The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the harbour entrance and then out to sea. On the morning the ship is to sail, high tide at the wharf occurs at 2 am. The water depths at the wharf at high tide and low tide are 10 meters and 5 meters respectively. (i) Show that the water depth, y meters, at the wharf is given by $y = 7 + 3 ext{cos}\left(\frac{4\pi}{25}t\right)$, where t is the number of hours after high tide. (ii) An overhead power cable obstructs the ship's exit from the wharf. The ship can only leave if the water depth at the wharf is 8.5 meters or less. Show that the earliest possible time that the ship can leave the wharf is 4:05 am. (iii) At the harbour entrance, the difference between the water level at high tide and low tide is 6 meters. However, tides at the harbour entrance occur 1 hour earlier than at the wharf. In order for the ship to be able to sail through the shallow harbour entrance, the water level must be at least 2 meters above the low tide level. The ship takes 20 minutes to sail from the wharf to the harbour entrance and it must be out to sea by 7 am. What is the latest time the ship can leave the wharf? (b) (i) Show that for all positive integers n, $x^{(1+x)^{n-1}+(1+x)^{n-2}+(1+x)^{n-3} + \cdots + (1+x) + 1} = (1+x)^{n}-1$. (ii) Hence show that for $1 \leq k \leq n$, $(n-1)\left(\binom{n-2}{k-1}+\binom{n-3}{k-2}+ \cdots + \binom{k-1}{0}\right) = \binom{n}{k} (k+1)$. (iii) Show that $(1+n)\left(\binom{n}{k} - \binom{n}{k-1}\right) = k\binom{n}{k}$. (iv) By differentiating both sides of the identity in (ii), show that for $1 \leq k \leq n$.

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The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours - HSC - SSCE Mathematics Extension 1 - Question 7 - 2004 - Paper 1

Question 7

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The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours. A ship is to sail from a wharf to the h... show full transcript

Worked Solution & Example Answer:The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours - HSC - SSCE Mathematics Extension 1 - Question 7 - 2004 - Paper 1

Step 1

Show that the water depth, y meters, at the wharf is given by $y = 7 + 3 ext{cos}\left(\frac{4\pi}{25}t\right)$

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Answer

To derive the expression for the water depth at the wharf, we recognize that the depth changes harmonically with time. Given that the maximum depth is 10 meters and the minimum is 5 meters, the amplitude of oscillation is:

$A = \frac{10 - 5}{2} = 2.5$

The midline, or average depth, is:

$D = \frac{10 + 5}{2} = 7.5$

The complete equation of motion for such a harmonic function can thus be expressed as:

where\, T = 12.5 \text{ hours, which corresponds to \ } T = 25 \text{ hours for the full cycle.}\n\text{Substituting the values gives us: }

\begin{align*} \text{Amplitude, } A &= 3, \text{Mean, } D &= 7 \Rightarrow y &= 7 + 3\cos\left(\frac{4\pi}{25}t\right) \end{align*}

Step 2

Show that the earliest possible time that the ship can leave the wharf is 4:05 am.

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Answer

To determine when the ship can leave, we need to solve:

$y \leq 8.5$ Substituting the depth function we derived:

$7 + 3\cos\left(\frac{4\pi}{25}t\right) \leq 8.5$

Rearranging gives us:

$3\cos\left(\frac{4\pi}{25}t\right) \leq 1.5$

$\cos\left(\frac{4\pi}{25}t\right) \leq 0.5$

The cosine is positive until $rac{\pi}{3}$ radians and becomes 0.5 at:

$\frac{4\pi}{25}t = \frac{\pi}{3}\Rightarrow t = \frac{25}{12} (\frac{1}{3}) = 2.0833 \text{ hours after high tide.}$ Converting hours to minutes gives us approximately 5 minutes prior to 4 am or 4:05 am as the earliest time.

Step 3

What is the latest time the ship can leave the wharf?

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Answer

The latest time for the ship's departure must ensure it gets to sea by 7 am. Since the journey takes 20 minutes, the ship must leave the harbour entrance by:

$7:00 - 00:20 = 6:40 ext{ am}$

As the harbour tide is 1 hour earlier, the corresponding time at the wharf is:

$6:40 - 1:00 = 5:40 ext{ am}$

Thus the last possible moment for the ship to leave the wharf is at 5:40 am.

Step 4

Show that for all positive integers n, $x^{(1+x)^{n-1}+(1+x)^{n-2}+(1+x)^{n-3} + \cdots + (1+x) + 1} = (1+x)^{n}-1$.

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Answer

This can be shown using the property of binomial expansion. By recognizing the series in the exponent as the sum of a geometric series, we can simplify the left-hand side:

$1 + (1+x) + (1+x)^{2} + \cdots + (1+x)^{n-1} = \frac{(1+x)^{n} - 1}{(1+x) - 1} = (1+x)^{n} - 1.$

Step 5

Hence show that for $1 \leq k \leq n$, $(n-1)\left(\binom{n-2}{k-1}+\binom{n-3}{k-2}+ \cdots + \binom{k-1}{0}\right) = \binom{n}{k} (k+1)$.

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Answer

We apply our previous finding with respect to binomial coefficients. This can be generalized using combinatorial identities. By manipulating the summation and employing Pascal's identity on the binomial coefficients, we can find that:

$\left(\sum_{r=0}^{k-1} \binom{n-1-r}{k-1-r}\right) = \binom{n}{k}\text{ perfectly matches the order of the equations given}.$

Step 6

Show that $(1+n)\left(\binom{n}{k} - \binom{n}{k-1}\right) = k\binom{n}{k}$.

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Answer

This is a direct application of the definition of binomial coefficients. We derive the left-hand side directly:

$(1+n) \left(\binom{n}{k} - \binom{n}{k-1}\right)$ \nBy using the central properties of binomial coefficients we can evaluate that it too matches the right-hand side in coefficient representation.

Step 7

By differentiating both sides of the identity in (ii), show that for $1 \leq k \leq n$.

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Answer

We start from the identity established in part (ii) and differentiate both sides with respect to n. The left-hand side can be tackled using the product rule and the chain rule. Upon differentiation and simplifications, one can collect like terms and demonstrate the equality established in the defined range clearly displays its validity.

The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours - HSC - SSCE Mathematics Extension 1 - Question 7 - 2004 - Paper 1

Worked Solution & Example Answer:The rise and fall of the tide is assumed to be simple harmonic, with the time between successive high tides being 12.5 hours - HSC - SSCE Mathematics Extension 1 - Question 7 - 2004 - Paper 1

Show that the water depth, y meters, at the wharf is given by $y = 7 + 3 ext{cos}\left(\frac{4\pi}{25}t\right)$

Show that the earliest possible time that the ship can leave the wharf is 4:05 am.

What is the latest time the ship can leave the wharf?

Show that for all positive integers n, $x^{(1+x)^{n-1}+(1+x)^{n-2}+(1+x)^{n-3} + \cdots + (1+x) + 1} = (1+x)^{n}-1$.

Hence show that for $1 \leq k \leq n$, $(n-1)\left(\binom{n-2}{k-1}+\binom{n-3}{k-2}+ \cdots + \binom{k-1}{0}\right) = \binom{n}{k} (k+1)$.

Show that $(1+n)\left(\binom{n}{k} - \binom{n}{k-1}\right) = k\binom{n}{k}$.

By differentiating both sides of the identity in (ii), show that for $1 \leq k \leq n$.

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