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

(a) For m

The minutes in the 6-hour period between first low tide and first high tide

From the given information:

• Low tide height = 0.50 m
• High tide height = 3.76 m
• The function given is:
  $$H(t)=1.63\sin (0.513(t-8.20))+2.13$$
• A sine function oscillates between low tide and high tide in a half of its period.

Step 1: Find the period of the function

The general form of a sine function is:

$$H(t)=a\sin (b(t-c))+d$$

The period of a sine function is given by:

$$\frac{2\pi }{b}$$

For this function,

$b=0.513$

, so:

$$\text{Period}=\frac{2\pi }{0.513}\approx 12.24\text{ hours}\mathrm{.}$$

Step 2: Find the time between low tide and high tide

Since a full period covers a full oscillation from low tide to high tide and back, the time between low tide and high tide is a half of the period:

$$\frac{12.24}{2}=6.12\text{ hours}\mathrm{.}$$

Converting 0.12 hours into minutes:

$$0.12\times 60=7.2\approx 7\text{ minutes}\mathrm{.}$$

Thus, the time is 6 hours and 7 minutes. Hence m=7.

(b) Finding the time the water is below 1m

We need to determine how long

$H(t)<1$

Step 1: Solve for  $t$

when

$H(t)=1$

We solve:

$$1.63\sin (0.513(t-8.20))+2.13=1.$$

Rearrange:

$$1.63\sin (0.513(t-8.20))=-\mathrm{1.13.}$$ $$$$ $$\sin (0.513(t-8.20))=\frac{-1.13}{1.63}\approx -\mathrm{0.693.}$$

Find the reference angle:

$$\theta =\arcsin (0.693)\approx {43.89}^{\circ }\mathrm{.}$$

Convert to radians:

$$\theta \approx 0.766\text{ radians}\mathrm{.}$$

Since the sine function is negative in the third and fourth quadrants:

$$0.513(t-8.20)=\pi +0.766\text{or}0.513(t-8.20)=2\pi -\mathrm{0.766.}$$

Solving these gives two values for

$t$

Therefore, t = 15.81 and t = 18.96. The duration is found by subtracting them.

 = 3.14 hours per cycle

(c) Rate of change of height at  $t=13$

The rate of change is given by the derivative:

$$H^{\mathrm{\prime }}(t)=\frac{d}{dt}(1.63\sin (0.513(t-8.20))+2.13)\mathrm{.}$$

Using the chain rule:

$$H^{\mathrm{\prime }}(t)=1.63\times 0.513\cos (0.513(t-8.20))\mathrm{.}$$ $$$$ $$H^{\mathrm{\prime }}(t)=0.835\cos (0.513(t-8.20))\mathrm{.}$$

Substituting

$t=13$

:

$$H^{\mathrm{\prime }}(13)=0.835\cos (0.513(13-8.20))\mathrm{.}$$

Computing:

$$0.513\times 4.8=\mathrm{2.46.}$$ $$$$ $$\cos (2.46)\approx -\mathrm{0.77.}$$ $$$$ $$H^{\mathrm{\prime }}(13)=0.835\times (-0.77)\approx -\mathrm{0.642.}$$

So the rate of change is:

$$\boxed{{\displaystyle -0.642}}\text{ m/h}$$

Part (d): Find the values of \(a\), \(b\), \(c\), and \(d\).

Step 1: Identify the parameters from the tide information.
The function for the height of water is given by \( h(t) = a \sin(b(t – c)) + d \). We know the following:
– At low tide (02:41), \( h(2.6833) = 0.40 \) m (since 02:41 is 2 hours and 41 minutes after midnight, which is \( 2 + \frac{41}{60} \approx 2.6833 \)).
– At high tide (09:02), \( h(9.0333) = 2.74 \) m (since 09:02 is 9 hours and 2 minutes after midnight, which is \( 9 + \frac{2}{60} \approx 9.0333 \)).

