# tftgthtftgtht = = + + rijk cos ,sin , tttt = r 2 cos cos , 2 cos sin ,sin ttt = + + r

advertisement ```Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
13.1 Curves in Space and their Tangents
Vector-Valued Function
● vector function for short, sometimes called space curves
● takes one or more real variables and returns a vector
● we’ll focus on single variable ( t ) vector functions that
give three-dimensional vectors
represents time in most applications
r ( t ) = f ( t ) , g (t ) , h (t ) = f (t ) i + g (t ) j + h ( t ) k
real valued functions called
component functions of r
We want to look at the calculus of these vector functions.
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Interesting vector valued functions:
r ( t ) = cos ( t ) ,sin ( t ) , t
called the circular helix
For a nice animation go to :
http://math.bu.edu/people/paul/225/circular_helix.html
r (t ) =
( 2 + cos ( )) cos t , ( 2 + cos ( ) ) sin ( t ) ,sin ( )
3t
2
3t
2
called the trefoil knot
For a nice animation go to :
http://math.bu.edu/people/paul/225/trefoil_knot.html
3t
2
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Limit of Vector Functions
r ( t ) = f ( t ) , g (t ) , h (t )
Limit of a vector function - taken component-wise
lim r ( t ) = lim f ( t ) , lim g ( t ) , lim h ( t )
t →a
provided the component
function limits exist
t →a
t →a
t →a
Let r ( t ) =
sin ( t ) 2t
, e , ln (1 − t ) . Find lim r ( t )
t →0
t
sin ( t )
lim r ( t ) = lim
, lim e2t , lim ln (1 − t )
t →0
t →0
t →0
t →0
t
1
1
0
lim r ( t ) = 1,1, 0 = i + j
t →0
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Differentiation of Vector Functions
r ( t ) = f ( t ) , g (t ) , h (t )
Derivative of a vector function - taken component-wise
r′ ( t ) = f ′ ( t ) , g ′ ( t ) , h′ ( t )
provided the component
function are differentiable
Let r ( t ) = sin −1 t , 1 − t 2 , ln (1 + 3t ) . Find r′ ( t )
r′ ( t ) =
r′ ( t ) =
1
, 1 1− t 2
1− t
2
(
2
1
1− t2
,
−t
1− t 2
,
−1/2
1
) ( −2t ) , 1 + 3t (3)
3
1 + 3t
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
r ( t ) is the position vector
r′ ( t ) is the tangent vector
r′ ( t ) will now be called the velocity vector, v ( t )
r′ ( t ) is called the speed of the particle
r′′ ( t ) is called the acceleration vector, a ( t )
( 0, 2 )
t =0
r ( t ) = sin t , 2 cos t
v ( t ) = r ′ ( t ) = cos t , −2 sin t
a ( π6 )
a ( t ) = r′′ ( t ) = − sin t , −2 cos t
(1, 0)
r ′ ( t ) = cos 2 t + 4sin 2 t
π 
r  =
6
1
2
, 3
π 
v  =
6
t=
3
2
, −1
2.
3.
4.
5.
6.
π
2
π 
a  = − 1 ,− 3
2
6
Differentiation Rules:
1.
v ( π6 )
d
u ( t ) + v ( t )  = u′ ( t ) + v′ ( t )
dt 
d
cu ( t )  = cu′ ( t )
dt 
d
 f ( t ) u ( t )  = f ′ ( t ) u ( t ) + f ( t ) u′ ( t )
dt 
d
u ( t ) ⋅ v ( t )  = u′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v′ ( t )
dt 
d
u ( t ) &times; v ( t ) = u′ ( t ) &times; v ( t ) + u ( t ) &times; v′ ( t )
dt 
d
 u ( f ( t ) )  = f ′ ( t ) u′ ( f ( t ) )

dt 
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Show that if r ( t ) = c (constant), then r′ ( t ) ⊥ r ( t ) for all t.
Assume r ( t ) = c.
2
The fact that r ( t ) = r ( t ) ⋅ r ( t ) , gives us r ( t ) ⋅ r ( t ) = c 2 .
Taking the derivative of both sides gives
d
r ( t ) ⋅ r ( t )  = 0.
dt 
Using the properties of differentiation, we get r ′ ( t ) ⋅ r ( t ) + r ( t ) ⋅ r ′ ( t ) = 0.
The dot product is commutative, so r ′ ( t ) ⋅ r ( t ) + r ′ ( t ) ⋅ r ( t ) = 0
or 2 r′ ( t ) ⋅ r ( t )  = 0.
So, r′ ( t ) ⋅ r ( t ) = 0 and thus r ′ ( t ) ⊥ r ( t ) ( for all t )
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Most vector functions do not have constant
magnitude.
The helix example from earlier: r ( t ) = cos ( t ) ,sin ( t ) , t
r ( t ) = cos 2 ( t ) + sin 2 ( t ) + t 2
r (t ) = 1 + t 2
It’s magnitude isn’t constant (it depends on t).
Consider its derivative however: r′ ( t ) = − sin ( t ) , cos ( t ) ,1
r′ ( t ) = sin 2 ( t ) + cos 2 ( t ) + 1 = 2
r′ ( t ) has constant magnitude, so r′ ( t ) ⊥ r ′′ ( t )
So for the helix, the velocity vector will always be orthogonal to the acceleraton vector
( don't think that this always happens )
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Show that if r ( t ) is a vector function such that r′′ ( t ) exists, then
d
r ( t ) &times; r′ ( t )  = r ( t ) &times; r′′ ( t )
dt 
Using the properties of differentiation, we get
d
r ( t ) &times; r ′ ( t )  = r ′ ( t ) &times; r ′ ( t ) + r ( t ) &times; r′′ ( t )
dt 
anytime you cross a vector
with itself, you get the zero vector
So,
d
r ( t ) &times; r ′ ( t )  = r ( t ) &times; r ′′ ( t ) .
dt 
When taking the derivative of the cross product of a vector function and its velocity,
you can just take the cross product of the vector function and its acceleration.
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
Geometry of the derivative
secant vector
vector : z
r (t + h ) − r (t )
P
Q
r (t )
r (t + h )
y
Tangent vector
z
x
P
r (t )
r (t + h ) − r (t )
h →0
h
r′ ( t ) = lim
tangent vector
Q
r (t + h ) − r (t )
h
r (t + h )
Unit tangent vector
y
T (t ) =
x
r′ ( t )
r′ ( t )
( it's the velocity vector with magnitude 1)
Math 114 – Rimmer
13.1 Vector Functions (Space Curves) and Their
Tangents
The tangent line to a smooth curve
r ( t ) = f ( t ) , g ( t ) , h ( t ) = f ( t ) i + g ( t ) j + h ( t ) k at t = t0
passes through the point ( f ( t0 ) , g ( t0 ) , h ( t0 ) ) ,
and has the same direction as the velocity vector at t0
f ′ ( t0 ) , g ′ ( t0 ) , h ′ ( t0 ) .
So the parametric equations of the tangent line are:
x =
f ( t0 ) +
y =
z =
g ( t0 ) +
h ( t0 ) +
f ′ ( t0 ) t
g ′ ( t0 ) t
h′ ( t0 ) t
```