Section 3.1 Calculus of Vector

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 R1 : 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; 2t; 2 ¡ ti
¿
À
p
1
~q (t) =
; ln (t) ; 2 ¡ t
t¡1
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; 2t; 2 ¡ ti
is de…ned for all t: Its component form is
x = 1+t
y = 2t
z = 2 ¡ t:
1
The graph is a straight line with a direction h1; 2; ¡1i passing through (1; 0; 2) :
Example 1.1. Find the domain of
¿
À
p
1
~r (t) =
; ln (t) ; 2 ¡ t :
t¡1
So
Sol: We know that
µ
¶
1
D
= ft 6= 1g = (¡1; 1) [ (1; 1)
t¡1
D (ln (t)) = ft > 0g = (0; 1)
¡p
¢
D 2 ¡ t = ft · 2g = (¡1; 2]:
µ
¶
¡p
¢
1
D (~r) = D
\ D (ln (t)) \ D 2 ¡ t
t¡1
= ((¡1; 1) [ (1; 1)) \ (0; 1) \ (¡1; 2]
= ((¡1; 1) [ (1; 1)) \ (0; 2]
= (0; 1) [ (1; 2]:
)
(
°
1
O
2
Limits of vector-valued functions are de…ned through components:
For any vector-valued function ~r (t) = hf (t) ; g (t) ; h (t)i ; the limit
D
E
lim ~r (t) = lim f (t) ; lim g (t) ; lim h (t)
t!a
t!a
t!a
t!a
exists if and only if the limits of all three components exist.
Example 1.2. Consider
~r (t) = h2 cos t; sin t; ti :
2
(a) Find
lim ~r (t) ; lim ~r (t) :
t!0
t!¼ =2
(b) Discuss and sketch its graph.
Solution: (a)
D
E
lim ~r (t) = lim (2 cos t) ; lim sin t; lim t
t!0
t!0
t!0
t!0
= h2; 0; 0i
¿
À
lim ~r (t) = lim (2 cos t) ; lim sin t; lim t
t!¼ =2
t!¼ =2
t!¼ =2
t!¼ =2
D
¼E
= 0; 1;
:
2
(b) Let us …rst take a look at the projection of the curve onto xy ¡ plane
x = 2 cos t
y = sin t:
We know that its graph is a ellipse
t=
¼
is the angle to x ¡ axis
4
In 3D; as t increases from t = 0; the curve starting at (2; 0; 0) on xy-plane,
moves in the way that its …rst two component (x; y) moving along the ellipse
in the above …gure counter-clockwise while its z¡ component increases linearly, as if we raise vertically the ellipse. The curve is on the elliptic cylinder,
and is called elliptic helix.
3
De…nition. For any vector-valued function ~r (t) = hf (t) ; g (t) ; h (t)i ;
if the limit of the di¤erence quotation
~r (t0 + h) ¡ ~r (t0 )
h!0
h
lim
exists, we say ~r (t) is di¤erentiable at t = t0 : In this case, we call the limit
the derivative at t = t0 and denote it by ~r0 (t0 ) or
d~r
~r (t0 + h) ¡ ~r (t0 )
(t0 ) = ~r0 (t0 ) = lim
:
h!0
dt
h
We can show that ~r (t) is di¤erentiable at t = t0 if and only if all three
components are di¤erentiable and
~r0 (t0 ) = hf 0 (t0 ) ; g 0 (t0 ) ; h0 (t0 )i :
The derivative vector for any t; ~r0 (t) ; is again a vector-valued function.
Higher order derivatives are then de…ned accordingly. For instance,
~r00 (t) = hf 00 (t) ; g 00 (t) ; h00 (t)i
Geometrically,
~r (t0 + h) ¡ ~r (t0 )
represents the vector from ~r (t0 ) to ~r (t0 + h) : So for any small h > 0;
~r (t0 + h) ¡ ~r (t0 )
h
4
r (t 0+h) – r(t0 )
r (t0 )
r (t0+h)
is a normalized (otherwise, the length of ~r (t0 + h) ¡ ~r (t0 ) would be a very
small) secant direction. Therefore, the limit vector is "tangent" to the curve
at t = t0 :
De…nition. We call
~r0 (t0 ) = hf 0 (t0 ) ; g 0 (t0 ) ; h0 (t0 )i
the tangent vector of the parametric curve ~r (t) at t = t0 ; and
~r0 (t0 )
T~ (t0 ) = 0
j~r (t0 )j
the unit tangent vector. A curve ~r (t) is called smooth if ~r0 (t) exists and
~r0 (t) 6= ~0:
Example 1.3. Consider a circular helix
~r (t) = hcos t; sin t; ti :
Find ~r0 (t) ; T~ (t) ; and ~r00 (t) : Find also ~r0 (0) ; T~ (0) :
5
Solution:
~r0 (t) = h¡ sin t; cos t; 1i
~r00 (t) = h¡ cos t; ¡ sin t; 0i
1
T~ (t) = 0
~r0 (t)
j~r (t)j
1
1
=p 2
h¡ sin t; cos t; 1i = p h¡ sin t; cos t; 1i
2
2
sin t + cos t + 1
0
~r (0) = h0; 1; 1i
1
T~ (0) = p h0; 1; 1i :
2
Properties of derivatives: ¸ is a scalar constant, f (t) is a scalar
function.
