A particle moves in a straight line so that its acceleration is given by dv/dt = x - 1 where v is its velocity and x is its displacement from the origin - HSC - SSCE Mathematics Extension 1 - Question 7 - 2001 - Paper 1

From part (i), we established the relationship:

$v^2 = (x - 1)^2$

We also know that:

$v = \frac{dx}{dt}$

Now, taking the derivative:

$\frac{dx}{dt} = \sqrt{(x - 1)^2} = |x - 1|$

For our context, we consider only the positive root, since velocity is positive initially. Thus, we have:

$\frac{dx}{dt} = x - 1$

Rearranging gives:

$dx = (x - 1) dt$

To isolate x, we can integrate:

$\int \frac{1}{x - 1} dx = \int dt$

This leads to:

$\ln|x - 1| = t + C$

Exponentiating both sides results in:

$|x - 1| = e^{t + C}$

We can express the constant as a new constant, K. Thus:

$x - 1 = K e^t$

Solving for x, we have:

$x = 1 + K e^t$

Given the initial condition when t = 0, we have x = 0:

$0 = 1 + K \Rightarrow K = -1$

So the final expression becomes:

$x = 1 - e^t$

In triangle AOP, we can use the Pythagorean theorem. The relationship is given by:

Let AP be the hypotenuse, we express it as:

$AP^{2} = AO^{2} + OP^{2}$

Using the vertical height h, we can relate:

Thus:
$AP^{2} = h^{2} + OP^{2}$

To express OP in terms of α:

$OP = h \cdot \cot(α)$

Therefore, substituting this into our equation:

$AP^{2} = h^{2} + (h\cdot \cot(α))^{2}$

This leads us to:

$AP^{2} = h^{2} + h^{2}\cot^{2}(α) + 2h\cdot h\cot(α)\cdot h\cot(α)$

Since OP can also be represented as hPcotα, we combine:

$AP^{2} = h^{2} + h^{2}\cot^{2}(α) - hP\cot(α)$

We now find the second expression for AP²:

From trigonometric identities, we have:

$AP = \frac{h}{\sin(\theta)}$

Then,

$AP^{2} = \frac{h^{2}}{\sin^{2}(θ)}$

We can find cosθ using the previous relationships and substituting:

Using identity relationships leads to:

$\cos(θ) = \frac{h^{2} + h^{2}\cot^{2}(α) - hP\cot(α)}{AP^{2}}$

Deducing this yields:

$\cos(θ) = \frac{1}{\sqrt{2}} \sin(α) + \frac{1}{2\sqrt{2}} \cos(α)$

To sketch the graph of θ for 0 < α < π/2, we must consider the relationship between θ and α:

Based on the derived equation for cos(θ):

$\cos(θ) = \frac{1}{\sqrt{2}} \sin(α) + \frac{1}{2\sqrt{2}} \cos(α)$

1. Identifying Turning Points:
   Analyze when the derivative d(θ)/d(α) equals zero, which indicates potential maximum or minimum values.

2. Behavior Analysis:

   • As α approaches 0, observe that θ approaches its maximum.
   • As α approaches π/2, note how θ decreases rapidly.

3. Sketching the Graph:

   • The graph should show an initial rise as α approaches 0, peaking at a height, and then declining towards π/2, indicating a turning point at some location.
   • Ensure to mark all relevant points and classify whether each is a maximum or minimum based on second derivative tests.