v  10 , v (2)  v (2)  1
j
3
k
0  0, 0,  6
0 6 0
 


v ( 2 )  v ( 2)
v3

6
 10 3

6
3 10

 0.1897
50
10 10
-7-
3 10
 0.1897 _____
50
2. Find the curvature of y  x 3at ( 1,1)
Method 2.

y 
1   y 
2 3/ 2

6x
  
1  3x

3/ 2
2 2
answer: __

6
1  3 
2 3/ 2

3 10
 0.1897 _____
50
6
10 
3/ 2

6
3 10

50
10 10
3. Find the osculating circle for y  1  x 2 at (0, 1)

y 
1   y 
2 3/ 2

2
1   2 x 
2 3/ 2

2
1  0 
2 3/ 2
 2  radius  
1
2



Let r (t )  t , 1  t 2 , 0 , v (t )  1,  2t , 0 , a (t )  0,  2, 0
 at (0, 1) t  0 because t  x  0



 r (0)  0, 1, 0 , v (t )  1, 0, 0 , a (t )  0,  2, 0
i
j k
 
 
 
 v  a  0 and v  a  1 0 0  2k  v  a  2
0 2 0
 
 
va 2
v a 0
 aT 
  0 and a N 
 2
v
1
v
1


 a  aT T
0,  2, 0  0
Therefore, N 

 0,  1, 0   j
a N
2
 
 1
 

In the formula   r   N ,   h, k , 0 , r  0,1, 0 , and  N  0, 1, 0 ,
2
Therefore, the center is given by



  r   N , h, k , 0  0,1, 0 
1
1
0, 1, 0  0, , 0
2
2
The osculating circle has center (0, ½ ) and radius ½ .
The equation of a circle  x  h    y  k 
2
2
2
1
1

r x y  
2
4

2
2
1
1

answer: __________ x   y    _________
2
4

2
-8-
2
4. . Find the osculating circle y  e x at ( 0,1)

y 
1   y 
2 3/ 2

ex
 
3/ 2
x 2
1  e




Let r (t )  t , e t , 0 , v (t )  1, e t , 0

e0
 
3/ 2
0 2
1  e



, a (t )  0, e t , 0

1
2 2
 radius   2 2
 at (0, 1) t  0 because t  x  0
 1, 1, 0



 r (0)  0, 1, 0 , v (t )  1, 1, 0 , a (t )  0, 1, 0 , v  2 , T 
2
i j k
 
 
 
 v  a  1 and v  a  1 1 0  k  v  a  1
0 1 0
 
 
va
v a
1
1
 aT 

and a N 

v
v
2
2
1, 1, 0
 0, 1, 0  1

 a  aT T
2
2  2  0, 1, 0  1, 1, 0 
Therefore, N 



1
a N
2 

2

2
1 1
N
 1, 1, 0 
,
,0
2
2 2

1 1
and  N  2 2
,
, 0   2,2, 0
2 2
Therefore, h, k , 0  0,1, 0   2,2, 0   2,3, 0





  r   N ,   h, k , 0 , r  0,1, 0
The center of the osculating circle is  2,3, the radius is 2 2
and the equation is  x  22   y  32  8
-9-
5. Curves in space are generally represented by parametric equations, or equivalently, by a

vector-values function r ( t ) . The moving trihedral consists of the three mutually orthogonal

 

(perpendicular) unit vectors T , N , and B where T is the unit tangent vector,


  
N is the principal normal vector. B is called the binormal vector and B  T  N
Find the curvature and the binormal vector at t=1 for the curve

r ( t )  2ti  t 2 j  (1  t )k
Solution:



r (t )  2t , t 2 , 1  t , v (t )  2, 2 t ,  1 , a (t )  0, 2 , 0

2 2 1



 at t  1, r (1)  2, 1, 0 , v (1)  2, 2 ,  1 , a (t )  0, 2 , 0 , T  , , 
3 3 3
i
j
k
0 2
0
 
 
 
v  9  3, v  a  4, v  a  2 2  1  2, 0, 4 , v  a  20  2 5
2 5
 0.165634665
27
4 2 2 1
8 10 4
0, 2, 0   , , 
 , ,
 
 

va 2 5
v a 4
3 3 3 3
9 9 9
aT 
 , a N

N

v
3
v
3
2 5
2 5
3
3

4
5
2
N 
,
,
3 5 3 5 3 5
 
2 5
27
  
 B T  N 
answer : curvature 
i
2
3
4

3 5
j
2
3
5
3 5
k
1
9
18

,0,
3
9 5 9 5
2
3 5
binormal vector = __
-10-

1
2
, 0,
5
5
1
2
, 0,
5
5
____