Question
A rotating sprinkler is at a fixed point S .
It waters all points inside and on a circle of radius 20 metres.
Point S is 14 metres from the edge of a path which runs in a north-south direction.
The edge of the path intersects the circle at points A and B .
This information is shown in the following diagram.
(a) Show that AB = 28.57, correct to four significant figures.
The sprinkler rotates at a constant rate of one revolution every 16 seconds.
(b) Show that the sprinkler rotates through an angle of $\frac{\pi}{8}$ radians in one second.
Let T seconds be the time that [AB] is watered in each revolution.
(c) Find the value of T.
Consider one clockwise revolution of the sprinkler.
At $t = 0$, the water crosses the edge of the path at $A$.
At time $t$ seconds, the water crosses the edge of the path at a movable point $D$ which is a distance $d$ metres south of point $A$.
Let $\alpha = \angle ASD$ and $\beta = \angle SAB$, where $\alpha, \beta$ are measured in radians.
This information is shown in the following diagram.
(d) Write down an expression for $\alpha$ in terms of $t$.
It is known that $\beta = 0.7754$ radians, correct to four significant figures.
(e) By using the sine rule in $\Delta ASD$, show that the distance, $d$, at time $t$, can be modelled by
$d(t) = \dfrac{20 \sin(\frac{\pi t}{8})}{\sin(2.37 – \frac{\pi t}{8})}$
A turtle walks south along the edge of the path.
At time $t$ seconds, the turtle’s distance, $g$ metres south of $A$, can be modelled by
$g(t) = 0.05t^2 + 1.1t + 18$, where $t \geq 0$.
(f) At $t = 0$, state how far south the turtle is from $A$.
Let $w$ represent the distance between the turtle and point $D$ at time $t$ seconds.
(g) (i) Use the expressions for $g(t)$ and $d(t)$ to write down an expression for $w$ in terms of $t$.
(ii) Hence find when and where on the path the water first reaches the turtle.
▶️Answer/Explanation
Detail Solution: –
Let’s dive into this sprinkler problem step by step, solving each part systematically. We’re dealing with a circle, a line (the path), and some rotational motion, so geometry and trigonometry will be our tools. I’ll assume a coordinate system to make the math concrete, then proceed with the calculations.
Part (a): Show that AB = 28.57, correct to four significant figures
Imagine the sprinkler at point S, fixed at the origin (0, 0) for simplicity. Its watering range forms a circle of radius 20 meters. The path runs north-south, and its edge is 14 meters from S. Let’s place the path’s edge along the vertical line \( x = 14 \), since it’s 14 meters from S and runs north-south (a vertical line in a 2D plane).
The circle’s equation, centered at (0, 0) with radius 20, is:
\[ x^2 + y^2 = 20^2 = 400 \]
The path’s edge is the line \( x = 14 \). To find points A and B (where the path intersects the circle), substitute \( x = 14 \) into the circle’s equation:
\[ 14^2 + y^2 = 400 \]
\[ 196 + y^2 = 400 \]
\[ y^2 = 400 – 196 = 204 \]
\[ y = \pm \sqrt{204} \]
Calculate \( \sqrt{204} \):
\[ \sqrt{204} = \sqrt{4 \cdot 51} = 2 \sqrt{51} \]
Using a calculator, \( \sqrt{51} \approx 7.141428 \), so:
\[ 2 \sqrt{51} \approx 2 \cdot 7.141428 \approx 14.282856 \]
Thus, the intersection points are approximately:
– A at (14, \( \sqrt{204} \)) ≈ (14, 14.282856)
– B at (14, \( -\sqrt{204} \)) ≈ (14, -14.282856)
The distance AB is the vertical distance between these points, i.e., the difference in their y-coordinates:
\[ AB = \sqrt{204} – (-\sqrt{204}) = 2 \sqrt{204} \]
\[ 2 \sqrt{204} = 2 \cdot 2 \sqrt{51} = 4 \sqrt{51} \]
Now compute numerically:
\[ 4 \sqrt{51} \approx 4 \cdot 7.141428 \approx 28.565712 \]
To four significant figures:
\[ 28.565712 \approx 28.57 \]
This matches the target value of 28.57. To confirm geometrically, consider the chord length formula for a circle. The perpendicular distance from the center (0, 0) to the line \( x = 14 \) is 14 meters. Half the chord length is the horizontal leg of a right triangle with hypotenuse 20 (radius) and leg 14:
\[ \text{Half-chord} = \sqrt{20^2 – 14^2} = \sqrt{400 – 196} = \sqrt{204} \]
\[ AB = 2 \sqrt{204} \approx 28.57 \]
The calculation holds, so AB = 28.57 meters, as required.
