How many nine-letter arrangements can be made using the letters of the word ISOSCELES? A particle moves in a straight line and its position at time t is given by x = 4sin t = 2 t + \frac{\pi}{3} - HSC - SSCE Mathematics Extension 1 - Question 3 - 2003 - Paper 1

To demonstrate that the particle is undergoing simple harmonic motion (SHM), we need to express its position equation.

The given equation is:

$x = 4 \sin \left(2\pi t + \frac{\pi}{3}\right)$

In the form of SHM, the standard equation is:

$x = A\sin(\omega t + \phi)$

where (A) is the amplitude, (\omega) is the angular frequency, and (\phi) is the phase constant. Here:

• Amplitude (A = 4)
• Angular frequency (\omega = 2\pi)
• Phase constant (\phi = \frac{\pi}{3})

Since this matches the format of SHM, we conclude that the particle is indeed undergoing simple harmonic motion.

The amplitude of the motion can be directly observed from the equation:

$x = 4\sin(2\pi t + \frac{\pi}{3})$

Here, the amplitude (A) is the coefficient of the sine function. Therefore, the amplitude is:

(A = 4).

The speed of the particle is determined by its maximum when the derivative of the position with respect to time is at its peak. The velocity function is obtained by differentiating the position function:

$v(t) = \frac{dx}{dt} = 4 \cdot 2\pi \cos(2\pi t + \frac{\pi}{3}) = 8\pi \cos(2\pi t + \frac{\pi}{3})$

For maximum speed, we set (\cos(2\pi t + \frac{\pi}{3}) = 1). This occurs when:

$2\pi t + \frac{\pi}{3} = 2k \pi$

for any integer (k). Solving for (t):

ightarrow t = k - \frac{1}{6}$$ The first instance occurs when \(k = 0\): thus: $$t = -\frac{1}{6} + \text{(adjusting for time starting at 0)}\implies t = 0$$ Since we are looking for the first positive time, incrementing to \(k = 1\): $$t = 1 - \frac{1}{6} = \frac{5}{6}$$. The maximum speed is first reached at \(t = \frac{5}{6}s\).

To find the probability of rolling a sum of 5 with a pair of dice, we achieve this through listing the possible outcomes. The pairs of numbers on two dice that result in a sum of 5 are:

• (1, 4)
• (2, 3)
• (3, 2)
• (4, 1)

There are a total of 4 successful outcomes. The total number of possible outcomes when tossing two dice is: $6 \times 6 = 36$

Thus, the probability (P(Sum = 5)) is:

$P(Sum = 5) = \frac{4}{36} = \frac{1}{9}.$

To find the probability of getting a sum of 5 at least twice when tossing a pair of dice 7 times, we first note the probability of getting a sum of 5 in one roll is (p = \frac{1}{9}) and hence the probability of not getting a sum of 5 is (q = 1 - p = \frac{8}{9}.

This scenario can be modeled using the binomial distribution. We calculate:

• The probability of getting the sum of 5 exactly (k) times can be computed using the binomial probability formula:

$P(X = k) = {n \choose k} p^k q^{n-k}$

Setting n = 7, we look for (k \geq 2).

The required probability:

$P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Calculating:

• For (k = 0): $P(X = 0) = {7 \choose 0} \left(\frac{1}{9}\right)^0 \left(\frac{8}{9}\right)^7 = 1 \cdot 1 \cdot \left(\frac{8}{9}\right)^7$

• For (k = 1): $P(X = 1) = {7 \choose 1} \left(\frac{1}{9}\right)^1 \left(\frac{8}{9}\right)^6 = 7 \cdot \left(\frac{1}{9}\right) \cdot \left(\frac{8}{9}\right)^6$

Thus: $P(X \geq 2) = 1 - \left(\frac{8}{9}\right)^7 - 7 \cdot \left(\frac{1}{9}\right) \cdot \left(\frac{8}{9}\right)^6$

To prove the statement by induction:

Base case (n = 1):

For (n = 1): $\frac{1}{1 \times 3} = \frac{1}{3}$ And the right-hand side: $\frac{1}{2 \cdot 1 + 1} = \frac{1}{3}$

The base case holds.

Inductive step: Assume it holds for some (n = k): $\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \cdots + \frac{1}{(2k-1)(2k+1)} = \frac{k}{2k+1}$

Prove for n = k + 1: $\frac{k}{2k+1} + \frac{1}{(2(k+1)-1)(2(k+1)+1)} = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)}$

Finding a common denominator: $= \frac{k(2k+3) + 1}{(2k+1)(2k+3)}\rightarrow\frac{2k^2 + 3k + 1}{(2k+1)(2k+3)}$

Now: $2k^2 + 3k + 1 = 2k^2 + 2k + k + 1 = \frac{(k+1)}{2(k+1)+1}$

Thus, it holds. Hence true by induction.