Home / IB DP Math AA Topic: SL 2.7 :Solving quadratic equations : IB Style Questions HL Paper 2

IB DP Math AA Topic: SL 2.7 :Solving quadratic equations : IB Style Questions HL Paper 2

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Question

Sule Skerry and Rockall are small islands in the Atlantic Ocean, in the same time zone.

On a given day, the height of water in metres at Sule Skerry is modelled by the function $H(t) = 1.63 \sin(0.513(t-8.20)) + 2.13$, where $t$ is the number of hours after midnight.

The following graph shows the height of the water for 15 hours, starting at midnight.

At low tide the height of the water is 0.50m. At high tide the height of the water is 3.76m.

All heights are given correct to two decimal places.

(a) The length of time between the first low tide and the first high tide is 6 hours and m minutes. Find the value of m to the nearest integer. 
(b) Between two consecutive high tides, determine the length of time, in hours, for which the height of the water is less than 1 metre. 
(c) Find the rate of change of the height of the water when t = 13 , giving your answer in metres per hour.

On the same day, the height of water at the second island, Rockall, is modelled by the function $h(t) = a \sin(b(t-c)) + d$, where $t$ is the number of hours after midnight, and $a, b, c, d > 0$.

The first low tide occurs at 02:41 when the height of the water is 0.40 m.

The first high tide occurs at 09:02 when the height of the water is 2.74 m.

(d) Find the values of $a, b, c$ and $d$.

When $t = T$, the height of the water at Sule Skerry is the same as the height of the water at Rockall for the first time.

(e) Find the value of $T$.

▶️Answer/Explanation

Detailed Solution

(a) Finding 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)=1.63sin⁡(0.513(t−8.20))+2.13H(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)=asin⁡(b(t−c))+dH(t) = a \sin(b(t – c)) + d

The period of a sine function is given by:

2πb\frac{2\pi}{b}

For this function,

b=0.513b = 0.513

, so:

Period=2π0.513≈12.24 hours.\text{Period} = \frac{2\pi}{0.513} \approx 12.24 \text{ hours}.
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:

12.242=6.12 hours.\frac{12.24}{4} = 3.06 \text{ hours}.

Converting 0.12 hours into minutes:

0.12×60=7.2≈7 minutes.0.06 \times 60 = 3.6 \approx 4 \text{ minutes}.

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 tt

 

when

H(t)=1H(t) = 1

 

We solve:

1.63sin⁡(0.513(t−8.20))+2.13=1.1.63 \sin(0.513(t – 8.20)) + 2.13 = 1.

Rearrange:

1.63sin⁡(0.513(t−8.20))=−1.13. 1.63 \sin(0.513(t – 8.20)) = -1.13. sin⁡(0.513(t−8.20))=−1.131.63≈−0.693.\sin(0.513(t – 8.20)) = \frac{-1.13}{1.63} \approx -0.693.

Find the reference angle:

θ=arcsin⁡(0.693)≈43.89∘.\theta = \arcsin(0.693) \approx 43.89^\circ.

Convert to radians:

θ≈0.766 radians.\theta \approx 0.766 \text{ radians}.

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

0.513(t−8.20)=π+0.766or0.513(t−8.20)=2π−0.766.0.513 (t – 8.20) = \pi + 0.766 \quad \text{or} \quad 0.513 (t – 8.20) = 2\pi – 0.766.

Solving these gives two values for

tt

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=13t = 13

 

The rate of change is given by the derivative:

H′(t)=ddt(1.63sin⁡(0.513(t−8.20))+2.13).H'(t) = \frac{d}{dt} \left( 1.63 \sin(0.513(t – 8.20)) + 2.13 \right).

Using the chain rule:

H′(t)=1.63×0.513cos⁡(0.513(t−8.20)). H'(t) = 1.63 \times 0.513 \cos(0.513(t – 8.20)). H′(t)=0.835cos⁡(0.513(t−8.20)).H'(t) = 0.835 \cos(0.513(t – 8.20)).

Substituting

t=13t = 13

:

H′(13)=0.835cos⁡(0.513(13−8.20)).H'(13) = 0.835 \cos(0.513(13 – 8.20)).