Part (b): Show that the sprinkler rotates through an angle of \(\frac{\pi}{8}\) radians in one second
The sprinkler completes one full revolution every 16 seconds. One revolution is \( 2\pi \) radians. To find the angular speed, divide the total angle by the time:
\[ \text{Angular speed} = \frac{\text{Total angle}}{\text{Time}} = \frac{2\pi}{16} = \frac{\pi}{8} \text{ radians per second} \]
In one second, it rotates:
\[ \frac{\pi}{8} \cdot 1 = \frac{\pi}{8} \text{ radians} \]
This is exactly what we needed to show. The sprinkler’s constant rate of one revolution per 16 seconds gives an angular velocity of \(\frac{\pi}{8}\) radians per second, confirming the result.
Part (c): Find the value of T
T is the time per revolution that the segment AB is watered. As the sprinkler rotates, its water stream sweeps a circular sector. AB, a vertical chord at \( x = 14 \), is watered when the sprinkler’s radius (a line of length 20) intersects points A and B. We need the time it takes to rotate between these two positions.
First, find the angle between the radii SA and SB. Place S at (0, 0), A at (14, \( \sqrt{204} \)), and B at (14, \( -\sqrt{204} \)). Compute the central angle \(\theta\) between SA and SB using the dot product of vectors SA and SB:
– \( SA = (14, \sqrt{204}) \)
– \( SB = (14, -\sqrt{204}) \)
Dot product:
\[ SA \cdot SB = 14 \cdot 14 + \sqrt{204} \cdot (-\sqrt{204}) = 196 – 204 = -8 \]
Magnitudes:
\[ |SA| = |SB| = 20 \] (since both are radii)
\[ \cos\left(\frac{\theta}{2}\right) = \frac{\text{Distance from S to path}}{\text{Radius}} = \frac{14}{20} = 0.7 \]
Alternatively, full angle \(\theta\):
\[ \cos(\theta) = \frac{SA \cdot SB}{|SA| |SB|} = \frac{-8}{20 \cdot 20} = \frac{-8}{400} = -0.02 \]
\[ \theta = \arccos(-0.02) \approx 1.5908 \text{ radians} \]
Or use the half-angle:
\[ \frac{\theta}{2} = \arccos(0.7) \approx 0.7954 \]
\[ \theta = 2 \cdot 0.7954 \approx 1.5908 \text{ radians} \]
Angular speed is \(\frac{\pi}{8} \approx 0.3927\) radians per second. Time T:
\[ T = \frac{\theta}{\text{Angular speed}} = \frac{1.5908}{0.3927} \approx 4.05 \text{ seconds} \]
Check: In a full revolution (16 seconds), it sweeps \(2\pi \approx 6.2832\) radians. The fraction of time AB is watered:
\[ \frac{\theta}{2\pi} \cdot 16 = \frac{1.5908}{6.2832} \cdot 16 \approx 4.05 \]
T ≈ 4.05 seconds, which is consistent.
Let’s dive into this sprinkler-and-turtle adventure step by step, solving each part methodically.
Part (d): Expression for \(\alpha\) in terms of \(t\)
We’re dealing with a sprinkler that rotates clockwise, completing one revolution over some period. At \(t = 0\), the water is at point \(A\), and at time \(t\), it’s at point \(D\), which is \(d\) meters south of \(A\). The angle \(\alpha = \angle ASD\) is what we need to express in terms of \(t\). Since the sprinkler rotates, \(\alpha\) represents the angular position of the water stream relative to some fixed line, likely from point \(S\) (the sprinkler’s location) through points \(A\) and \(D\).
The problem doesn’t explicitly state the sprinkler’s rotation period, but part (e) gives us a clue with the expression \(\frac{\pi t}{8}\), suggesting the angular speed might be tied to that. Assume the sprinkler completes one full revolution ( \(2\pi\) radians) in \(T\) seconds, so the angular speed is \(\frac{2\pi}{T}\) radians per second. Starting at \(t = 0\) at point \(A\), and moving clockwise, the angle \(\alpha\) increases with time. If \(\alpha = 0\) at \(t = 0\) (when the water is at \(A\)), then for a clockwise rotation:
\[
\alpha = \omega t
\]
where \(\omega\) is the angular speed. From part (e), we’ll see \(\frac{\pi t}{8}\), implying the period \(T = 16\) seconds (since \(2\pi = \omega \cdot 16\), so \(\omega = \frac{2\pi}{16} = \frac{\pi}{8}\)). Thus:
\[
\alpha = \frac{\pi}{8} t
\]
This makes sense for a uniform rotation, and we’ll confirm it aligns with part (e).
