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

Starting from the derived expression:

1. We already have: $v = x - 1$

2. Using the relationship of velocity: $v = \frac{dx}{dt}$

   Thus, we can write: $\frac{dx}{dt} = x - 1$

3. Rearranging gives: $\frac{dx}{x - 1} = dt$

4. Integrate both sides: [ \int \frac{1}{x - 1} , dx = \int dt ]

5. This results in: $\ln |x - 1| = t + C$

   Solving for $x$:

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

   We can simplify to: $x = e^{t + C} + 1$

6. With an initial condition (for, say, $t=0$), we can find $C$, which will allow us to express $x$ in terms of $t$.

In triangle $AOP$:

1. Using the Pythagorean theorem, we can express: $AP^2 = AO^2 + OP^2$

2. From the geometry:

   • $AO = h$ (height of the tower),
   • $OP = h \cot \alpha$ (from the definition of cotangent).

3. Therefore: $AP^2 = h^2 + (h \cot \alpha)^2$

4. Recognizing that $HP = OP \cos \alpha$, we substitute: $= h^2 (1 + \cot^2 \alpha) - HP \cot \alpha$

5. Simplifying gives: $AP^2 = h^2 + h^2 \cot^2 \alpha - HP \cot \alpha.$

To find a second expression for $AP^2$, consider the relationship in triangle $ATP$:

1. Again apply the Pythagorean theorem: $AP^2 = AT^2 + TP^2$

2. Using the known angles and respective lengths, and the height $h$: $AT = h \cot \theta$

3. Relation then gives: $AP^2 = h^2 \cot^2 \theta + TP^2$

4. By equating both expressions for $AP^2$, we can set: $h^2 + h^2 \cot^2 \alpha - HP \cot \alpha = h^2 \cot^2 \theta + TP^2$

5. From this equality and considering the trigonometric identities, deduce: [ \cos \theta = \frac{1}{\sqrt{2}} \sin \alpha + \frac{1}{2\sqrt{2}} \cot \alpha. ]

To sketch the graph of $\theta$ as a function of $\alpha$:

1. Consider the behaviour as $\alpha \to 0$ and $\alpha \to \frac{\pi}{2}$.

   • As $\alpha \to 0$, we expect $\theta$ to approach a certain limit.
   • As $\alpha \to \frac{\pi}{2}$, analyze the asymptotic behaviour which predicts $\theta \to \infty$.

2. Identify local maxima and minima by checking the first derivative through critical points, determining turning points where $\frac{d\theta}{d\alpha} = 0$.

3. Plot the identified points, computing value trends between to form the full graph, showing regions of increase and decrease clearly.

4. Classify turning points as local minima/maxima based on changes in the derivative sign around those points.