IBDP Maths AA: Topic : SL 3.8: Solving trigonometric equations: IB style Questions HL Paper 2

Detailed Solution

Part (a): Show that \( AB = 28.57 \), correct to four significant figures

Step 1: Set up the coordinate system
Place the sprinkler \( S \) at the origin \( (0, 0) \). The circle watered by the sprinkler has radius 20 meters, so its equation is:

\[
x^2 + y^2 = 20^2 = 400
\]

The path runs in the north-south direction (parallel to the \( y \)-axis) and is 14 meters from \( S \). Since \( S \) is at \( (0, 0) \), the path’s equation is:

\[
x = 14
\]

The path intersects the circle at points \( A \) and \( B \), as shown in the diagram, with \( A \) above \( S \) and \( B \) below.

Step 2: Find the coordinates of points \( A \) and \( B \)
Substitute \( x = 14 \) into the circle’s equation to find the \( y \)-coordinates of \( A \) and \( B \):

\[
14^2 + y^2 = 400
\]

\[
196 + y^2 = 400
\]

\[
y^2 = 204
\]

\[
y = \pm \sqrt{204} = \pm \sqrt{4 \cdot 51} = \pm 2\sqrt{51}
\]

Since \( \sqrt{51} \approx 7.141 \), we have:

\[
y \approx \pm 14.282
\]

– Point \( A \) (above the origin, positive \( y \)): \( (14, 2\sqrt{51}) \)
– Point \( B \) (below the origin, negative \( y \)): \( (14, -2\sqrt{51}) \)

Step 3: Calculate the distance \( AB \)
The distance \( AB \) is the difference in the \( y \)-coordinates (since \( x = 14 \) for both points):

\[
AB = 2\sqrt{51} – (-2\sqrt{51}) = 4\sqrt{51}
\]

Numerically:

\[
\sqrt{51} \approx 7.141, \quad 4 \times 7.141 = 28.564
\]

To four significant figures:

\[
AB = 28.57
\]

This matches the required value, so the result is confirmed.

Part (b): Show that the sprinkler rotates through an angle of \( \frac{\pi}{8} \) radians in one second

The sprinkler completes one revolution every 16 seconds. One revolution is \( 2\pi \) radians. The angular speed \( \omega \) (in radians per second) is:

\[
\omega = \frac{\text{Total angle in one revolution}}{\text{Time for one revolution}} = \frac{2\pi}{16} = \frac{\pi}{8}
\]

Thus, the sprinkler rotates through an angle of \( \frac{\pi}{8} \) radians in one second, as required.

Part (c): Find the value of \( T \), the time that \( [AB] \) is watered in each revolution

Step 1: Understand what it means for \( [AB] \) to be watered
The sprinkler rotates at a constant angular speed, sending water out to a radius of 20 meters. As it rotates, the water stream sweeps across the circle. The segment \( [AB] \) lies on the path at \( x = 14 \). The sprinkler waters \( [AB] \) when the water stream intersects the line \( x = 14 \) between points \( A \) and \( B \). We need to find the time during one revolution (16 seconds) that the stream covers this segment.

Step 2: Set up the geometry with angles
Place \( S \) at the origin. The water stream rotates around \( S \), and at any time, it makes an angle \( \theta \) with the positive \( x \)-axis (let’s assume it starts at \( \theta = 0 \)). The position of the water stream at angle \( \theta \) is a ray from \( (0, 0) \) at angle \( \theta \), reaching the circle at:

\[
(20 \cos \theta, 20 \sin \theta)
\]

We need to find the angles \( \theta \) where this ray intersects the path \( x = 14 \). The equation of the ray can be parameterized as:

\[
x = r \cos \theta, \quad y = r \sin \theta, \quad 0 \leq r \leq 20
\]

Set \( x = 14 \):

\[
14 = r \cos \theta
\]

\[
r = \frac{14}{\cos \theta}
\]

For the point to be on the circle, \( r \leq 20 \):

\[
\frac{14}{\cos \theta} \leq 20
\]

\[
\cos \theta \geq \frac{14}{20} = 0.7
\]

\[
\theta \leq \cos^{-1}(0.7) \approx 0.795 \text{ radians}
\]