Part (e): Derive the expression for \(d(t)\)
We’re given \(\beta = \angle SAB = 0.7754\) radians and need to show, using the sine rule in \(\triangle ASD\), that:
\[
d(t) = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]
In \(\triangle ASD\), let:
– \(AS = 20\) meters (inferred from the numerator’s coefficient),
– \(AD = d\) (the distance from \(A\) to \(D\)),
– \(\angle ASD = \alpha\),
– \(\angle SAD = \theta\) (to be determined).
Since \(D\) is south of \(A\), and the sprinkler at \(S\) rotates, assume \(S\) is positioned such that \(A\) and \(D\) lie along a vertical path, and \(S\) is off to the side. \(\beta = \angle SAB\) is fixed, but we’re in \(\triangle ASD\). The angle at \(A\) in \(\triangle ASD\) (\(\angle SAD\)) depends on the sprinkler’s position.
The sum of angles in \(\triangle ASD\) is \(\pi\), so:
\[
\angle SDA = \pi – \alpha – \theta
\]
Using the sine rule:
\[
\frac{d}{\sin(\angle ASD)} = \frac{AS}{\sin(\angle SDA)}
\]
\[
\frac{d}{\sin(\alpha)} = \frac{20}{\sin(\pi – \alpha – \theta)}
\]
Since \(\sin(\pi – x) = \sin(x)\):
\[
\frac{d}{\sin(\alpha)} = \frac{20}{\sin(\alpha + \theta)}
\]
\[
d = 20 \cdot \frac{\sin(\alpha)}{\sin(\alpha + \theta)}
\]
From (d), \(\alpha = \frac{\pi t}{8}\). Now, relate \(\theta\) to \(\beta\). In \(\triangle SAB\), \(\beta = 0.7754\), but \(B\) isn’t defined here. Assume \(B\) is a reference point (possibly where the sprinkler points at the end of a cycle), and in \(\triangle ASD\), \(\theta = \angle BAS – \alpha\), where \(\angle BAS\) is the fixed angle from \(A\) to \(S\) relative to the path. However, the denominator suggests:
\[
\alpha + \theta = 2.37 – \frac{\pi t}{8}
\]
\[
\theta = 2.37 – \frac{\pi t}{8} – \alpha = 2.37 – \frac{\pi t}{8} – \frac{\pi t}{8} = 2.37 – \frac{2\pi t}{8} = 2.37 – \frac{\pi t}{4}
\]
But the expression uses \(\sin\left(2.37 – \frac{\pi t}{8}\right)\), so:
\[
d = 20 \cdot \frac{\sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]
Here, \(2.37\) radians (≈ 135.8°) seems to be \(\pi – \beta\) or a related angle. If \(\beta = 0.7754\), then \(\pi – 0.7754 \approx 2.3662\), close to 2.37. Adjusting for significant figures or a typo, assume the denominator’s angle is \(\pi – \beta – \alpha\):
\[
\pi – 0.7754 – \frac{\pi t}{8} \approx 2.3662 – \frac{\pi t}{8}
\]
The problem states 2.37, so:
\[
d = 20 \cdot \frac{\sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]
This holds with the given form, confirming \(\alpha = \frac{\pi t}{8}\).
Part (f): Turtle’s distance at \(t = 0\)
Given:
\[
g(t) = 0.05t^2 + 1.1t + 18
\]
At \(t = 0\):
\[
g(0) = 0.05 \cdot 0^2 + 1.1 \cdot 0 + 18 = 18
\]
The turtle is 18 meters south of \(A\).
Part (g): Distance \(w\) and when water reaches the turtle
(i) Expression for \(w\)
\(w\) is the distance between the turtle (at \(g(t)\)) and point \(D\) (at \(d(t)\)). Since both are distances south of \(A\) along the same path:
\[
w = |g(t) – d(t)|
\]
\[
w = \left| (0.05t^2 + 1.1t + 18) – \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \right|\]
(ii) Hence find when and where on the path the water first reaches the turtle.
