MA231 Vector Analysis
Example Sheet 1: Hints and partial solutions
2010, term 1
Stefan Adams
A3 (a) (i) ∇f= (2y, 2x+2y), (ii) ∇f= (ycos(πy), x cos(πy)−πxy sin(πy). (b) D(−1,5)f(2,3) =
(6,10) ·(−1,5) = 44.
A4 (a) R1
0t√4+9t2dt. (b) Use parameterisations of the two parts of the curve: α(t)=(t, 0), for
t∈[−2,2], and β(t) = (2 cos t, 2 sin t), for t∈[0, π]. The line integral along αis zero. The
line integral along βgives the answer −32/3.
A5 (a) x2y+x3+C. (b) x2yz +xz +y+C. (c) Later in the course, when we have introduced
curls, we see this immediately from the fact that the curl of vis non-zero. For now, argue by
integration that it is impossible.
A6 (a) (−2/3,−2/3,1/3). (b) (4s2t−2t3,−st2,2t2−2s2)
√(4s2t−2t3)2+s2t4+(2t2−2s2)2.
A7 (a) ∂r/∂s = (1,0,2s)and ∂r/∂t = (0,1,1) gives
∂r/∂ s ×∂r/∂t
=√4s2+ 2. Thus
RSx dS =R1
0R1
−1s√4s2+ 2 dtds=√6−√2/3. (b) For example parameterise by x(s, t) =
(scos t, s sin t, s2)over s∈[0, L1/2], t ∈[0,2π]. The surface area is π
6((1 + 4L)3/2−1).
B 1- B 4 see details of solutions in supervision classes or support class.
C1 ∇T= (2x, 2y, 1) and D(cos θ,sin θ,1) T(1,1,0) = (2,2,1) ·(cos θ, sin θ, 1) = 1 +2 cosθ+ 2 sin θ.
This is maximised at θ=π/4.
C2 Let Cbe the straight line joining the origin to the point x0. The FCT for gradient vector fields
we have f(x0)−f(0) = RCv·ˆ
Tds=g(x0)−g(0).
C3 (b) u=∇1
2ln(1 + g2(x). (c) ∇φ(r2)=2φ0(r2)x. Find φso that 2φ0(r2) = rm, namely
φ(r2) = rm+2
m+2 if m6=−2and φ(r2) = log(r)if m=−2.
C4 The velocity of the tracer particle is u(t, x(t)). Differentiate in tusing the chain rule to find
the acceleration. To explain the notation, use ∇as if it were the vector (∂
∂x1,∂
∂x2,∂
∂x3)so that
the notation u· ∇ becomes P3
i=1 ui∂
∂xiwhich then acts on each component of u.
C5 (b) ∇f=ma constant. (c) ∇g=−3x/r5. (d) The dipole vector field is −3r−5(x·m)x+
r−3m.
C6 (a) The tangent vector is (1, g0(t)) which has length p1 + g0(t)2. (b) The unit normal vec-
tor is given by (∂g/∂ s,∂g/∂ s,1)
√1+(∂g/∂ s)2+(∂g/∂ t)2. Hence the area of the desired part of the graph is
R1
0R1
0p1+(∂g/∂ s)2+ (∂g/∂t)2dsdt.