Calculus of trigonometric functions: derivatives, integrals and harmonic motion modelling

Use $\int \cos(k x) \, dx = \frac{\sin(k x)}{k}$ with $k = 2$.

$\int_0^{\pi/2} \cos(2 x) \, dx = \left[ \frac{\sin(2 x)}{2} \right]_0^{\pi/2} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0 - 0 = 0$.

Markers reward the correct antiderivative, the bracket notation, and an evaluated answer (here exactly zero, since the integrand is symmetric about $\pi/4$).

Differentiate using the chain rule.

$h'(t) = 1.5 \cos\left( \frac{\pi t}{6} \right) \cdot \frac{\pi}{6} = \frac{\pi}{4} \cos\left( \frac{\pi t}{6} \right)$.

At $t = 2$: $h'(2) = \frac{\pi}{4} \cos\left( \frac{\pi}{3} \right) = \frac{\pi}{4} \cdot \frac{1}{2} = \frac{\pi}{8} \approx 0.393$ m/hr.

The water is rising at about $0.39$ metres per hour at $t = 2$ hours.

Markers expect the chain rule applied correctly, the standard angle evaluated exactly, and units in the final answer.

Differentiate each term with the chain rule. For a linear inside argument $kx$, the derivative of $\sin(kx)$ is $k\cos(kx)$ and of $\cos(kx)$ is $-k\sin(kx)$:

Combine.

Marker's note: one mark for $6\cos(2x)$ (the chain-rule factor $2$ carried through), one for $5\sin(5x)$ with the correct sign. Dropping the inside factor, or losing the double minus on the cosine term, is the usual slip.

Integrate each standard form. Recall $\int \sin(kx)\,dx = -\frac{1}{k}\cos(kx) + C$ and $\int \sec^2 x\,dx = \tan x + C$:

Combine with a single constant.

Marker's note: one mark for $-\frac{1}{3}\cos(3x)$ (dividing by the inside coefficient and keeping the minus sign), one for $\tan x + C$. Forgetting the $\frac{1}{3}$ or omitting $+C$ loses a mark.

Part (a): read the parameters of $A\sin(\omega t)$. Here $A = 0.4$ and $\omega = 880\pi$, so the amplitude is $0.4$ mm and the period is

Part (b): velocity is the derivative of displacement.

Marker's note: one mark for amplitude $0.4$ mm, one for period $\frac{1}{440}$ s (using $T = \frac{2\pi}{\omega}$), one for $v(t) = 352\pi\cos(880\pi t)$ with the chain-rule factor carried through. A period of $\frac{2\pi}{880\pi} = \frac{1}{440}$ (not $\frac{1}{880}$) is the point to check.

Rewrite the squared trig function with a double-angle identity. There is no direct antiderivative of $\sin^2 x$, so use $\sin^2 x = \frac{1 - \cos 2x}{2}$:

Substitute the limits. At the upper limit, $2x = \frac{2\pi}{3}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$:

Marker's note: one mark for applying the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$, one for the correct antiderivative $\frac{x}{2} - \frac{\sin 2x}{4}$, one for the exact value $\frac{\pi}{6} - \frac{\sqrt{3}}{8}$. Trying to integrate $\sin^2 x$ directly is the trap.

Part (a): amplitude and centre from the peak and trough. The amplitude is half the peak-to-trough range and the centre is the average of the two:

Part (b): period from the minimum-to-maximum gap. A minimum at $t = 4$ and the next maximum at $t = 16$ are half a cycle apart, so

Part (c): build the model, then differentiate. With period $24$, $\omega = \frac{2\pi}{24} = \frac{\pi}{12}$. Taking the centre-crossing rise at $t = 10$ (midway between the min at $t = 4$ and the max at $t = 16$), the temperature is

Differentiate:

At $t = 10$ the cosine argument is $0$, so $\cos 0 = 1$ and

Marker's note: one mark for amplitude $0.4$ and centre $36.8$, one for period $24$ hours, one for a correct model or $\omega = \frac{\pi}{12}$, one for the rate $\frac{\pi}{30}$ degrees per hour. The temperature climbs fastest exactly as it crosses its mean, which is why $t = 10$ gives the maximum rate.

Part (a): read the model $h(t) = 8 + 2.5\cos\!\left( \frac{\pi t}{6} \right)$. The centre depth is $8$ and the amplitude is $2.5$, so

High tide is at $t = 0$ (cosine at its peak). The first low tide is half a period later; the period is $T = \frac{2\pi}{\pi/6} = 12$ hours, so low tide is at $t = 6$ hours.

Part (b): solve $h(t) = 7$.

Let $\theta = \frac{\pi t}{6}$. For $0 \le t \le 12$ we have $0 \le \theta \le 2\pi$. The solutions of $\cos\theta = -0.4$ in this range are

Convert back with $t = \frac{6\theta}{\pi}$:

Between these times the depth is below $7$ m, so the ferry cannot enter for

Part (c): differentiate twice.

Since $h(t) - 8 = 2.5\cos\!\left( \frac{\pi t}{6} \right)$, this is exactly

as required. The acceleration of the water level is proportional to the displacement from the mean depth of $8$ m and directed back towards it, so the tide is simple harmonic motion.

Marker's note: one mark for the max and min depths, one for the first low tide at $t = 6$ h; one for reducing to $\cos\theta = -0.4$, one for both times $t \approx 3.79$ and $t \approx 8.21$ h, one for the duration $\approx 4.42$ h; one for the differentiation showing $h''(t) = -\frac{\pi^2}{36}(h(t) - 8)$ with the simple-harmonic interpretation. Keeping $\theta$ in radians and finding the second solution as $2\pi - \arccos(-0.4)$ is where marks are won or lost.

Part (a): displacement is the integral of velocity. Integrate $v(t) = 6\cos(2t)$:

Apply the initial condition $x(0) = 1$. Since $\sin 0 = 0$, we get $C = 1$, so

Part (b): acceleration is the derivative of velocity.

From part (a), $x(t) - 1 = 3\sin(2t)$, so $\sin(2t) = \frac{x(t) - 1}{3}$. Substitute:

as required. The motion is simple harmonic about the centre $x = 1$.

Part (c): the displacement oscillates about $x = 1$ with amplitude $3$. Since $-1 \le \sin(2t) \le 1$, the displacement ranges over

Distance from the origin is $|x(t)|$, and the extremes are $|{-2}| = 2$ and $|4| = 4$. The greatest distance from the origin is therefore $4$ metres.

Marker's note: one mark for integrating to $3\sin(2t) + C$, one for $C = 1$ giving $x(t) = 3\sin(2t) + 1$; one for $a(t) = -12\sin(2t)$ and one for the substitution showing $a = -4(x - 1)$; one for the greatest distance $4$ m (comparing $|{-2}|$ and $|4|$, not just quoting the amplitude). The trap in (c) is stopping at the amplitude $3$ instead of measuring from the origin.