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Section 3.1 Calculus of Vector-Functions
De…nition. A vector-valued function is a rule that assigns a vector to each member in a subset of R 1 : In other words, a vector-valued function is an ordered triple of functions, say f (t) ; g (t) ; h (t) ; and can be expressed as
~r (t) = hf (t) ; g (t) ; h (t)i :
For instance,
~r (t) = h1 + t; 2 t; 2 ¡ ti
~q (t) =
t ¡ 1
; ln (t) ;
p 2 ¡ t
À
are vector-valued functions. The domain of a vector-valued function is a subset of all real number at which the function is well-de…ned, i.e.,
Domain of ~r (t) = ft j ~r (t) = hf (t) ; g (t) ; h (t)i is de…nedg
= ft j each of f (t) ; g (t) ; h (t) is de…nedg = ft j f (t) is de…nedg \ ft j g (t) is de…nedg \ ft j g (t) is de…nedg :
So D (~r) = D (f ) \ D (g) \ D (h) : Any vector-valued function ~r (t) = hx; y; zi may be written in terms of its components as
x = f (t) y = g (t) z = h (t) :
Thus, the graph of a vector-valued function is a parametric curve in space. For instance, the function
~r (t) = h1 + t; 2 t; 2 ¡ ti
is de…ned for all t: Its component form is
x = 1 + t y = 2t z = 2 ¡ t:
The graph is a straight line with a direction h 1 ; 2 ; ¡ 1 i passing through (1; 0 ; 2) : Example 1.1. Find the domain of
~r (t) =
t ¡ 1
; ln (t) ;
p 2 ¡ t
À
Sol: We know that
D
μ 1 t ¡ 1
= ft 6 = 1g = (¡1; 1) [ (1; 1 )
D (ln (t)) = ft > 0 g = (0; 1 ) D
¡p 2 ¡ t
= ft · 2 g = (¡1; 2]:
So
D (~r) = D
μ 1 t ¡ 1
\ D (ln (t)) \ D
¡p 2 ¡ t
= ((¡1; 1) [ (1; 1 )) \ (0; 1 ) \ (¡1; 2]
= ((¡1; 1) [ (1; 1 )) \ (0; 2]
= (0; 1) [ (1; 2]:
O 1 2
Limits of vector-valued functions are de…ned through compo- nents:
For any vector-valued function ~r (t) = hf (t) ; g (t) ; h (t)i ; the limit
lim t!a ~r (t) =
D
lim t!a f (t) ; lim t!a g (t) ; lim t!a h (t)
E
exists if and only if the limits of all three components exist. Example 1.2. Consider
~r (t) = h2 cos t; sin t; ti :
De…nition. For any vector-valued function ~r (t) = hf (t) ; g (t) ; h (t)i ; if the limit of the di¤erence quotation
lim h! 0
~r (t 0 + h) ¡ ~r (t 0 ) h
exists, we say ~r (t) is di¤erentiable at t = t 0 : In this case, we call the limit the derivative at t = t 0 and denote it by ~r 0 (t 0 ) or
d~r dt
(t 0 ) = ~r 0 (t 0 ) = lim h! 0
~r (t 0 + h) ¡ ~r (t 0 ) h
We can show that ~r (t) is di¤erentiable at t = t 0 if and only if all three components are di¤erentiable and
~r 0 (t 0 ) = hf 0 (t 0 ) ; g 0 (t 0 ) ; h 0 (t 0 )i :
The derivative vector for any t; ~r 0 (t) ; is again a vector-valued function. Higher order derivatives are then de…ned accordingly. For instance,
~r 00 (t) = hf 00 (t) ; g 00 (t) ; h^00 (t)i
Geometrically, ~r (t 0 + h) ¡ ~r (t 0 )
represents the vector from ~r (t 0 ) to ~r (t 0 + h) : So for any small h > 0 ;
~r (t 0 + h) ¡ ~r (t 0 ) h
r (t 0 )
r (t 0 +h)
r (t 0 +h) – r(t 0 )
is a normalized (otherwise, the length of ~r (t 0 + h) ¡ ~r (t 0 ) would be a very small) secant direction. Therefore, the limit vector is "tangent" to the curve at t = t 0 : De…nition. We call
~r 0 (t 0 ) = hf 0 (t 0 ) ; g 0 (t 0 ) ; h^0 (t 0 )i
the tangent vector of the parametric curve ~r (t) at t = t 0 ; and
T^ ~ (t 0 ) = ~r^
(^0) (t 0 ) j~r 0 (t 0 )j
the unit tangent vector. A curve ~r (t) is called smooth if ~r 0 (t) exists and ~r 0 (t) 6 = ~ 0 : Example 1.3. Consider a circular helix
~r (t) = hcos t; sin t; ti :
Find ~r 0 (t) ; T~ (t) ; and ~r 00 (t) : Find also ~r 0 (0) ; T~ (0) :
All above properties can be veri…ed by direction computations. As in the case of one variable functions, derivative ~r 0 (t 0 ) measures the rate (vector) at which function ~r (t) changes across t = t 0 : Thus
j~r 0 (t 0 )j is the magnitude of the rate of change T^ ~ (t 0 ) is the direction of change.