Computing:

0.513×4.8=2.46. 0.513 \times 4.8 = 2.46. cos⁡(2.46)≈−0.77. \cos(2.46) \approx -0.77. H′(13)=0.835×(−0.77)≈−0.642.H'(13) = 0.835 \times (-0.77) \approx -0.642.

So the rate of change is:

−0.642 m/h.\boxed{-0.642} \text{ m/h}.
(d) Finding a,b,c,da, b, c, d

 

for Rockall

The function for Rockall is:

h(t)=asin⁡(b(t−c))+d.h(t) = a \sin(b(t – c)) + d.
  • Low tide at
    t=2.683t = 2.683

     

    (02:41) with

    h=0.40h = 0.40

    .

  • High tide at
    t=9.033t = 9.033

     

    (09:02)

    with

    h=2.74h = 2.74

    .

Step 1: Find aa

 

The maximum and minimum heights give:

d+a=2.74,d−a=0.40.d + a = 2.74, \quad d – a = 0.40.

Solving:

2d=3.14⇒d=1.57. 2d = 3.14 \quad \Rightarrow \quad d = 1.57. a=2.74−1.57=1.17.a = 2.74 – 1.57 = 1.17.
Step 2: Find bb

 

The time between low tide and high tide is:

9.033−2.683=6.35 hours.9.033 – 2.683 = 6.35 \text{ hours}.

Since this is a half of the period:

2πb=2×6.35=12.70. \frac{2\pi}{b} = 4 \times 6.35 = 25.40. b=2π 12.70

 

≈0.494.b = \frac{2\pi}{25.40} \approx 0.247.
Step 3: Find cc

 

Substituting the known values a, b and d in the equation , we get:

h(t)=1.17sin⁡(0.247(2.683-c))+1.57=0.4
Therefore, c = 5.86
(e) Find TT

 

when

H(T)=h(T)H(T) = h(T)

 

Solve:

1.63sin⁡(0.513(T−8.20))+2.13=1.17sin⁡(0.247(T–5.86))+1.57.1.63 \sin(0.513(T – 8.20)) + 2.13 = 1.17 \sin(0.247(T + 16.39)) + 1.57.

Numerical methods or graphing is required. Estimating gives

T≈4.16T \approx 5.3

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
$m = (6.12396… – 6) \times 60 = 7.43773…$
$m = 7$

(b) attempt to solve $H(t) = 1$
3.56919… OR 6.70684… OR 15.8171… OR 18.9547…
(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
d = 1.57
attempt to find time between low and high tide in hours
6 hours and 21 minutes = 6.35 hours
(period =) 12.7

$b = \frac{2\pi}{12.7} = 0.494739…$
b = 0.495$\left [ =\frac{60\pi }{381} \right ]$

attempt to find mean of low and high tide times OR substitute values of a known point
$c = \frac{1}{2}(2\frac{41}{60} + 9\frac{2}{60})$OR eg 0.40 = 1.17 \sin(0.495 (2.68333… – c)) + 1.57
c = 5.85833…
c = 5.86

METHOD 2

a = 1.17
d = 1.57
substituting at least one point into h(t)
$1.17 \sin(b(2\frac{41}{60}-c))+1.57=0.4$ OR $1.17 \sin(b(9\frac{2}{60}-c))+1.57=2.74$
$b(2\frac{41}{60}-c)=\frac{\pi}{2}(-1.57)$ AND $b(9\frac{2}{60}-c)=\frac{\pi}{2}(=1.57)$

accept any angles of the form $-\frac{\pi}{2}+c\pi k$ and $\frac{\pi}{2}+c\pi k$

EITHER

use of graph or table to find their intersection

OR

attempt to solve their equations simultaneously
$\frac{2\frac{41}{60}-c}{\frac{2}{60}-c}=-1$

THEN

$c = 5.85833…$
$c = 5.86$
$b = 0.494739…$
$b = 0.495$

(e) attempt to find point of intersection of two graphs
$T = 4.16292…$ OR $T = 4.16417…$ 
$T = 4.16$

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