Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 1
Due: at the end of the tutorial session Tuesday/Thursday, 26/28 January 2016
Name and student number:
Consider the vector function (with values in R3 )
r(t) = ln(2 − t) i + (1 + t) j −
(t − 2)2
k
4
1. Find the domain D(r) of the vector function r(t).
Solution: The domain D(r) of r(t) is the intersection of domains of its component functions.
2
Since D(ln(2 − t)) = (−∞, 2), D(1 − t) = (−∞, ∞) and D(− (t+2)
) = (−∞, ∞), one gets
4
D(r) = (−∞, 2),
that is the vector function r(t) is defined for t < 2.
2. Find
(a) the derivative dr/dt,
(b) the norm ||dr/dt||
(c) the unit tangent vector T for all values of t in D(r).
Simplify the expressions obtained.
Hint: use the formula a2 + 2ab + b2 = (a + b)2 .
Solution:
(a)
1
t
dr
=(
, 1, 1 − ).
dt
t−2
2
(b) The magnitude of this vector is
s
s
2
2
2
2
dr
1
t
1
1 2−t
2−t
2
|| || =
+1 + 1−
=
+2
+
dt
t−2
2
2−t
2−t 2
2
s
2
1
2−t
1
2−t
=
+
=
+
,
2−t
2
2−t
2
because t < 2.
(c) The unit tangent vector is
dr
2
4 − 2t
(2 − t)2
dt
T = dr = −
,
,
.
6 − 4t + t2 6 − 4t + t2 6 − 4t + t2
|| dt ||
1
3. Find the vector equation of the line tangent to the graph of r(t) at the point P0 (0, 2, − 14 )
on the curve.
Solution: The point P0 (0, 2, − 14 ) on the curve corresponds to t = 1. We find
r0 = r(1) = 2j −
1
k,
4
v0 =
dr
1
(1) = −i + j + k .
dt
2
Thus the tangent line equation is
1
r = r0 + (t − 1) v0 = (1 − t) i + (t + 1) j + (2t − 3) k .
4
Note that the same line is also described by the following equation which is obtained from the
one above by the shift of the parameter t: t → t + 1
1
r = r0 + t v0 = −t i + (t + 2) j + (2t − 1) k .
4
4. Find the arc length of the graph of r(t) if −2 ≤ t ≤ 1.
Solution: The arc length of the graph of r(t) is given by the definite integral
Z 1 Z 1
dr
1
2−t
(2 − t)2 1
15
|| || dt =
+
dt = − ln(2 − t) −
=
+ 2 ln 2 .
L=
−2
dt
2−t
2
4
4
−2
−2
5. Find a positive change of parameter from t to s where s is an arc length parameter of
the curve having r(1) as its reference point.
Solution: The arc length parameter s can be found as follows
Z t
Z t
dr
1
2−u
(2 − u)2 t
t2 − 4t + 3
s=
|| || du =
+
du = − ln(2 − u) −
=
−
−ln(2−t) .
1
du
2−u
2
4
4
1
1
2