Note that since
j~r (t)j =
p ~r (t) ¢ ~r (t) = (~r (t) ¢ ~r (t))
1 2
we have
d dt
j~r (t)j =
(~r (t) ¢ ~r (t)) ¡^
(^12) (~r (t) ¢ ~r (t))
0
j~r (t)j ¡^1 (~r 0 (t) ¢ ~r (t) + ~r (t) ¢ ~r 0 (t))
~r 0 (t) ¢ ~r (t) j~r (t)j
This shows that in general,
d dt
j~r (t)j 6 = j~r 0 (t)j ;
i.e.,
Rate of change for j~r (t)j 6 = Magnitude of rate of change for ~r (t) :
In physics, if ~r (t) describes the position of a moving object, then
~v (t) = ~r 0 (t) is velocity ¿ (t) = j~v (t)j is speed ~a (t) = ~v 0 (t) = ~r 00 (t) is acceleration.
De…nition. Integrals, inde…nite and de…nite, are de…ned accordingly: Z ~r (t) dt =
¿Z
f (t) dt;
Z
g (t) dt;
Z
h (t) dt
À
Z (^) b
a
~r (t) dt =
¿Z (^) b
a
f (t) dt;
Z (^) b
a
g (t) dt;
Z (^) b
a
h (t) dt
À
Note that for inde…nite integrals, we always end up a constant vector C~ = hC 1 ; C 2 ; C 3 i:
Z ~r (t) dt =
¿Z
f (t) dt;
Z
g (t) dt;
Z
h (t) dt
À
+ C:~
Example 1.4. Consider
~r (t) =
1 + t 3 ; te ¡t^ ; sin (2t)
Find (a) ~r 0 (t) ; and (b) equations of the tangent at t = 0: Solution: (a)
~r 0 (t) =
3 t 2 ; e ¡t^ ¡ te¡t^ ; 2 cos (2t)
(b) The tangent line passes through the terminal point of the vector ~r (0) = h 1 ; 0 ; 0 i ;i.e., passing through (1; 0 ; 0) with direction
~r 0 (0) = h 0 ; 1 ; 2 i :
So the equations are
x = 1 y = t z = 2t:
Example 1.5. Find (a)
R
~r (t) dt and (b)
R º
0 ~r^ (t)^ dt^ if
~r (t) =
2 cos t; sin t; 3 t 2
Solution: (a) Z ~r (t) dt =
¿Z
2 cos tdt;
Z
sin tdt;
Z
3 t 2 dt
À
2 sin t + C 1 ; ¡ cos t + C 2 ; t 3 + C (^3)
2 sin t; ¡ cos t; t 3
+ C~
where C~ = hC 1 ; C 2 ; C 3 i is an arbitrary constant vector.
- Find (i) unit tangent T~ at given point and (ii) equation of tangent line to the curve at that point.
(a) ~r (t) = h 2 e¡t^ cos t; e ¡t^ sin t; e ¡t^ i ; (2; 0 ; 1) (b) ~r (t) =
ln t; 2
p t; t^2
- The angle between two curves at a point of intersection is deined as the angle between their tangents. Find the point of intersection and the angle between ~r 1 = ht; 1 ¡ t; 3 + t^2 i and ~r 2 = h 3 ¡ s; s ¡ 2 ; s^2 i :
