Question
The diagram shows points in a park viewed from above, at a specific moment in time.
The distance between two trees, at points A and B, is 6.36m.
Odette is playing football in the park and is standing at point O, such that OA = 25.9m and OÂB = \(125^0\).
(a) Calculate the area of triangle AOB.
Odette’s friend, Khemil, is standing at point K such that he is 12m from A and KÂB = \(45^0\)
(b) Calculate Khemil’s distance from B.
XY is a semicircular path in the park with centre A, such that KÂY = \(45^0\). Khemil is standing on the path and Odette’s football is at point X. This is shown in the diagram below.
The length KX = 22.2m, KÔX = \(53.8^0\) and OK̂X = \(51.1^0\).
(c) Find whether Odette or Khemil is closer to the football.
Khemil runs along the semicircular path to pick up the football.
(d) Calculate the distance that Khemil runs.
Answer/Explanation
Answer:
(a) attempt to use area of triangle formula
\(\frac{1}{2} \times 25.9 \times 6.36 \times sin (125^0)\)
67.5 \(m^2\) (67.4700… \(m^2\))
(b) attempt to use cosine rule
(BK=) \(\sqrt{12^2 + 6.36^2 – 2 \times 12 \times 6.36 \times cos 45^0}\)
8.75 (m) (8.74738… (m))
(c) METHOD 1
attempt to use sine rule with measurements from triangle OKX
\(\frac{OX}{sin 51.1^o} = \frac{22.2}{sin 53.8^o}\)
(OX=) 21.4 (m) (21.4099…)(m)
(21.4 (m) < 22.2 (m))
METHOD 2
sketch of triangle OXK with vertices, angles and lengths
\(51.1^o\) is smallest angle in triangle OXK
opposite side (OX) is smallest length
therefore Odette is closest
(d) attempt to use length of arc formula
\(\frac{135}{360} \times 2 \pi \times 12 \)
28.3 (m) (9 \(\pi\), 28.2743…) (m)
Question
A scientist is conducting an experiment on the growth of a certain species of bacteria.
The population of the bacteria, P, can be modelled by the function
P(t) = 1200 × \(k^t\), t ≥ 0.
where t is the number of hours since the experiment began, and k is a positive constant.
(a) (i) Write down the value of P(0).
(ii) Interpret what this value means in this context.
3 hours after the experiment began, the population of the bacteria is 18 750.
(b) Find the value of k
(c) Find the population of the bacteria 1 hour and 30 minutes after the experiment began.
The scientist conducts a second experiment with a different species of bacteria.
The population of this bacteria, S, can be modelled by the function
S(t) = 5000 × \(1.65^t\)
, t ≥ 0 ,
where t is the number of hours since both experiments began.
(d) Find the value of t when the two populations of bacteria are equal.
It takes 2 hours and m minutes for the number of bacteria in the second experiment to reach 19 000.
(e) Find the value of m, giving your answer as an integer value.
Answer/Explanation
Answer:
(a) (i) 1200
(ii) the initial population of the bacteria
(b) 1200 × \(k^3\) = 18750
(k=) 2.5
(c) 1200 × \(2.5^{1.5}\)
4740 (4743.41…)
(d) equating P (t) and S (t) OR equating each function to a common variable
1200 × \(2.5^t\) = 5000 × \(1.65^t\) ; 1200 × \(2.5^t\) = x and 5000 × \(1.65^t\) = x
t = 3.43 (hours) (3.43456…)
(e) METHOD 1
5000 × \(1.65^t\) = 19000
(t=) 2.66586… OR (t – 2 ) 0.66586… (seen)
multiplying by 60 seen to convert to minutes
(m = 39.9521…)
(m=) 40 (minutes) OR 2 hours and 40 minutes
METHOD 2
equating an expression for S (t) to 19000
expressing t as 2 + \(\frac{m}{60}\)
5000 \(\times 1.65^{2 + \frac{m}{60}}\) = 19000
2 + \(\frac{m}{60}\) = 2.66586…
(m=) 40 (minutes) OR 2 hours and 40 minutes