1. Addition:
(~u (t) + ~v (t))0 = ~u0 (t) + ~v0 (t)
2. Scalar function multiplication:
(f (t) ~u (t))0 = f (t) ~u0 (t) + f 0 (t) ~u (t)
3. Scalar (constant) multiplication:
(¸~u (t))0 = ¸~u0 (t)
4. Dot product:
(~u (t) ¢ ~v (t))0 = ~u0 (t) ¢ ~v (t) + ~u (t) ¢ ~v0 (t)
5. Cross product:
(~u (t) £ ~v (t))0 = ~u0 (t) £ ~v (t) + ~u (t) £ ~v 0 (t)
6. Chain rule:
d
~u (f (t)) =
dt
µ
d~u
dt
¶
(f (t))
6
df
(t) = ~u0 (f (t)) f 0 (t) :
dt
All above properties can be veri…ed by direction computations.
As in the case of one variable functions, derivative ~r0 (t0 ) measures the
rate (vector) at which function ~r (t) changes across t = t0 : Thus
j~r0 (t0 )j is the magnitude of the rate of change
T~ (t0 ) is the direction of change.
Note that since
j~r (t)j =
we have
p
1
~r (t) ¢ ~r (t) = (~r (t) ¢ ~r (t)) 2
1
0
d
1
j~r (t)j = (~r (t) ¢ ~r (t))¡ 2 (~r (t) ¢ ~r (t))
dt
2
1
= j~r (t)j¡1 (~r0 (t) ¢ ~r (t) + ~r (t) ¢ ~r0 (t))
2
~r0 (t) ¢ ~r (t)
=
:
j~r (t)j
This shows that in general,
d
j~r (t)j 6= j~r0 (t)j ;
dt
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) = ~r0 (t) is velocity
À (t) = j~v (t)j is speed
~a (t) = ~v0 (t) = ~r00 (t) is acceleration.
De…nition. Integrals, inde…nite and de…nite, are de…ned accordingly:
¿Z
À
Z
Z
Z
~r (t) dt =
f (t) dt; g (t) dt; h (t) dt
¿Z b
À
Z b
Z b
Z b
~r (t) dt =
f (t) dt;
g (t) dt;
h (t) dt :
a
a
a
7
a
~ =
Note that for inde…nite integrals, we always end up a constant vector C
hC1 ; C2 ; C3 i:
¿Z
À
Z
Z
Z
~
~r (t) dt =
f (t) dt; g (t) dt; h (t) dt + C:
Example 1.4. Consider
®
~r (t) = 1 + t3 ; te¡t ; sin (2t) :
Find (a) ~r0 (t) ; and (b) equations of the tangent at t = 0:
Solution: (a)
®
~r0 (t) = 3t2 ; e¡t ¡ te¡t ; 2 cos (2t) :
(b) The tangent line passes through the terminal point of the vector ~r (0) =
h1; 0; 0i ;i.e., passing through (1; 0; 0) with direction
~r0 (0) = h0; 1; 2i :
So the equations are
x=1
y=t
z = 2t:
R¼
~r (t) dt and (b) 0 ~r (t) dt if
®
~r (t) = 2 cos t; sin t; 3t2 :
Example 1.5. Find (a)
Solution: (a)
Z
where
~r (t) dt =
R
¿Z
2 cos tdt;
Z
sin tdt;
Z
2
3t dt
À
®
= 2 sin t + C1 ; ¡ cos t + C2 ; t3 + C3
®
~
= 2 sin t; ¡ cos t; t3 + C
~ = hC1 ; C2 ; C3 i is an arbitrary constant vector.
C
8
(b) According to Fundamental Theorem of Calculus,
Z ¼
®
~r (t) dt = 2 sin t; ¡ cos t; t3 jt=¼
t=0
0
®
= 2 sin ¼ ; ¡ cos ¼ ; ¼ 3 ¡ h2 sin 0; ¡ cos 0; 0i
®
= 0; 2; ¼ 3
Homework:
1. Find domain and limit.
¿
À
t¡1 p
2
(a) ~r (t) =
; t; sin (¼ (t + 1)) ; lim ~r (t) =?
t!1
t+1
¿
À
2 ln t
(b) ~r (t) = arctan t; e¡t ;
; lim ~r (t) =?
t!1
t
2. Sketch the curve. Indicate with an arrow the direction in which t
increases.
(a) ~r (t) = hcos 2t; t; sin 2ti
(b) ~r (t) = h1 + 2t; t; ¡3ti
3. Find a vector equation for the curve of intersection of two surfaces.
(a) x2 + y2 = 4 and z = xy
(b) z = 2x2 + y2 and y = x2
4. Find the derivative.
D
E
3
(a) ~r (t) = t2 ; cos 3t; e¡t
³
´
(b) ~r (t) = t2~a £ et ~b + 2t ~c ; ~a; ~b; and ~c are three constant vectors.
5. Find the integral.
R
(a) hsin ¼ t; cos ¼t; e2t i dt
´
R1³
(b) 0 8t2 ~i + 9t2 ~j + 25t4 ~k dt
9
6. Find (i) unit tangent T~ at given point and (ii) equation of tangent line
to the curve at that point.
(a) ~r (t) = h2e¡t cos t; e¡t sin t; e¡t i ; (2; 0; 1)
p
®
(b) ~r (t) = ln t; 2 t; t2 ; (0; 2; 1)
7. 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 ~r1 = ht; 1 ¡ t; 3 + t2 i and ~r2 = h3 ¡ s; s ¡ 2; s2 i :
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