Instructor: Longfei Li
Math 243 Lecture Notes
13.2 Derivatives and Integrals of Vector Functions
Derivatives
Definition: The derivative of a vector function r(t) is defined as:
dr
r(t + h) − r(t)
= r0 (t) = lim
h→0
dt
h
Geometric understanding:
r0 is called the tangent vector to the curve defined by r at the point P
Tangent Line: the tangent line of a curve C at point P is defined to be the line through P parallel
to the tangent vector r0 at P
Unit Tangent Vector:
T(t) =
r0 (t)
|r0 (t)|
Theorem: If r(t) =< f (t), g(t), h(t) >, where f, g and h are differentiable functions, then
r0 (t) =< f 0 (t), g 0 (t), h0 (t) >
Proof:
1
r(t + ∆t) − r(t)
∆t
< f (t + ∆t), g(t + ∆t), h(t + ∆t) > − < f (t), g(t), h(t) >
= lim
∆t→0
∆t
f (t + ∆t) − f (t) g(t + ∆t) − g(t) h(t + ∆t) − h(t)
= lim <
,
,
>
∆t→0
∆t
∆t
∆t
g(t + ∆t) − g(t)
h(t + ∆t) − h(t)
f (t + ∆t) − f (t)
, lim
, lim
>
=< lim
∆t→0
∆t→0
∆t→0
∆t
∆t
∆t
=< f 0 (t), g 0 (t), h0 (t) >
r0 (t) = lim
∆t→0
Example:
(a) Find the derivative of r(t) = (1 + t3 )i + te−t j + sin 2tk
(b) Find the unit tangent vector at the point where t = 0
Solution:
(a) Differentiate component-wisely:
r0 (t) = 3t2 i + (e−t − te−t )j + 2 cos 2tk
(b) when t = 0, the point is (1, 0, 0), and
r0 (0) = j + 2k
So the unit tangent vector at the point (1, 0, 0) is
√
√
r0 (0)
j + 2k
5
2 5
T= 0
= √
j+
k
=
|r (0)|
5
5
5
Example: Find parametric equations for the tangent line to the helix with parametric equations
x = 2 cos t, y = 2 sin t, z = t
π
at the point (0, 2, ).
2
Solution:
The tangent vector of the helix is
r0 (t) =< −2 sin t, 2 cos t, 1 >
π
π
The point (0, 2, ) corresponds to t = . So at the given point, the tangent vector is
2
2
π
r0 ( ) =< −2, 0, 1 >
2
π
The tangent line is through the point (0, 2, ) parallel the the tangent vector < −2, 0, 1 >, so parametric
2
equations are
π
x = −2t, y = 2, z = + t
2
Second Derivatives: The second derivative of a vector function r(t) is defined as the derivative of
r0 (t), i.e. r00 (t) = (r0 (t))0
Theorem:
2
1.
d
[u(t) + v(t)] = u0 (t) + v0 (t)
dt
2.
d
[cu(t)] = cu0 (t)
dt
3.
d
[f (t)u(t)] = f 0 (t)u(t) + f (t)u0 (t)
dt
4.
d
[u(t) · v(t)] = u0 (t) · v(t) + u(t) · v0 (t)
dt
5.
d
[u(t) × v(t)] = u0 (t) × v(t) + u(t) × v0 (t)
dt
6.
d
[u(f (t))] = f 0 (t)u0 (f (t)) (Chain Rule)
dt
The theorem can be proved directly from definitions.
Example: Show that if |r(t)| = c (constant), then r0 (t) is orthogonal to r(t) for all t.
Proof:
|r(t)|2 = r(t) · r(t) = c2
So
0 = [r(t) · r(t)]0 = r0 (t) · r(t) + r(t) · r0 (t) = 2r0 (t) · r(t)
⇒ r0 (t) · r(t) = 0
Therefore, r0 (t) is always orthogonal to r(t)
Integrals
Z
b
Z
Z
b
Z
g(t)dt j +
f (t)dt i +
r(t)dt =
a
b
a
a
b
h(t)dt k
a
Remark: Integrate component-wisely
Fundamental Theorem of Calculus can be extended to vector functions:
Z b
r(t)dt = R(b) − R(a)
a
here R(t) is the antiderivative of r(t), i.e. R0 (t) = r(t).
3