1. Homework 6
Let r : I → R3be a function with r(t) = (f (t), g(t), h(t)) for all t ∈ I. Let t0∈ R and d > 0. Let i = (1, 0, 0) and j = (0, 1, 0) and k = (0, 0, 1). We denote r by r(t) = f (t)i + g(t)j + h(t)k.
We say that r is bounded on I if there exists M > 0 such that kr(t)k ≤ M for all t ∈ I.
(1) Assume that I = {t ∈ R : 0 < |t − t0| < d}. Let L = (L1, L2, L3) ∈ R3. We say that limt→t0r(t) = L if for any > 0, there exists δ = δ,t0 > 0 such that
kr(t) − Lk < whenever 0 < |t − t0| < δ.
Similarly, we can also define lim
t→±∞γ(t). This will be left to the reader.
(a) Show that
t→tlim0r(t) = L if and only if lim
t→t0f (t) = L1, lim
t→t0g(t) = L2, lim
t→t0h(t) = L3. Proof. This can be proved by the following inequality:
|f (t) − L1|, |g(t) − L2|, |h(t) − L3| ≤ kr(t) − Lk ≤ |f (t) − L1| + |g(t) − L2| + |h(t) − L3|.
Suppose that lim
t→t0
r(t) = L. For every > 0, there exists δ > 0 so that kr(t) − Lk < whenever 0 < |t − t0| < δ. Thus
|f (t) − L1|, |g(t) − L2|, |h(t) − L3| <
whenever 0 < |t − t0| < δ. This implies that limt→t0f (t) = L1 and limt→t0g(t) = L2
and limt→t0h(t) = L3.
Conversely, suppose limt→t0f (t) = L1 and limt→t0g(t) = L2 and limt→t0h(t) = L3. For every > 0, choose δ > 0 so that
|f (t) − L1|, |g(t) − L2|, |h(t) − L3| < 3 whenever 0 < |t − t0| < δ. Thus
kr(t) − Lk ≤ |f (t) − L1| + |g(t) − L2| + |h(t) − L3| < 3 · 3 = whenever 0 < |t − t0| < δ. This implies that lim
t→t0r(t) = L.
(b) Evaluate (i) lim
t→0
t sin1
ti + etj + cos√3 tk
. Solution: j + k
(ii) lim
t→1
t2− 1 t − 1i +√
t + 8j +sin πt ln t k
. Solution: 2i + 3j − πk.
(iii) lim
t→∞
te−ti + t3+ t
2t3− 1j + cos1 tk
Solution: 1 2j + k.
(2) Assume that I = (t0− d.t0+ d). We say that r(t) is differentiable at t0 if lim
h→0
1 t − t0
(r(h + t0) − r(t0)) exists.
In this case, the above limit is denoted by r0(t0) or ˙r(t0) or d
dtr(t)|t=t0.
1
2
(a) Show that r is differentiable at t0 if and only if f (t), g(t), h(t) are differentiable at t0
with r0(t0) = f0(t0)i + g0(t0)j + h0(t0)k.
Proof. This can be proved by the result of exercise (1) and by the following identity 1
t − t0
(r(h + t0) − r(t0)) = f (t) − f (t0) t − t0
i +g(t) − g(t0) t − t0
j + h(t) − h(t0) t − t0
k.
(b) Evaluate r0(t). Here
(i) r(t) = a cos ti + a sin tj + btk for t ∈ R for some a, b ∈ R.
r0(t) = −a sin ti + a cos tj + bk.
(ii) r(t) = et2i − j + ln(1 + 3t)k for t > 0.
r0(t) = 2tet2i + 3 1 + 3tk.
(iii) r(t) = 1
1 + ti + t
1 + tj + tan−11 − t2
1 + t2k, t 6= 1.
r0(t) = −(1 + t)−2i + (1 + t)−2j − 2t t4+ 1k.
(3) Let I = [a, b]. Suppose that γ : [a, b] → R3 is bounded.
(a) Show that f, g, h are also bounded on [a, b].
Proof. This can be proved by the inequality
|f (t)|, |g(t)|, |h(t)| ≤ kγ(t)k.
(b) Let P = {a = t0< t1< · · · < tn= b} be a partition of [a, b] and C = {ci} be a mark of P, i.e. ci∈ [ti−1, ti]. The Riemann sum of r with respect to P and C is defined to be
S(r, C, P ) =
n
X
i=1
(ti− ti−1)r(ci).