Since \( \theta \) can be positive or negative (depending on the direction of rotation), the angles are:

\[
\theta = \pm \cos^{-1}(0.7)
\]

At these angles, the \( y \)-coordinate is:

\[
r = \frac{14}{\cos \theta} = 20 \quad (\text{at the boundary})
\]

\[
y = r \sin \theta = 20 \sin \theta
\]

\[
\sin \theta = \sqrt{1 – (0.7)^2} = \sqrt{1 – 0.49} = \sqrt{0.51} \approx 0.714
\]

\[
y = 20 \times 0.714 \approx 14.28
\]

This matches the \( y \)-coordinates of \( A \) and \( B \), confirming that \( \theta = \cos^{-1}(0.7) \) corresponds to point \( A \), and \( \theta = -\cos^{-1}(0.7) \) corresponds to point \( B \).

Step 3: Calculate the angle between the positions
The sprinkler waters \( [AB] \) as it rotates from \( \theta = -\cos^{-1}(0.7) \) to \( \theta = \cos^{-1}(0.7) \). The total angle swept is:

\[
\cos^{-1}(0.7) – (-\cos^{-1}(0.7)) = 2 \cos^{-1}(0.7) \approx 2 \times 0.795 = 1.59 \text{ radians}
\]

Step 4: Calculate the time \( T \)
The angular speed is \( \frac{\pi}{8} \) radians per second. The time to rotate through an angle of \( 2 \cos^{-1}(0.7) \) is:

\[
T = \frac{\text{Angle swept}}{\text{Angular speed}} = \frac{2 \cos^{-1}(0.7)}{\frac{\pi}{8}} = \frac{2 \cos^{-1}(0.7) \cdot 8}{\pi}
\]

\[
\cos^{-1}(0.7) \approx 0.795, \quad 2 \cos^{-1}(0.7) \approx 1.59
\]

\[
T \approx \frac{1.59 \cdot 8}{\pi} \approx \frac{12.72}{3.1416} \approx 4.05 \text{ seconds}
\]

This is the time per revolution that \( [AB] \) is watered.

Let’s analyze and solve the given parts:

(d) Expression for

in terms of

The sprinkler rotates clockwise. Let

be the angular velocity of the sprinkler (in radians per second). Since

represents the angle

, we assume the sprinkler starts from some initial position and rotates at a constant angular velocity.

Since angular displacement is given by:

$$\alpha = \omega t$$

The angular speed is \( \

= frac{\pi}{8} \) radians per second

this provides the required expression for

in terms of

as 

(e) To show that \( d(t) = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \).
applies sine rule in △ASD
$\frac{d}{\sin \alpha} = \frac{20}{\sin ADS}$
Finding 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) The turtle’s distance south of point \( A \) at time \( t \) seconds is given by:

\[
g(t) = 0.05t^2 + 1.1t + 18
\]

At \( t = 0 \):

\[
g(0) = 0.05 \cdot 0^2 + 1.1 \cdot 0 + 18 = 18
\]

So, at \( t = 0 \), the turtle is 18 meters south of point \( A \).

(g)(i) Use the expressions for \( g(t) \) and \( d(t) \) to write down an expression for \( w \) in terms of \( t \).

The distance \( w \) is defined as the distance between the turtle and point \( D \) at time \( t \). Since both the turtle and point \( D \) are on the path south of \( A \):
 The turtle’s position is at \( g(t) \) meters south of \( A \).
 Point \( D \) is at \( d(t) \) meters south of \( A \), where:

\[
d(t) = \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)}
\]

The distance \( w \) is the absolute difference between their positions:

\[
w(t) = |g(t) – d(t)|
\]

Substitute the expressions:

\[
w(t) = \left| 0.05t^2 + 1.1t + 18 – \frac{20 \sin\left(\frac{\pi t}{8}\right)}{\sin\left(2.37 – \frac{\pi t}{8}\right)} \right|
\]

This is the expression for \( w \) in terms of \( t \).

(g)(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………………………………….

Solution: –

(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 = $\frac{\pit}{8}$

(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)