Step 2: Set up the equations based on the tide information.
From the low tide:
\[
0.40 = a \sin(b(2.6833 – c)) + d \quad (1)
\]
From the high tide:
\[
2.74 = a \sin(b(9.0333 – c)) + d \quad (2)
\]

Step 3: Determine the amplitude \(a\) and vertical shift \(d\).
The difference between high tide and low tide gives us the amplitude:
\[
a = \frac{2.74 – 0.40}{2} = \frac{2.34}{2} = 1.17
\]
The average height (vertical shift) is:
\[
d = \frac{2.74 + 0.40}{2} = \frac{3.14}{2} = 1.57
\]

Step 4: **Substitute \(a\) and \(d\) back into the equations.**
Substituting \(a\) and \(d\) into equation (1):
\[
0.40 = 1.17 \sin(b(2.6833 – c)) + 1.57
\]
\[
-1.17 = 1.17 \sin(b(2.6833 – c))
\]
\[
\sin(b(2.6833 – c)) = -1 \quad (3)
\]
From equation (2):
\[
2.74 = 1.17 \sin(b(9.0333 – c)) + 1.57
\]
\[
1.17 = 1.17 \sin(b(9.0333 – c))
\]
\[
\sin(b(9.0333 – c)) = 1 \quad (4)
\]

Step 5: Solve for \(b\) and \(c\).
From equation (3), since \(\sin\) reaches -1 at \(-\frac{\pi}{2} + 2k\pi\) for integers \(k\):
\[
b(2.6833 – c) = -\frac{\pi}{2} + 2k\pi
\]
From equation (4), since \(\sin\) reaches 1 at \(\frac{\pi}{2} + 2m\pi\) for integers \(m\):
\[
b(9.0333 – c) = \frac{\pi}{2} + 2m\pi
\]

Step 6: Set the equations for \(b\) and \(c\).
We can set \(k = 0\) and \(m = 0\) for the first occurrence:
\[
b(2.6833 – c) = -\frac{\pi}{2} \quad (5)
\]
\[
b(9.0333 – c) = \frac{\pi}{2} \quad (6)
\]

Step 7: Solve equations (5) and (6) simultaneously.
From equation (5):
\[
c = 2.6833 + \frac{\pi}{2b}
\]
Substituting \(c\) into equation (6):
\[
b(9.0333 – (2.6833 + \frac{\pi}{2b})) = \frac{\pi}{2}
\]
\[
b(6.35 – \frac{\pi}{2b}) = \frac{\pi}{2}
\]
\[
6.35b – \frac{\pi}{2} = \frac{\pi}{2}
\]
\[
6.35b = \pi
\]
\[
b = \frac{\pi}{6.35} \approx 0.495
\]

Step 8: Substitute \(b\) back to find \(c\).
Substituting \(b\) back into the equation for \(c\):
\[
c = 2.6833 + \frac{\pi}{2 \cdot 0.495} \approx 2.6833 + 3.18 \approx 5.8633
\]

Summary of Values
– \(a = 1.17\)
– \(b \approx 0.495\)
– \(c \approx 5.8633\)
– \(d = 1.57\)

 Part (e): Find the value of \(T\).

when

$H(T)=h(T)$

Solve:

$$1.63\sin (0.513(T-8.20))+2.13=1.17\sin (0.247(T–5.86))+\mathrm{1.57.}$$

Numerical methods or graphing is required. Estimating gives

$T\approx 4.16$

hours.

————Markscheme—————–

solution:-

(a) EITHER
attempt to find value of t for the first low tide OR the first high tide 
$11.2619…− 5.13801…$
$= 6.12396…$
OR
attempt to find half of the period
$ \frac{1}{2}\times \frac{2\pi}{0.513} $
$=6.12396… $
THEN
$m =(6.12396 … − 6 ) ×60= 7.437 73…$

$m = 7 $
(b) attempt to solve H t( ) = 1 
$3.56919… OR 6.70684…  OR 15.8171… OR 18.95 74…$
$(6.70684…− 3.56919…=)3.13764…$
$= 3.14 (hours)$ 
(c) recognition that H ‘ ( 13) is required 
$=−0.650622…$
$= −0.651 (m/h)$

(d) METHOD 1
a = 1.17 A1
d =1.57

METHOD 2
a = 1.17 A1
d = 1.57 A1
substituting at least one point into h(t)