ASSIGNMENT 1 - SOLUTIONS
1. Compute
lim
t→0
e2t − 1 1 − cos t
2
,
t
+
2
.
,
t
t2
Solution. We have
e2t − 1 1 − cos t
e2t − 1
1 − cos t
1
2
2
lim
,
, t +2 = lim
, lim
, lim (t +2) = (2, , 2),
t→0
t→0
t→0
t→0
t
t2
t
t2
2
since
e2t − 1
2e2t
1 − cos t
sin t
1
lim
= lim
= 2,
lim
= lim
= .
t→0
t→0 1
t→0
t→0 2t
t
t2
2
2. What planar curve does the vector-valued function
r(t) = (2 cos t, 3 sin t),
0 ≤ t ≤ 2π,
represent? Find the unit tangent vector T at the point r( π2 ) = (0, 3) and find the
equation of the tangent line at this point.
Solutions. The functions x(t) = 2 cos t, y(t) = 3 sin t satisfy
x2 (t) y 2 (t)
+
= 1, 0 ≤ t ≤ 2π.
4
9
This shows that the curve is an ellipse. We have
r0 (t) = (−2 sin t, 3 cos t).
Hence r0 ( π2 ) = (−2, 0) and
π r0 ( π )
(−2, 0)
T
= 0 π2 =
= (−1, 0).
2
kr ( 2 )k
2
The tangent line contains the point (0, 3) and is parallel to the vector (−1, 0), and
therefore it has the parametric equation (x, y) = (−t, 3), t ∈ R (this is the line y = 3).
3. Find the arc length of the portion of cycloid given by
π
3π
r(t) = (t − sin t, 1 − cos t),
≤t≤
.
2
2
Solution. The arc length is
Z 3π/2 Z 3π/2 q
dr L=
(1 − cos t)2 + sin2 t dt
du =
du
π/2
π/2
Z 3π/2 p
Z 3π/2 p
=
1 − 2 cos t + cos2 t + sin2 t dt =
2(1 − cos t) dt
π/2
Z
3π/2
=
π/2
π/2
r
t
4 sin ( ) dt =
2
2
Z
3π/2
π/2
t
2| sin( )| dt = 2
2
1
Z
3π/2
π/2
t
sin( ) dt
2
2
ASSIGNMENT 1 - SOLUTIONS
√
√
h
√
2
2
t i3π/2
= −4(−
= 2 −2 cos( )
) − 4(−
) = 4 2.
2 t=π/2
2
2
4. Find an arc length parametrization for the circular helix
r(t) = (2 cos t, 2 sin t, t)
with reference point r(0).
Solution. An arc length parameter is
Z t Z t
dr s=
k(−2 sin u, 2 cos u, 1)k du
du =
0 du
0
Z tq
√
=
4( sin2 u + cos2 u) + 1 du = 5 t.
0
Then t =
√s
5
and an arc length parametrization with reference point r(0) is
s s s
r(s) = (2 cos √ , 2 sin √ , √ ).
5
5
5
5. What curve does the vector-valued function
r(t) = (t, −t2 ),
t ∈ R,
represent? Find the unit normal vector N , the curvature κ and the osculating circle
at the point r(0).
Solution. From x = t, y = −t2 it follows that y = −x2 which is a parabola. We
compute r0 (t) = (1, −2t). The unit tangent vector at t is
T (t) =
Then
(1, −2t)
r0 (t)
=√
.
kr0 (t)k
1 + 4t2
−4t
−2
,
, T 0 (0) = (0, −2).
(1 + 4t2 )3/2 (1 + 4t2 )3/2
The unit normal vector at t = 0 is given by
T 0 (0)
N (0) =
= (0, −1).
kT 0 (0)k
The curvature at t = 0 is
kT 0 (0)k
2
k(0) = 0
= = 2.
kr (0)k
1
1
The radius of the osculating circle is ρ = k(0)
= 21 and the centre is at r(0) + ρN (0) =
(0, 0) + 21 (0, −1) = (0, − 21 ).
T 0 (t) =