MATH 251 – LECTURE 6
JENS FORSGÅRD
http://www.math.tamu.edu/~jensf/
This week: 11.6–7, 12.1–3
webAssign: 11.6, 12.1, 12.3, due 2/8 11:55 p.m.
Next week: 12.4–6
webAssign: 12.4–6, opens 2/8 12 a.m.
Help Sessions:
M W 5.30–8 p.m. in BLOC 161
Office Hours:
BLOC 641C
M 12:30–2:30 p.m.
W 2–3 p.m.
or by appointment.
Vector functions
A vector valued function (or simply vector function) is a function
r(t) = hf (t), g(t), h(t)i = f (t)i + g(t)j + h(t)k.
Example 1. Each parametric representation of a line is a vector function: r(t) = r0 + tv.
Example 2. Let r(t) = hcos(t), sin(t), t/10i.
0.5
0.0
1.0
0.5
-0.5
-1.0
0.0
-0.5
-0.5
0.0
0.5
1.0
-1.0
Vector functions - Domains
The domain of a space curve is the set of all parameters t for which all the function r(t) is defined. That is, all
parameters for which all of the function f (t), g(t), and h(t) are defined.
√
t
i.
Exercise 3. Find the domain of the vector function r(t) = h 5 + t, t2 + 1, t−1
Vector functions - Limits
Let r(t) = hf (t), g(t), h(t)i, then
D
lim r(t) = lim f (t), lim g(t), lim h(t)
t→a
t→a
Exercise 4. Let
r(t) =
Compute limt→0 r(t).
t→a
2
E
t→a
1
sin(t) sin (t)
,
,
arctan
.
t
t(et − 1)
t2
Vector functions - Continuity
A vector function r(t) is continuous at a if limt→a r(t) = r(a). A vector function r(t) is continuous at a if and
only if all component functions f, g, and h are continuous at a.
If r(t) is continuous for all points in its domain, then the image of r(t) (or its range) is said to be a space curve.
Example 5. Let r(t) = ht, t2, e−ti.
Space curves - Derivatives
The derivative r0(t) of a vector function r(t) is defined as
r(t + h) − r(t)
dr
= lim
,
dt h→0
h
provided that the limit exists.
If r(t) = hf (t), g(t), h(t)i where f, g, and h are differentiable, then r0(t) = hf 0(t), g 0(t), h0(t)i.
Exercise 6. Let r(t) = hcos(2t), t, sin(2t)i. Compute r0(t).
Space curves - Derivatives
The vector r0(t) is the tangent vector of the curve C at the point r(t).
Exercise 7. Find a parametric representation of the tangent line to the curve r(t) = hcos(2t), t, sin(2t)i at the
point r(π/4) = hcos(π/2), π/4, sin(π/2)i = h0, π/4, 1i.
Space curves - Derivatives
Exercise 8. At what point does the curves r1(t) = het, t2 + 1, 3ti and r2(s) = h1, cos(s), s2i intersect? Find
their angle of intersection.
Space curves - Derivatives
Suppose r(t), r1(t), and r2(t) are differentiable vector function, that F (t) is a differentiable function, and let k
be a scalar. Then
(1)
d
dt (r1 (t)
+ r2(t)) = r01(t) + r02(t)
(2)
d
dt kr(t)
= kr0(t)
(3)
d
dt F (t)r(t)
(4)
d
dt (r1 (t)
· r2(t)) = r01(t) · r2(t) + r1(t) · r02(t)
(5)
d
dt (r1 (t)
× r2(t)) = r01(t) × r2(t) + r1(t) × r02(t)
(6)
d
dt r(F (t))
= F 0(t)r(t) + F (t)r0(t)
= F 0(t)r0(F (t)).
Space curves - Integrals
The integral
Rb
a
r(t) dt is defined component wise as
Z b
Z b
Z b
Z b
r(t) dt =
f (t) dt,
g(t) dt,
h(t) dt .
a
a
a
a
The Fudamental Theorem of Calculus is still true! That is, if R(t) is an antiderivative of r(t), then
Z b
r(t) dt = [R(t)]t=b
t=a = R(b) − R(a)
a
Exercise 9. Compute the integral of r(t) = ht, sin(t), eti over the interval [0, 4].
Space curves - Arc length
The length of a space curve r(t) = hf (t), g(t), h(t)i as a ≤ t ≤ b is
Z bp
Z b
|r0(t)| dt =
L=
f 0(t)2 + g 0(t)2 + h0(t)2 dt
a
a
Exercise 10. Find the length of the curve r(t) = hcos(t), sin(t), ti as 0 ≤ t ≤ 1.
Space curves - Arc length
√
√ 3
2
Exercise 11. Compute the length of the curve r(t) = h 2 t, 3 t , 2 t i as 0 ≤ t ≤ 1.