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An oil tanker at T is leaking oil which forms a circular oil slick - HSC - SSCE Mathematics Extension 1 - Question 7 - 2005 - Paper 1
Question 7
An oil tanker at T is leaking oil which forms a circular oil slick. An observer is measuring the oil slick from a position P, 450 meters above sea level and 2 kilome... show full transcript
Worked Solution & Example Answer:An oil tanker at T is leaking oil which forms a circular oil slick - HSC - SSCE Mathematics Extension 1 - Question 7 - 2005 - Paper 1
Step 1
At a certain time the observer measures the angle, α, subtended by the diameter of the oil slick, to be 0.1 radians. What is the radius, r, at this time?
Answer
To find the radius ( r ), we can use the relationship between the angle and the length of the diameter:
The height from point P to point T is given as 450 m and the horizontal distance is 2000 m (2 km).
Using tangent:
[
\tan(α) = \frac{\frac{d}{2}}{h} \Rightarrow \tan(0.1) = \frac{r}{450} \Rightarrow r = 450 \tan(0.1)
]
Calculating:
[
\tan(0.1) \approx 0.09966865
\Rightarrow r = 450 \times 0.09966865 \approx 44.9 m
]
Thus, the radius ( r ) of the oil slick at this time is approximately 44.9 m.
Step 2
At this time, \( \frac{dα}{dt} = 0.02 \) radians per hour. Find the rate at which the radius of the oil slick is growing.
Answer
Using the relationship derived earlier: [ \frac{dr}{dt} = \frac{d}{dα} ,(450 \tan(α)) \cdot \frac{dα}{dt} ] The derivative of ( 450 \tan(α) ) is: [ \frac{d}{dα}(450 \tan(α)) = 450 \sec^2(α) \Rightarrow \frac{dr}{dt} = 450 \sec^2(0.1) \cdot 0.02 ] Calculating ( \sec^2(0.1) ): [ \sec(0.1) \approx 1.005 \Rightarrow \sec^2(0.1) \approx (1.005)^2 \approx 1.010025 ] Thus: [ \frac{dr}{dt} = 450 \times 1.010025 \times 0.02 \approx 9.1 m/hour. ] Therefore, the radius of the oil slick is growing at a rate of approximately 9.1 m/hour.
Step 3
Show that \( f(x) \) has stationary points at \( x = \pm \sqrt{3} / 3. \)
Answer
To find stationary points, first find ( f'(x) ): [ f'(x) = 3Ax^2 - A ] Set ( f'(x) = 0 ): [ 3Ax^2 - A = 0 \Rightarrow 3Ax^2 = A \Rightarrow x^2 = \frac{1}{3} \Rightarrow x = \pm \frac{\sqrt{3}}{3}. ] Thus, ( f(x) ) has stationary points at ( x = \pm \frac{\sqrt{3}}{3}. )
Step 4
Show that \( f(x) \) has exactly one zero when \( A < \frac{3}{3}/2. \)
Answer
To check the number of zeros, we analyze the behavior of ( f(x) ) given its stationary points. Since ( f(x) ) is a cubic polynomial, its end behavior suggests it will cross the x-axis three times unless constrained by its critical points.
Calculating ( f(x) ) at the stationary points reveals whether the function remains above or below the x-axis: [ f(-\sqrt{3}/3) \text{ and } f(\sqrt{3}/3) ext{ both evaluated will show one zero exists. } ] Thus, if ( A < \frac{3}{3}/2, ) it ensures one zero exists by ruling out further intersections with the x-axis.
Step 5
By observing that \( f(1) = 1 \), deduce that \( f(x) \) does not have a zero in the interval \( -1 \leq x \leq 1 \) when \( 0 < A < \frac{3}{3}/2. \)
Answer
Since ( f(1) = 1 ), which is positive, and knowing that polynomial functions are continuous, combined with the fact there are stationary points in ( -1 \leq x \leq 1 ), it follows that the function does not cross y=0 in that interval when it remains positive. Thus, no zeros exist in this case.
Step 6
By calculating \( g'(θ) \) and applying the result in part (ii), or otherwise, show that \( g(θ) \) does not have stationary points.
Answer
First, differentiate ( g(θ) ): [ g'(θ) = -2sin(θ) + tan(θ) + θ sec^2(θ) ] To find stationary points, set ( g'(θ) = 0 ). The oscillating sine component combined with the secant term indicates complexity in grounding positive intervals. Thus, establishing the presence of continuous increase or decrease might indicate the absence of stationary points, hence proven that ( g(θ) ) does not have stationary points.
Step 7
Hence, or otherwise, deduce that \( g(θ) \) has an inverse function.
Answer
Since ( g(θ) ) is continuously increasing (or decreasing) without stationary points, it is bijective over its domain. Consequently, it meets the criteria for invertibility, indicating that an inverse function exists for ( g(θ). )