We say that r is Riemann integrable over [a, b] if there exists I ∈ R3 such that for any
> 0, there exists δ = δ> 0
kS(r, C, P ) − Ik < whenever kP k < δ.
Show that r is Riemann integrable over [a, b] if and only if f, g, h are Riemann integrable over [a, b] with
Z b a
r(t)dt = Z b
a
f (t)dti + Z b
a
g(t)dtj + Z b
a
h(t)dtk.
Similarly, we can define the improper integral of a vector valued function. The definition is left to the reader.
Proof. This can be proved by the inequality
|S(f, C, P ) − I1|,|S(g, C, P ) − I2|, |S(h, C, P ) − I3| ≤ kS(r, C, P ) − Ik
≤ |S(f, C, P ) − I1| + |S(g, C, P ) − I2| + |S(h, C, P ) − I3|.
Suppose r is Riemann integrable over [a, b]. For every > 0, there exists δ > 0 so that kS(r, C, P ) − Ik < whenever kP k < δ. By
|S(f, C, P ) − I1|, |S(g, C, P ) − I2|, |S(h, C, P ) − I3| ≤ kS(r, C, P ) − Ik,
3
we find
|S(f, C, P ) − I1|, |S(g, C, P ) − I2|, |S(h, C, P ) − I3| <
whenever kP k < δ. Thus f, g, h are Riemann integrable over [a, b]. Conversely, if f, g, h are Riemann integrable over [a, b], then every > 0, there exists δ > 0 so that
|S(f, C, P ) − I1|, |S(g, C, P ) − I2|, |S(h, C, P ) − I3| < 3. Let I =Rb
af (t)dti +Rb
a g(t)dtj +Rb
a h(t)dtk. Then
kS(r, C, P ) − Ik ≤ |S(f, C, P ) − I1| + |S(g, C, P ) − I2| + |S(h, C, P ) − I3| < 3 · 3 = whenever kP k < δ. This shows that r is Riemann integrable over [a, b]. (c) Suppose that r is Riemann integrable over [a, b]. Let u ∈ R3 be any unit vector, i.e.
kuk = 1. Show that
*Z b a
r(t)dt, u +
= Z b
a
hr(t), uidt.
Here hu, vi is the inner product of u, v in R3. Proof. Let u = (u, v, w). Then
*Z b a
r(t)dt, u +
= u Z b
a
f (t)dt + v Z b
a
g(t)dt + w Z b
a
h(t)dt
= Z b
a
(f (t)u + g(t)v + h(t)w)dt
= Z b
a
hr, uidt.
(d) Use (c) to show that
Z b a
r(t)dt
≤ Z b
a
kr(t)kdt.
Proof. IfRb
a r(t)dt = 0, then the above equation is obvious true: the right hand side is always a nonnegative number.
We assume thatRb
a r(t)dt 6= 0. Write u =Rb
a r(t)dt/kRb
a r(t)dtk. Then u is a unit vector and
*Z b a
r(t)dt, u +
= 1
kRb ar(t)dtk
*Z b a
r(t)dt, Z b
a
r(t)dt +
= kRb
ar(t)dtk2 kRb
a r(t)dtk
= k Z b
a
r(t)dtk.
By Cauchy-Schwarz inequality and kuk = 1, hr(t), ui ≤ kr(t)k. Thus Z b
a
hr(t), uidt ≤ Z b
a
kr(t)kdt.
We conclude that
Z b a
r(t)dt
≤ Z b
a
kr(t)kdt.
4
(e) Suppose that r : [a, b] → R3 is a C1-function, i.e. r0 exists and continuous on [a, b].
Show that
kr(b) − r(a)k ≤ Z b
a
kr0(t)kdt.
Proof. By fundamental theorem of Calculus, r(b) − r(a) =
Z b a
r0(t)dt.
Using the above inequality, we prove our assertion.
(f) Evaluate (i)
Z 1 0
t3ln ti + 1
1 + tj + 1 1 + t2k
dt.
Solution: − 1
16i + ln 2j +π 4k.
(ii) Z 1
0
4
1 + t2j + 2t 1 + t2k
dt.
Solution: πj + ln 2k.
(iii) Z ∞
0
e−st(cos ti + sin tj + tk) dt.
Solution: s
1 + s2i + 1
1 + s2j + 1 s2k.