The water reaches the turtle when the turtle and point \( D \) are at the same position, i.e., when \( w = 0 \):
\[
|g(t) – d(t)| = 0
\]
\[
g(t) = d(t)
\]
\[
0.05t^2 + 1.1t + 18 = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]
This equation is transcendental due to the sine functions, so let’s solve it by testing values and analyzing behavior. First, let’s define the function:
\[
f(t) = 0.05t^2 + 1.1t + 18 – \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]
We need \( f(t) = 0 \). Let’s evaluate at key points (\( t \geq 0 \)):
At \( t = 0 \):
\[
g(0) = 18
\]
\[
d(0) = \frac{20 \sin(0)}{\sin(2.37)} = 0
\]
\[
f(0) = 18 – 0 = 18 > 0
\]
At \( t = 3 \):
\[
g(3) = 0.05 \cdot 3^2 + 1.1 \cdot 3 + 18 = 0.45 + 3.3 + 18 = 21.75
\]
\[
\frac{\pi \cdot 3}{8} = \frac{3\pi}{8} \approx 1.1781, \quad 2.37 – \frac{3\pi}{8} \approx 2.37 – 1.1781 = 1.1919
\]
\[
\sin\left(\frac{3\pi}{8}\right) \approx 0.9239, \quad \sin(1.1919) \approx 0.9298
\]
\[
d(3) \approx \frac{20 \cdot 0.9239}{0.9298} \approx \frac{18.478}{0.9298} \approx 19.87
\]
\[
f(3) \approx 21.75 – 19.87 = 1.88 > 0
\]
At \( t = 4 \)
\[
g(4) = 0.05 \cdot 4^2 + 1.1 \cdot 4 + 18 = 0.8 + 4.4 + 18 = 23.2
\]
\[
\frac{\pi \cdot 4}{8} = \frac{\pi}{2} \approx 1.5708, \quad 2.37 – \frac{\pi}{2} \approx 2.37 – 1.5708 = 0.7992
\]
\[
\sin\left(\frac{\pi}{2}\right) = 1, \quad \sin(0.7992) \approx 0.7157
\]
\[
d(4) \approx \frac{20 \cdot 1}{0.7157} \approx 27.94
\]
\[
f(4) \approx 23.2 – 27.94 = -4.74 < 0
\]
A root exists between \( t = 3 \) and \( t = 4 \), since \( f(t) \) changes from positive to negative.
At \( t = 3.5 \):
\[
g(3.5) = 0.05 \cdot (3.5)^2 + 1.1 \cdot 3.5 + 18 = 0.05 \cdot 12.25 + 3.85 + 18 = 0.6125 + 3.85 + 18 = 22.4625
\]
\[
\frac{\pi \cdot 3.5}{8} = \frac{3.5\pi}{8} \approx 1.3744, \quad 2.37 – 1.3744 = 0.9956
\]
\[
\sin(1.3744) \approx 0.9808, \quad \sin(0.9956) \approx 0.8388
\]
\[
d(3.5) \approx \frac{20 \cdot 0.9808}{0.8388} \approx \frac{19.616}{0.8388} \approx 23.39
\]
\[
f(3.5) \approx 22.4625 – 23.39 \approx -0.9275 < 0
\]
At \( t = 3.4 \):
\[
g(3.4) = 0.05 \cdot (3.4)^2 + 1.1 \cdot 3.4 + 18 = 0.05 \cdot 11.56 + 3.74 + 18 = 0.578 + 3.74 + 18 = 22.318
\]
\[
\frac{\pi \cdot 3.4}{8} \approx 1.335, \quad 2.37 – 1.335 = 1.035
\]
\[
\sin(1.335) \approx 0.972, \quad \sin(1.035) \approx 0.860
\]
\[
d(3.4) \approx \frac{20 \cdot 0.972}{0.860} \approx 22.60
\]
\[
f(3.4) \approx 22.318 – 22.60 \approx -0.282 < 0
\]
At \( t = 3.3 \):
\[
g(3.3) = 0.05 \cdot (3.3)^2 + 1.1 \cdot 3.3 + 18 = 0.05 \cdot 10.89 + 3.63 + 18 = 0.5445 + 3.63 + 18 = 22.1745
\]
\[
\frac{\pi \cdot 3.3}{8} \approx 1.295, \quad 2.37 – 1.295 = 1.075
\]
\[
\sin(1.295) \approx 0.962, \quad \sin(1.075) \approx 0.879
\]
\[
d(3.3) \approx \frac{20 \cdot 0.962}{0.879} \approx 21.89
\]
\[
f(3.3) \approx 22.1745 – 21.89 \approx 0.2845 > 0
\]
The root is between \( t = 3.3 \) and \( t = 3.4 \). Let’s try \( t = 3.35 \):
\[
g(3.35) = 0.05 \cdot (3.35)^2 + 1.1 \cdot 3.35 + 18 = 0.05 \cdot 11.2225 + 3.685 + 18 = 0.561125 + 3.685 + 18 = 22.246125
\]
\[
\frac{\pi \cdot 3.35}{8} \approx 1.315, \quad 2.37 – 1.315 = 1.055
\]
\[
\sin(1.315) \approx 0.967, \quad \sin(1.055) \approx 0.870
\]
\[
d(3.35) \approx \frac{20 \cdot 0.967}{0.870} \approx 22.22
\]
\[
f(3.35) \approx 22.246 – 22.22 \approx 0.026 > 0
\]
The root is between 3.35 and 3.4, very close to 3.35. Let’s approximate \( t \approx 3.36 \):
\[
g(3.36) \approx 22.258, \quad d(3.36) \approx 22.26
\]
When: \( t \approx 3.36 \) seconds.
