1
Section 10.4: Tangent Vectors and Normal Vectors
Practice HW from Larson Textbook (not to hand in)
p. 637 # 3-9 odd, 25-35 odd (just find T(t) and N(t))
Note: Sometimes, it is convenient to normalize a vector tangent to a vector valued
function. This gives the unit tangent vector.
Unit Tangent Vector
Given a vector function r on an open interval I, the unit tangent vector T(t) is given by
T ( t) 
r (t)
1

r (t) where r ( t )  0
|| r (t) || || r (t) ||
Example 1: Find the unit tangent vector T(t) to the curve given by r(t ) = t i + t 2 j
when t = 1.
Solution:
2
3
Example 2: Find the unit tangent vector T(t) for r(t ) = 2 cos t i + sin 6t j + 2t k at

t .
6
Solution: The unit tangent vector is given by the formula
T ( t) 
r (t)
1

r (t) .
|| r (t) || || r (t) ||
Using r(t ) = 2 cos t i + sin 6t j + 2t k, we see that
r ( t )  2 sin t i  6 cos 6t j  2k   2 sin t , 6 cos 6t , 2 
and
|| r ( t ) || (2 sin t ) 2  (6 cos 6t) 2  (2) 2  4 sin 2 t  36 cos2 6t  4
Thus,
T ( t) 
At t 
1
r (t) 
|| r (t) ||

6


4 sin 2 t  36 cos 2 6t  4
 2 sin t , 6 cos 6t, 2 
, we have

T( ) 
6
1
1
4 sin 2 ( / 6)  36 cos 2 (6 / 6)  4
 2 sin(  / 6) , 6 cos( 6 / 6), 2 
1
1
 2( ) , 6( 1), 2  (Note that sin(  / 6)  1 / 2 , cos(6/6)  cos  -1)
2
(1 / 2) 2  36( 1) 2  4
1
 1,  6, 2 
1
4( )  36(1)  4
4
1

 1,  6, 2 
1  36  4
1

 1 ,  6, 2 
41
1
6
2
 
,
,

41
41 41
(continued on next page)
4
The following graph plots in 3D space the vector function r(t) and the corresponding unit

tangent vector T(t) evaluated at t  .
6
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5
Calculating the Unit Normal Vector
In describing motion, sometimes it can be an advantage to calculate a vector orthogonal
to the unit tangent vector T of the vector function r(t ) = f (t) i + g (t) j + h (t) k.
z
r (t )
y
x
One method is the use the fact that the T is a unit vector. Then, we know that
T(t )  T(t ) || T(t ) ||2  1
If we use the product rule and differentiate both sides of this equation, we obtain
T(t )  T(t )  T(t )  T(t )  0
or
2T(t )  T(t )  0 .
Thus,
T(t )  T(t )  0
Hence, the derivative of T, T ' , is orthogonal to T. Normalizing T will give the principal
unit normal vector N at t.
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Principal Unit Normal Vector N
For the vector valued function r(t ) = f (t) i + g (t) j + h (t) k with unit tangent vector T,
the unit normal vector, assuming T(t )  0 , is given by
N( t ) 
T(t )
1

T(t )
|| T(t ) || || T(t ) ||
Note: Unless the function r(t) is simple, the unit normal vector N can be very computationally
intensive. Maple can be a very convenient tool for simplifying the computations.
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Example 3: Find the unit normal vector N for the vector to the curve given by
r(t ) = t i + t 2 j when t = 1.
Solution:
8
9
t
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10
Example 4: Find the unit normal vector N for the vector r (t )  t , 5 sin t , 5 cos t  . Use
your result to find the unit normal vector at the point ( ,0,5)
Solution: To find the unit normal vector N, we must first find the unit tangent vector T.
Recall that the formula for the unit tangent vector T is given by
T(t) 
1
r (t )
|| r (t ) ||
For r (t )  t , 5 sin t , 5 cos t  , we have r (t )  1, 5 cos t , - 5 sin t  and
|| r (t ) || (1) 2  (5 cos t ) 2  ( 5 sin t ) 2  1  25 cos 2 t  25 sin 2 t
 1  25(cos 2 t  sin 2 t )  1  25(1)  26
Thus,
T( t) 
1
1
1
5
5
r (t ) 
 1, 5 cos t ,  5 sin t 
,
cos t , 
sin t 
|| r (t ) ||
26
26 26
26
To get the unit normal vector, we use the formula
N( t ) 
Using T(t ) 
1
26
,
5
26
cos t , 
T(t )
1

T(t )
|| T(t ) || || T(t ) ||
5
26
sin t  , we see that T(t )  0,
5
26
sin t , 
5
26
and
2
2




5
5
25 2
25
| T(t ) | (0) 2   
sin t    
cos t   0 
sin t 
cos 2 t
26
26
26
26





Then,
25
(sin 2 t  cos 2 t ) 
26
25
(1) 
26
5
26
(continued on next page)
cos t 
11
N (t ) 
1
T(t ) 
| T(t ) |
1
1
5
 0,
5
26
sin t , 
5
26
cos t 
26


26
5
5
 0,
sin t , 
cos t 
5
26
26
26
26
5
26
5
 0,
 (
) sin t ,
 (
) cos t 
5
5
5
26
26
 0, sin t , cos t 
We last want to evaluate N(t )  0, sin t , cos t  at the point ( ,0,5) . We need to
find the value of t that produces this point. Since r (t )  t , 5 sin t , 5 cos t  , t   since
and 5 sin t  5 sin   5(0)  0 and 5 cos t  5 cos   5(1)  5 . Thus,
Normal Vector at Po int
( ,0,5)
 N( )  0, sin  , cos   0,0,  1  0,0,1 
t 
The following shows a framed graph of r (t )  t , 5 sin t , 5 cos t  (in green) with the unit
1
5
5
tangent vector T(t ) 
,
cos t , 
sin t  (in red) and unit normal vector
26 26
26
N(t )  0, sin t , cos t  (in blue) displayed at the point ( ,0,5) .
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