Calc 3 Lecture Notes
Section 11.5
Page 1 of 11
Section 11.5: Tangent and Normal Vectors
Big idea: There are three mutually orthogonal vectors to any curve in 3D that can be calculated
(fairly) easily in terms of the vector-valued function that traces out the curve.
Big skill: You should be able to compute unit tangent, normal, and binormal vectors to the curve
traced out by a given a vector-valued function, compute the radius of curvature and center of the
osculating circle, and compute components of acceleration parallel to and perpendicular to a
curve.
Motivation: Our usual x-, y-, and z-axes represent a fixed coordinate frame not always
convenient for investigating the behavior of a moving object.
Occasionally it is desirable to have a “moving coordinate frame” in which one of the “coordinate
axes” always points in the direction of motion of the object. Think: airplanes or roller coasters.
The moving coordinate frame will still be a “rectangular” coordinate frame, consisting of three
mutually perpendicular “axes” defined by unit vectors; it’s just that these vectors will rotate as
the object moves along its trajectory.
So, given the position vector r(t), how do we compute three mutually perpendicular unit vectors
at any time t, one of which points in the direction of motion?
We already have one such vector:
First Local Unit Vector: The Unit Tangent Vector
r  t 
T t  
r  t 
Now, need to find two other unit vectors perpendicular to T(t) (and to each other).
We can get one of them by recall Theorem 2.4 in Section 11.2:
r  t  is constant if and only if r(t) and r(t) are orthogonal for all t.
The above statement applies to any arbitrary vector function r(t), including T(t). Since T  t  is
constant (because it’s a unit vector), T(t) must be perpendicular to T(t) for all t.
If you don’t buy that argument, we can prove T(t) and T(t) are orthogonal by showing
T(t)T(t) = 0:
Calc 3 Lecture Notes
Section 11.5
Page 2 of 11
r  t  d  r  t  
 

r  t  dt  r  t  
r  t   1 d
1 
d


r  t   r  t 
r  t  
dt
r  t   r  t  dt


r  t   1
2 d


r  t   r  t  1 r  t 
r  t  
dt
r  t   r  t 


r  t   r  t 
r  t  d






r
t

r
t




2
dt
r  t   r  t 


r
t





r  t   r  t 
r  t  1
1/ 2 d







r
t

r
t
r
t

r
t












2

2
dt 
r  t   r  t 


r
t





r  t   r  t 
r  t  1 1







r
t

r
t

r
t

r
t










2



2
r  t   r  t 
r
t




r t 


r  t   r  t  r  t   r  t   r  t   




3
r  t   r  t 


r
t



r  t   r  t  r  t   r  t   r  t   r  t  


2
4
r  t 
r  t 
T  t   T  t  

r  t   r  t 
r  t 
2

r  t 
2
 r  t   r  t  
r  t 
4
0
Phew! That stunk! Here is a shorter proof:
T t   T t   T t   1
2
d
d
T  t   T  t    1
dt
dt


T t   T t   T t   T t   0
2T  t   T  t   0
T  t   T  t   0
This sidetrack is meant to motivate you to begin to remember vector identities, theorems, and
tricks, instead of just cranking through the derivatives…
Calc 3 Lecture Notes
Section 11.5
Page 3 of 11
Anyway, we’ve shown in agonizing detail that T(t) and T(t) are orthogonal, so we just need to
normalize it to make it a unit vector:
Second Local Unit Vector: The Principal Unit Normal Vector (Definition 5.1)
T  t 
N t  
T  t 
So, in which of the infinitely many directions that are perpendicular to T(t) does N(t) point?
By the Chain Rule, T  t  
dT dT ds
.

