1. **Problem Statement:** We have a water wheel with radius $r=3$ m, submerged depth $d=0.5$ m, rotating every $T=40$ s. We want to: a) Draw the height graph of a point $P$ on the wheel for 2 cycles starting at the lowest point. b) Find the equation of this height as a function of time $t$. c) Find the height of point $P$ at $t=32$ s. 2. **Understanding the problem:** - The wheel rotates with period $T=40$ s, so angular velocity is $\omega=\frac{2\pi}{T}=\frac{2\pi}{40}=\frac{\pi}{20}$ radians per second. - The height of point $P$ varies sinusoidally as the wheel rotates. - The lowest point is submerged $0.5$ m below water, so the vertical position of the wheel's center is $3 - 0.5 = 2.5$ m above water. 3. **Height function formula:** The height $h(t)$ of point $P$ above water is given by: $$ h(t) = \text{vertical center} + r \sin(\theta(t)) $$ where $\theta(t)$ is the angle of rotation at time $t$. Since the wheel starts at the lowest point at $t=0$, the angle $\theta(0) = -\frac{\pi}{2}$ (point $P$ is at bottom). So, $$ \theta(t) = \omega t - \frac{\pi}{2} = \frac{\pi}{20}t - \frac{\pi}{2} $$ 4. **Equation of height:** $$ h(t) = 2.5 + 3 \sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) $$ 5. **Calculate height at $t=32$ s:** $$ \theta(32) = \frac{\pi}{20} \times 32 - \frac{\pi}{2} = \frac{32\pi}{20} - \frac{\pi}{2} = \frac{8\pi}{5} - \frac{\pi}{2} = \frac{16\pi}{10} - \frac{5\pi}{10} = \frac{11\pi}{10} $$ Then, $$ h(32) = 2.5 + 3 \sin\left(\frac{11\pi}{10}\right) = 2.5 + 3 \times (-0.3420) = 2.5 - 1.026 = 1.474 \text{ m} $$ **Final answers:** - a) The graph is a sinusoidal wave with period 40 s, amplitude 3 m, vertical shift 2.5 m, starting at minimum. - b) Equation: $h(t) = 2.5 + 3 \sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right)$ - c) Height at $t=32$ s is approximately $1.474$ m.