Where: At \( t = 3.36 \), the position is \( g(3.36) \approx 22.26 \) meters south of \( A \).
————Markscheme—————–
(a)METHOD 1
let M be the midpoint of [AB] and so AB = 2AM
attempts to use Pythagoras’ theorem to find AM² or AM
AM² = 20² – 14² (= 204) OR AM = √(20² – 14²) (= 14.2828… = √204 = 2√51)
recognizes that AB = 2AM
AB = 2 × 14.2828… (= 28.5657…) (= 2√204 = 4√51)
AB = 28.5657…
AB = 28.57 (m)
METHOD 2
let M be the midpoint of [AB] and so AB = 2AM
let θ = ∠ASM
θ = 0.795398… (= cos⁻¹(14/20))
attempts to use a valid trigonometric ratio
EITHER
$AM = 14\tan(0.795398…) (= 14.2828… = 14\tan(\cos^{-1}\frac{14}{20}))$
OR
$AM = 20\sin(0.795398…) (= 14.2828… = 20\sin(\cos^{-1}\frac{14}{20}))$
THEN
$AB = 28.5657…$
$AB = 28.57 (m)$
(b) EITHER
the sprinkler rotates through (an angle of) $2\pi$ (radians) every 16 seconds and
hence rotates through $\frac{2\pi}{16}$ (radians) in 1 second
OR
$(\frac{2\pi}{16} = n) \Rightarrow n = \frac{2\pi}{16} = \frac{\pi}{8}$
THEN
sprinkler rotates through an angle of $\frac{\pi}{8}$ radians in one second
(c) attempts to find 2θ where θ = ∠ASM
= 2(0.795398…) = 1.59079… = 2cos⁻¹\frac{14}{20}
uses $\frac{\theta}{t}$ (rad/s) or similar to form an equation involving T
$\frac{2\pi}{16} = \frac{1.59079…}{T}$ OR $\frac{2\pi}{16} = \frac{2cos^{-1}\frac{14}{20}}{T}$
T = 4.05093… (= $\frac{1.59079…}{\frac{2\pi}{16}}$ = $\frac{2cos^{-1}\frac{14}{20}}{\frac{2\pi}{16}}$)
T = 4.05 (s)
(d) α =$ \frac{\pi t}{8}$
(e) applies sine rule in △ASD
$\frac{d}{\sin \alpha} = \frac{20}{\sin ADS}$
attempts to find ADS in terms of α
ADS = π – β – α (= π – 0.7754 – α) (= 2.366… – α) (= 2.37 – α)
$d = \frac{20 \sin \alpha}{\sin(2.366… – \alpha)} = \frac{20 \sin \alpha}{\sin(2.37 – \alpha)}$ (accept $d = \frac{20 \sin \alpha}{\sin(\pi – \beta – \alpha)}$)
$d = \frac{20 \sin \frac{\pi}{8}}{\sin(2.37 – \frac{\pi}{8})}$
(f) 18 (m)
(g) (i) $w = 0.05t^2 + 1.1t + 18 – \frac{20 \sin(\frac{\pi t}{8})}{\sin(2.37 – \frac{\pi t}{8})}$
(ii) attempts to solve $w = 0$ for $t$
$t = 3.34880… (12.7765…)$
$t = 3.35 (s)$
22.2444…
22.2 (m) (south of A)