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MATH172 May 2009 exam: model solutions and marking scheme
All problems are similar to homework and class examples, except where stated explicitly otherwise.
Section A
- Question Evaluate the vector c = 4 a − 3 b and its length |c|, if a = i + 2 j + 3 k and b = − 2 i + 4 k.
Answer c = 10i + 8j; |c| =
[4 marks]
- Question Evaluate the scalar product of a = i + 2 j + 3 k and b = − 2 i + 4 k.
Answer a · b = 10.
[4 marks]
- Question Evaluate the vector product of a = i + 2 j + 3 k and b = − 2 i + 4 k.
Answer a × b =
i j k 1 2 3 − 2 0 4
= 8 i − 10 j + 4 k
[6 marks]
- Question Find the velocity, speed and acceleration of a particle whose position vector at time t is
r(t) = 4t i + 3 cos t j + 3 sin t k.
Answer Velocity v =
dr dt
= 4 i − 3 sin t j + 3 cos t k. Speed |v| =
42 + 3^2 sin^2 t + 3^2 cos^2 t = 5. Acceleration
a = dv dt
= −3 cos t j − 3 sin t k.
[5 marks]
- Question Force F of magnitude 26 and of the same direction as vector − 12 i − 5 j, moves a particle from point (1, 0)(m) to point (0, 1). Find the work done.
Answer The unit vector in the direction of the force is F/|F| = (− 12 i − 5 j)/
122 + 5^2 = − 1213 i − 135 j. The force is F = 26
− 1213 i − 135 j
= − 24 i − 10 j. The vector of displacement is ∆r = (0, 1) − (1, 0) = (− 1 , 1). The work done is W = ∆r · F = (−24)(−1) + (−10)(1) = 14.
[6 marks]
- Question Find the gradient grad φ of the scalar field φ(x, y, z) = x^2 + y^3 + sin(2z).
Answer ∇φ = 2x i + 3y^2 j + 2 cos(2z) k
[6 marks]
- Question Find the general solution of the equation with separable variables dy dx
sin x y
Answer
y dy =
sin x dx, hence y^2 /2 = − cos x + C 1 and finally y = ±
C 2 − 2 cos x where C 2 = 2C 1.
[7 marks]
- Question Find the general solution of the linear differential equation dy dx
y x
= x^2.
Answer The integrating factor: I = exp
x
dx
= eln^ x^ = x so the equation can be re-written as x dy dx
- y = d dx (xy) = x^3 hence xy =
x^3 dx = 14 x^4 + C and y = 14 x^3 + C/x.
[8 marks]
- Question Find the general solution of the differential equation
d^2 y dx^2
dy dx
− 6 y = 0.
Answer The characteristic equation is λ^2 + λ − 6 = 0 which has the solutions λ 1 , 2 = −^1 ±
√1+ 2 =^ {−^3 ,^2 }. Hence the general solution is y = C 1 eλ^1 x^ + C 2 eλ^2 x= C 1 e−^3 x^ + C 2 e^2 x
[8 marks]
- Question Evaluate matrix products AB and B^2 for matrices A and B given by
A =
[
]
, B =
Answer AB =
[
]
; B^2 =
[6 marks] total for this part: 60
Section B
- (i) Question Find det A and AT^ where matrix A is given by
A =
Answer det A = −5, AT^ =
(ii) Question Show that matrix
B =
is the inverse of matrix A.
Answer BA = (^15)
⎦ (^) = I as required.
(iii) Question Hence or otherwise solve the system of simultaneous equations
x + y − 2 z = 1 , − 2 x − y + z = − 1 , − 2 y + z = 3.
Answer This system is equivalent to a matrix equation
Ax = b
where A is as defined above, x = (x, y, z)T^ and b = (1, − 1 , 3)T^ , so the solution can be obtained as
x = A−^1 b; ⎡ ⎣
x y z
that is x = 1, y = −2, z = −1.
Question Find the general solution of this differential equation at these values of parameters.
Answer The characteristic equation is λ^2 + 3λ + 2 = 0, which has two real roots λ 1 , 2 = {− 1 , − 2 }, and the complementary functions is zh = C 1 e−t^ + C 2 e−^2 t^.
The trial solution zp = A cos(2t) + B sin(2t). Thus z′ p = 2B cos(2t) − 2 A sin(2t), z p′′ = − 4 A cos(2t) − 4 B sin(2t), substitution into the equation gives
(− 4 A + 6B + 2A) cos(2t) + (− 4 B − 6 A + 2B) sin(2t) = 65 cos(2t)
which leads to the system
− 2 A + 6B = 65 − 6 A − 2 B = 0
the solution of which is A = − 13 /4, B = 39/4. Thus the general solution of the equation is
z = zp + zh = −
cos(2t) +
sin(2t) + C 1 e−t^ + C 2 e−^2 t
Question Hence determine the coordinate z(t) of the body at time t, if at time t = 0 it was resting at the origin, that is z(0) = 0, dz/dt(0) = 0.
Answer The corresponding velocity is
dz dt
sin(2t) +
cos(2t) − C 1 e−t^ − 2 C 2 e−^2 t
Substituting this into the initial conditions, we get
z(0) = C 1 + C 2 − 13 / 4 = 0 z′(0) = −C 1 − 2 C 2 + 39/ 2 = 0
the solution of which is C 1 = −13, C 2 = 65/4, and the ultimate answer is
z(t) = −
cos(2t) +
sin(2t) − 13 e−t^ +
e−^2 t
Question In particular, calculate the coordinate z of the body at the time t = 5π, accurate to four decimal places.
Answer z(5π) ≈ − 13 /4 = − 3 .2500. NB: if you do it using a calculator, you should take the value of π with a rather high precision. On the other hand, this answer can be obtained without calculator, if only you care to roughly estimate the value of e−^5 π^ and see that is far too small to influence the answer in the 4 d.p. requested.
[20 marks] total for this part: 80
