MA22S1: TUTORIAL 1 SOLUTIONS
1. Calculate the following limit:
lim hcos t, sin t, t ln ti
t→0+
Solution: We have limt→0+ cos t = cos 0 = 1 and limt→0+ sin t =
sin 0 = 0. Note that using L’Hôpital’s Rule we have
lim+ t ln t =
t→0
lim+
ln t
t→0
1
t
= lim+ −
t→0
1
t
1
t2
= lim+ −t = 0
t→0
Hence
lim hcos t, sin t, t ln ti =
t→0+
lim cos t, lim+ sin t, lim+ t ln t
t→0+
t→0
t→0
= h1, 0, 0i
2. Let C be the smooth space curve with vector equation
r(t) = h2 cos t, 2 sin t, ti, 0 ≤ t ≤ 4π
(a) Sketch the curve C.
(b) Find the unit tangent vector to C at the point P (2, 0, 0).
(c) Find the total length of the curve C.
(d) Reparametrise C with respect to arc length measure.
(e) If you started at the point P (2, 0, 0) and travelled a distance
√
5π along the curve, at what point on the curve would you be?
Solution:
(a) See Figure 1.
1
2
MA22S1: TUTORIAL 1 SOLUTIONS
2*cos(t), 2*sin(t), t
12
10
8
6
4
2
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
1
1.5
2
0.5
-0.5
0
-1
-1.5
-2
Figure 1. r(t) = h2 cos t, 2 sin t, ti, 0 ≤ t ≤ 4π (Helix)
(b) Note that r(0) = h2, 0, 0i and so the parameter value corresponding to the point P (2, 0, 0) is t = 0. First we compute the unit
tangent vector T(t):
r0 (t) = h−2 sin t, 2 cos t, 1i,
T(t) =
|r0 (t)| =
√
5
2
2
1
r0 (t)
= h− √ sin t, √ cos t, √ i
0
|r (t)|
5
5
5
Hence the unit tangent vector at P (2, 0, 0) is T(0) = h0, √25 , √15 i.
(c)
L=
Z
4π
0
|r (t)| dt =
0
Z
4π
√
√
5 dt = 4 5π
0
(d) The arc length function is
s(t) =
Z
t
0
0
|r (t)| dt =
Z t√
0
5 dt =
√
5t
MA22S1: TUTORIAL 1 SOLUTIONS
Thus t =
√s
5
3
and we can reparametrise the curve in terms of s:
s
s
s
r(s) = h2 cos √ , 2 sin √ , √ i,
5
5
5
√
0 ≤ s ≤ 4 5π
(e) Using the arc-length parametrisation, the point which lies a dis√
tance 5π along the curve has position vector
√
r( 5π) = h2 cos π, 2 sin π, πi = h−2, 0, πi