Question 7. [Maximum mark: 17]
Consider the following system of coupled differential equations.
\(\frac{dx}{dt}\)= −4x
\(\frac{dx}{dt}\)= 3x − 2y
a. Find the eigenvalues and corresponding eigenvectors of the matrix 
[6]
b. Hence, write down the general solution of the system. [2]
c. Determine, with justification, whether the equilibrium point (0, 0) is stable or unstable. [2]
d. Find the value of \(\frac{dy}{dx}\)
(i) at (4 , 0).
(ii) at (−4, 0). [3]
e. Sketch a phase portrait for the general solution to the system of coupled differential equations for −6 ≤ x ≤ 6, −6 ≤ y ≤ 6. [4]
▶️Answer/Explanation
(a) 
\( (-4-\lambda )(-2-\lambda )=0\)
\(\lambda =-4\) OR \(\lambda =-2 \)
\(\lambda =-4\) 
\(3x-2y=-4y\)
\(3x=-2y\)
possible eigenvector is \((_{3}^{-2})(or any real multiple) \lambda =-2\) 
\(x = 0, y = 1\) possible eigenvector is \((_{1}^{0}) (or any real multiple) 
(c) two (distinct) real negative eigenvalues (or equivalent (eg both \(e^{-4t}\rightarrow 0,e^{-2t}\rightarrow 0 as t\rightarrow \)∞ ⇒ stable equilibrium point
(d)\(\frac{dy}{dx}=\frac{3x-2y}{-4x}(i)(4,0)\Rightarrow \frac{dy}{dx}=-\frac{3}{4}(ii)(-4,0)\Rightarrow \frac{dy}{dx}=-\frac{3}{4}\)
(e) 
Question 6. [Maximum mark: 15]
A particle P moves along the x-axis. The velocity of P is v m s–1 at time t seconds, where
v = −2t2 + 16t − 24 for t ≥ 0.
a. Find the times when P is at instantaneous rest. [2]
b. Find the magnitude of the particle’s acceleration at 6 seconds. [4]
c. Find the greatest speed of P in the interval 0 ≤ t ≤ 6. [2]
d. The particle starts from the origin O. Find an expression for the displacement of P from O at time t seconds. [4]
e. Find the total distance travelled by P in the interval 0 ≤ t ≤ 4. . [3]
▶️Answer/Explanation
(a) solving v = 0 t= 2, t= 6
(b)use of power rule \(\frac{dv}{dt}=-4t+16 (t=6)\Rightarrow a=-8magnitude = 8 ms^{-2}\)
(c) using a sketch graph of \(v=24ms^{-1}\)
(d) \(x=\int vdt\) attempt at integration of
\(v -\frac{2t^{3}}{3}+8t^{2}-24t+c\)
attempt to find c (use of t x = 0 = x= 0)
\(c = 0\) \(x=-\frac{2t^{3}}{3}+8t^{2}-24t)\)
(e) \(\int_{0}^{4}|v|dt = 32 m\)
Question 6. [Maximum mark: 18]
An ice-skater is skating such that her position vector when viewed from above at time t
seconds can be modelled by
\((_{x}^{y})\) = 
with respect to a rectangular coordinate system from a point O, where the non-zero constants a and b can be determined. All distances are in metres.
a. Find the velocity vector at time t . [3]
b. Show that the magnitude of the velocity of the ice-skater at time t is given by
\(a^{ebt}\sqrt{(1+b^{2})}\) [4]
At time t = 0 , the displacement of the ice-skater is given by \((_{0}^{5})\) and the velocity of the
ice-skater is given by ( 5 ) .
c. Find the value of a and the value of b . [3]
d. Find the magnitude of the velocity of the ice-skater when t = 2 . [2] At a point P, the ice-skater is skating parallel to the y-axis for the first time.
e. Find OP. [6]
▶️Answer/Explanation
(a) use of product rule \((_{y}^{x})= (_{abe^{bt}sin t + ae^{bt}cost t}^{abe ^{bt}cost – ae ^{bt}sin t})\)
(b) \(|v|^{2}= x^{2}+y^{2}=[abe^{bt}cost – ae^{bt}sint]^{2}+ [abe^{bt}sint+ae^{bt}cost]^{2}\)
\(= [a^{2}sin^{2}t – 2a^{2}bsin t cost + a^{2}b^{2}cos^{2}t + a^{2}cos^{2}+ 2a^{2}bsint costt +a^{2}b^{2}sin^{2}t ]\) use of \(sin^{2}t + cos^{2}t=1\)
\(=a^{2}(b^{2}+1)e^{2bt}\) magnitude of velocity is \(ae^{bt}\sqrt{(1+ b^{2})}\)
(c) when t = 0, \(ae^{bt}cost = 5 a=5 abe^{bt}cost – ae^{bt}sint = -3.5 b = -0.7\ )
(d)\(5e^{-0.7\times 2}\sqrt{(1+(-0.7)^{2})}1.51(1.50504…)\)
(e) \(x=0 ae^{bt}(b cost – simt)=0 tant=b t = 2.53 (2.53086…)\) correct substitution of their t to find x or y
\(x = − 0.697 (- 0.696591…)\) and \(y = 0.488 (0.487614…)\)
use of Pythagoras / distance formula \(OP = 0.850 m (0.850297…)\)
Question
A particle moves in a straight line. The velocity, v ms–1 , of the particle at time t seconds is given by v (t) = t sin t – 3, for 0 ≤ t ≤ 10.
The following diagram shows the graph of v .
Find the smallest value of t for which the particle is at rest. [2]
Find the total distance travelled by the particle. [2]
Find the acceleration of the particle when t = 7 . [2]
▶️Answer/Explanation
Ans:
(a) recognising v= o t = 6.74416… = 6.74
= 37.0968…
= 37.1 (m)
(c)
recognizing acceleration at t = 7 is given by v′(7)
acceleration = 5.93430…
5.93
Question
Two boats A and B travel due north.
Initially, boat B is positioned 50 metres due east of boat A.
The distances travelled by boat A and boat B, after t seconds, are x metres and y metres respectively.
The angle θ is the radian measure of the bearing of boat B from boat A.
This information is shown on the following diagram.
Show that y = x + 50 cot θ. [1]
At time T , the following conditions are true.
Boat B has travelled 10 metres further than boat A. Boat B is travelling at double the speed of boat A.
The rate of change of the angle θ is –0.1 radians per second.
Find the speed of boat A at time T . [6]
▶️Answer/Explanation
Ans:
(a)
tan \(\Theta = \frac{50}{y-x}\)
OR
\(cot \Theta = \frac{y-x}{50 }\)
\(y= x+50 \: cot\Theta \)
(b)
attempt to differentiate with respect to t
\(\frac{dy}{dt}= \frac{dx}{dt}= – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)
attempt to set speed of B equal to double the speed of A
\(2 \frac{dx}{dt}=\frac{dx}{dt} – 50 (cosec\Theta )^{2}\frac{d\Theta }{dt}\)
\(\Theta = arctan 5 (=1.373…= 78.69..)\)
OR
\(cosec ^{2}\Theta = 1+ cot^{2}= 1+(\frac{1}{5})^2= \frac{26}{25}\)
\(\frac{dx}{dt}= -50(\frac{26}{25})\times -0.1\)
So the speed of boat A is 5.2 (ms-1)