A particle moves in a straight line - HSC - SSCE Mathematics Extension 1 - Question 6 - 2007 - Paper 1

The general form of simple harmonic motion has a period given by:

$T = \frac{2\pi}{\omega},$

where (\omega) is the angular frequency. In this case, we have:

$\omega = 2.$

Thus, the period is:

$T = \frac{2\pi}{2} = \pi.$

We already have:

$\frac{dx}{dt} = 2\left(\frac{3}{\text{sqrt{3}}}\text{cos}(2t) + 2\text{sin}(2t)\right).$

Using the identity for cosine, we can express this as:

$\frac{dx}{dt} = A \text{cos}(2t - \alpha)$

where $A = \sqrt{\left(\frac{3}{\text{sqrt{3}}}\right)^2 + 2^2}$ and (\alpha = \tan^{-1}\left(\frac{2}{\frac{3}{\text{sqrt{3}}}}\right).$$

Thus, we can determine the exact values for (A) and (\alpha).

To find when the particle is moving at 2 metres per second, we set:

$\frac{dx}{dt} = 2.$

This leads to solving:

$2\left(\frac{3}{\text{sqrt{3}}}\text{cos}(2t) + 2\text{sin}(2t)\right) = 2,$

which simplifies to $\text{cos}(2t) + \text{sin}(2t) = 1.$ Solving this equation within the bounds of $0 \leq t \leq \frac{\pi}{2}$ will yield the time values.

To verify that $f(x)$ is increasing, we compute the derivative:

$f'(x) = e^x + e^{-x}.$

Since both $e^x$ and $e^{-x}$ are always positive for all values of x, we determine that $f'(x) > 0$ for all x. Therefore, $f(x)$ is increasing for all values of x.

To find the inverse function, we set $y = f(x) = e^x - e^{-x}$. Rearranging gives:

$e^x = y + e^{-x}.$

Multiplying by $e^x$ results in:

$e^{2x} = ye^x + 1$ followed by substituting $y = 5$ and solving will yield the inverse function.

Given the equation $e^x - e^{-x} = 5$, we can rearrange to find:

$e^{2x} - 5e^x - 1 = 0.$

Solving this quadratic equation using the quadratic formula provides:

$e^x = \frac{5 + \text{sqrt{25 + 4}}}{2}.$

Taking logarithms yields $x = \ln(e^{x})$, hence the solution gives us the value correct to two decimal places.