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AMATH 231 ASSIGNMENT # 2 Solutions: Curves and Paths Fall
2014
Due Monday, September 22, 2014 at 2pm in box 7, slot 11 (A-M)
and 12 (N-Z), locatedacross from MC4066. Late assignments or
assignments submitted to the incorrect dropboxwill receive a grade
of zero. Write your solutions clearly and concisely. Marks will
bededucted for poor presentation and incorrect notation.
1. Consider a moving particle with position ~g(t) =(
2t1+t2
, 1t2
1+t2
)for time t [10, 10].
a) Show that the path of the particle lies on a circle. [2
marks]
b) Find the velocity and speed of the particle. [4 marks]
Solution:
a) Since we know the x(t) and y(t) co-ordinates we can square
them an obtain thatit is a constant, therefore the equation of a
circle,
4t2
(1 + t2)2+
1 2t2 + t4(1 + t2)2
=(1 + t2)2
(1 + t2)2= 1.
b) The velocity is obtained by computing the derivative,
~g(t) =(
2
1 + t2 4t
2
(1 + t2)2,2t
1 + t2 2t (1 t
2)
(1 + t2)2
)=
(2(1 + t2) 4t2
(1 + t2)2,2t(1 + t2 + 1 t2)
(1 + t2)2
)=
(2(1 t2)(1 + t2)2
,4t
(1 + t2)2
).
The speed is the magnitude of this,
~g(t) = 2
(1 t2)2 + 4t2(1 + t2)2
=2
1 + t2
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2. Let ~f(t) and ~g(t) be differentiable vector-value functions
of one variable and let h bea differentiable scalar-value function.
Then prove the following identities:
a) ddt
(h~f) = h ~f + h~f [2 marks]
b) ddt
(~f ~g) = ~f ~g + ~f ~g
c) ddt
(~f ~g) = ~f ~g + ~f ~g [2 marks]d) d
dt(~f(h(t))) = h(t)~f (h(t)) [2 marks]
Note that for a), b) and d) you can use tensor notation to make
things shorter.Solution:
a)d
dt(h~f) =
d
dt(h~fi) = h
fi + hf i = h ~f + h~f
b)d
dt(~f ~g) = d
dt(figi) = f
igi + fIg
i =
~f ~g + ~f ~g
c)
d
dt(~f ~g) = d
dt(f2g3 f3g2, f3g1 f1g3, f1g2 f2g1) ,
= (f 2g3 f 3g2, f 3g1 f 1g3, f 1g2 f 2g1) ,+ (f2g
3 f3g2, f3g1 f1g3, f1g2 f2g1) ,
= ~f ~g + ~f ~g
d)d
dt(~f(h(t))) =
d
dt(fi(h(t))) = h
(t)f i(h(t)) = h(t)~f (h(t)).
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3. Given that the Euclidean norm is defined as x2 = ~x ~x and
that ~x(t) is a C1 functionprove that
a) ddt
(~x(t)1) = ~x(t) ~x(t)/~x3 [2 marks]b) d
dt
(~x(t)~x(t)
)= (~x(t)~x
(t))~x(t)+(~x(t)~x(t))~x(t)~x3 [2 marks]
Solution:
a)
d
dt
(~x(t)1) = 1~x(t)2 ddt ~x(t) ,= 1~x(t)2
~x(t) ~x(t)~x2 ,
= ~x(t) ~x(t)
~x3
b)
d
dt
(~x(t)
~x(t))
= ~x(t)~x(t) ~x(t)
~x3 +~x(t)~x(t)
=(~x(t) ~x(t))~x(t) + (~x(t) ~x(t))~x(t)
~x3
4. The motion of a particle is described by the curve
~g(t) = (sin t t cos t, cos t+ t sin t, t2), 0 t 2pi.a) Show
that the particle moves on a quadratic surface. Describe and sketch
the
curve.
b) Calculate the distance travelled by the particle.
Solution:
a) Since x(t) = sin t t cos t, y(t) = cos t+ t sin t, z(t) = t2
we deduce the following,x2 + y2 = (sin t t cos t)2 + (cos t+ t sin
t)2
= sin2 t 2t sin t cos t+ t2 cos2 t+ cos2 t+ 2t cos t sin t+ t2
sin2 t,= 1 + t2 = 1 + z.
Therefore the surface can be described as x2 + y2 = 1 + z.
The surface starts at z = 1 where it is a point. As we move
upwards it takesthe form of a circle of radius 1 + z. This can be
described as a hyperboloid.
b) The distance is the norm of the displacement,
~g(t) =x2 + y2 + z2 =
1 + t2 + t2 =
1 + 2t2.
