MA22S1: TUTORIAL 2 SOLUTIONS
Let C be the smooth space curve with vector equation
r(t) = h2 cos t, 2 sin t, ti, 0 ≤ t ≤ 4π
1. Find the unit tangent vector, unit normal vector and binormal vector
to C at the point P (2, 0, 0).
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(0) at P ,
r0 (t) = h−2 sin t, 2 cos t, 1i
|r0 (t)| =
√
5
r0 (t)
=
T(t) = 0
|r (t)|
2
2
1
− √ sin t, √ cos t, √
5
5
5
1
2
T(0) = 0, √ , √
5
5
Next we compute the unit normal vector N(0) at P ,
0
T (t) =
2
2
− √ cos t, − √ sin t, 0
5
5
2
|T0 (t)| = √
5
N(t) =
T0 (t)
= h− cos t, − sin t, 0i
|T0 (t)|
N(0) = h−1, 0, 0i
1
2
MA22S1: TUTORIAL 2 SOLUTIONS
Finally we compute the binormal vector B(0) at P ,
j
k i
B(0) = T(0) × N(0) = 0 √2 √1 5
5 −1 0 0 2
1
2
1
= − √ j + √ k = 0, − √ , √
5
5
5
5
2. Find parametric equations for the tangent line and normal line to C
at P (2, 0, 0).
The tangent line passes through the point (2, 0, 0) and is parallel
to the tangent vector T(0) = h0, √25 , √15 i. Thus parametric equations
for the tangent line are,
x = x0 + t.a = 2,
2t
y = y0 + t.b = √ ,
5
t
z = z0 + t.c = √
5
The normal line passes through the point (2, 0, 0) and is parallel to
the normal vector N(0) = h−1, 0, 0i. Thus parametric equations for
the normal line are,
x = x0 + t.a = 2 − t,
y = y0 + t.b = 0,
z = z0 + t.c = 0
3. Find the curvature of C. What does this tell us about the space
curve?
κ(t) =
|T0 (t)|
2
=
0
|r (t)|
5
Since the curvature is constant we can conclude that the space curve
is changing direction at the same rate at every point.
4. Find the scalar equation for the normal plane at P (2, 0, 0).
The normal plane passes through (2, 0, 0) and is orthogonal to the
tangent vector T(0) = h0, √25 , √15 i. Thus a scalar equation for the
MA22S1: TUTORIAL 2 SOLUTIONS
normal plane is,
(x − x0 )a + (y − y0 )b + (z − z0 )c = 0 =⇒ 2y + z = 0
3