图中展示了一个叶片长度为 $25\text{ m}$ 的风力发电机。风力发电机的中心距离地面 $35\text{ m}$。在离开 $X$ 点 $t$ 秒后，特定叶片尖端距离地面的高度 $h\text{ m}$ 可建模为 $h = a \cos bt + c$，其中 $k$ 为常数。
风力发电机叶片的旋转速度为每 $3\pi$ 秒转 $1$ 圈。

1. Find the value of $c$ (the principal axis).

The centre of the wind turbine is $35\text{ m}$ above the ground, which represents the mean height of the blade tip.
$c = 35$

2. Find the value of $a$ (related to the amplitude).

The amplitude of the motion is the length of the blade, which is $25\text{ m}$.
At $t = 0$, the blade is at $X$, which is the lowest point of its path.
Minimum height $= 35 - 25 = 10\text{ m}$.
Substitute $t = 0$ into $h = a \cos bt + c$:
$h(0) = a \cos(0) + 35 = a + 35$
$a + 35 = 10$
$a = -25$

3. Find the value of $b$ using the period of the motion.

The blade completes $1$ revolution in $3\pi$ seconds, so the period is $3\pi$.
$\text{Period} = \frac{2\pi}{b}$
$\frac{2\pi}{b} = 3\pi$
$b = \frac{2\pi}{3\pi} = \frac{2}{3}$

1. State the equation of the graph using values from part (a)

From part (a), $a = -25$, $b = \frac{2}{3}$, and $c = 35$.
The equation is $h = -25 \cos\left(\frac{2}{3}t\right) + 35$.

2. Determine the key features of the graph (amplitude, period, and key points)

Amplitude $= 25$, principal axis is $h = 35$.
Period $= \frac{2\pi}{2/3} = 3\pi$.
The domain $0 < t < 6\pi$ covers exactly two full cycles.
Key points:

• $t = 0, h = 10$ (minimum, open circle)
• $t = 1.5\pi, h = 60$ (maximum)
• $t = 3\pi, h = 10$ (minimum)
• $t = 4.5\pi, h = 60$ (maximum)
• $t = 6\pi, h = 10$ (minimum, open circle)

3. Sketch the graph

Draw a smooth wave passing through the key points identified, ensuring the endpoints at $t=0$ and $t=6\pi$ are marked with open circles to reflect the strict inequalities $0 < t < 6\pi$.

1. Set up the equation using the model from part (a)

From part (a), the model is $h = -25 \cos\left(\frac{2}{3}t\right) + 35$.
Substitute $h = 42$:
$-25 \cos\left(\frac{2}{3}t\right) + 35 = 42$

2. Isolate the trigonometric term

$-25 \cos\left(\frac{2}{3}t\right) = 7$
$\cos\left(\frac{2}{3}t\right) = -0.28$

3. Solve for the first positive value of $t$

Since we want the first time, we find the smallest positive angle in the second quadrant:
$\frac{2}{3}t = \cos^{-1}(-0.28)$
$\frac{2}{3}t \approx 1.8546$
$t = 1.8546 \times \frac{3}{2}$
$t \approx 2.7819$
$t \approx 2.78\text{ s (to 3 s.f.)}$