dt
ds dt
Substituting into the definition of N(t):
T  t 
N t  
T  t 
dT ds
 ds dt
dT ds
ds dt
dT ds
 ds dt
dT ds
ds dt
1 dT
N t  
 ds
Where the facts that  
dT
ds ds
and
 because s  t   r  t   0 have been used.
ds
dt dt
Since N(t) points in the same direction as
dT
, N(t) points in the direction in which T(t)
ds
turns as arc length increases.
This implies that N(t) always points toward the concave side of the curve.
WinPlot can really help to visualize this…
Calc 3 Lecture Notes
Section 11.5
Page 4 of 11
Practice:
1. Find the unit tangent and principal unit normal vectors to the curve defined
by r  t   t 2 , t .
2. Find the unit tangent and principal unit normal vectors to the curve defined
by r  t   2 sin  t  ,  2 cos  t  , 0.5t .
Calc 3 Lecture Notes
Section 11.5
Page 5 of 11
If our curve is in three dimensions, we can get a third vector that is mutually orthogonal to T(t)
and N(t) using the vector cross product:
Third Local Unit Vector: The Binormal Vector (Definition 5.2)
B t   T t   N t 
This is a unit vector already, since
B  t   T  t   N  t   T  t  N  t  sin    11 sin  90   1
This triple of local, moving unit vectors T(t), N(t), and B(t) is known as the TNB frame, the
Frenet-Serret frame, or sometimes the moving trihedral. It’s important for differential
geometry, spacecraft navigation, nonlinear system analysis, magnetic field studies, and a host of
other applications (not one of which I could list right now…)
Practice:
1. Find T(t), N(t), and B(t) for the curve defined by r  t   2sin  t  , 2 cos  t  , 0.5t , and at
the point (2, -2, 0).
Calc 3 Lecture Notes
Section 11.5
Page 6 of 11
Normal Planes, Osculating Planes, and Osculating Circles
At each point on the curve, the plane defined by the normal vector N(t) and the binormal vector
B(t) is called the normal plane. It is the plane that contains all vectors perpendicular to the
tangent vector. Alternatively, the normal plane is the plane that passes through the specified
point on the curve and has T(t) as a normal vector.
The plane defined by the tangent vector T(t) and the principal unit normal vector N(t) is called
the osculating plane (for a 2-dimensional curve, the osculating plane is just the xy-plane). This
name arises from the fact that the osculating plane contains the osculating circle, which is the
largest circle that just “kisses” (or is tangent to) the curve at a given point…
So, recall that the curvature of a circle is the reciprocal of its radius, as was shown in Section
11.4 At some point P on our curve, suppose that we have computed the curvature  (and
suppose this curvature is nonzero). The circle whose radius is 1/  , lies in the osculating plane,
and has its center is located 1/  units away from P in the direction of the principal unit normal
is called the osculating circle for the curve at that point.
The osculating circle . . . .
 has the same curvature as the curve itself at P
 has the same tangent vector as the curve at P (namely, T is tangent to both).
 has a radius of 1/  , called the radius of curvature for the curve at P.
 is considered the circle of “best fit” for the curve at P.
1
 is centered at the terminal point of the vector r  t  
N t 
 t 
Practice:
1. Find the osculating circle for the parabola defined by r  t   t 2 , t .
Calc 3 Lecture Notes
Section 11.5
Page 7 of 11
Tangential and Normal Components of Acceleration
A simple picture in 2D will serve to introduce this topic. At any point along the path of an object
moving along a smooth curve, the acceleration vector a(t) can be decomposed into the sum of a
vector parallel to the unit tangent vector T(t) and a vector parallel to the principal unit normal
vector N(t):
Note:
aT  t    a  t   T  t   T  t 
aN  t    a  t   N  t   N  t 
… This is much shorter than what is about to come…
Thus, in general, we expect that part of the acceleration vector acts in the direction the object is
already moving (or in the opposite direction), tending to cause the object to change speed, while
part of the vector will act in a direction orthogonal to the direction of motion, tending to cause
the object to change direction.
Now wish to investigate this phenomenon more carefully; keep in mind that in 3D there is a third
direction as indicated by the binormal vector B(t).
Recall:
Therefore,
T t  
r  t 
.
r  t 
Thus,
r  t   T  t  r  t 
v  t   r  t   T  t  r  t 
Using the product rule, the acceleration vector a(t) = v(t) can be shown to be :
Calc 3 Lecture Notes
Section 11.5
Page 8 of 11
Recall that the principal unit normal vector N(t) is given by N  t  
T  t 
, and therefore
T  t 
T  t   T  t  N  t  .
Also, the Chain Rule tells us that
T  t  
dT
dT ds
dT ds



dt
ds dt
ds dt
This allows us to conclude that T  t   
ds
N  t  , and therefore our expression for acceleration
dt
above can be rewritten as
2
d 2s
 ds 
a t   2 T t      N t 
dt
 dt 
Comments/observations:
Above formula shows us how the acceleration vector a(t) can be decomposed into the sum of a
vector parallel to T(t) and a vector parallel to N(t).
Apparently, the acceleration vector always lies in the plane determined by T(t) and N(t);
there is no component of acceleration in the direction of the binormal B(t).
d 2s
The coefficient 2 is called the tangential component of acceleration, denoted aT , and
dt
measures how rapidly the speed is changing (if it’s negative, the object is slowing down).
2
 ds 
The coefficient    is called the normal component of acceleration, denoted a N , and
 dt 
measures how rapidly the object is turning.
Normal component of acceleration related to centripetal force. We see here that this force is
directly proportional to the curvature and the square of the speed.
Above formula often expressed in the form
a  t   aT T  aN N ,
where aT and a N are given by the formulas
aT 
d 2s
,
dt 2
 ds 
aN    
 dt 
2
Often wish to compute these quantities, so we can better understand how the object’s speed and
direction are changing.
Calc 3 Lecture Notes
Section 11.5
Computing aT is generally fairly easy; given r(t):
Find r(t)
ds
 r  t 
dt
d 2s
ds
Find aT  2 by differentiating
.
dt
dt
Compute
Page 9 of 11
Calc 3 Lecture Notes
Section 11.5
Page 10 of 11
Computing a N using the second of these two formulas is tedious because it requires us to
compute the curvature.
Easier way of computing a N comes from observing that the three vectors a(t), aTT, and aNN,
form a right triangle (recall that T(t) and N(t) are perpendicular):
By the Pythagorean Theorem,
Thus, we have the following alternative formula for a N :
Thus, once you have computed a(t), and the tangential component aT, you can compute aN using
the above formula (without having to compute the curvature).
2
 ds 
Added bonus: once we’ve found a N using this alternative formula, then, since a N     ,
 dt 
aN
we can compute the curvature using:  
2
 ds 
 
 dt 
Finally, worth noting that aT and a N , being, by definition, the components of the vector a(t) in
the directions of Tˆ (t ) and Nˆ (t ) , are given by the following:
aT  a  T ;
aN  a  N
These formulas, however, don’t generally provide us with easier ways of computing aT and a N
(recall that N̂ tends to be tedious to compute).
Calc 3 Lecture Notes
Section 11.5
Page 11 of 11
Example
Consider a moving object whose position at time t is given by the vector function

r (t )  t , t 2 , 3t , where distances are measured in feet and time in seconds.
(a) Compute the tangential and normal components of acceleration.
(b) Is the object speeding up or slowing down at t  2 ? At what rate?
(c) Compute an expression for the curvature  .