Standard +0.8 This requires two differentiations of products involving exponential and trigonometric functions (using product rule twice), then computing the magnitude of the resulting acceleration vector. While the calculus is A-level standard, the algebraic manipulation to simplify to the given form requires careful work with e^t(cos t - sin t) and e^t(sin t + cos t) terms, and recognizing that their squares sum neatly. The 'show that' format and 'fully justify' instruction indicate this is above-average difficulty, requiring sustained accurate calculation rather than novel insight.
17 A particle is moving such that its position vector, \(\mathbf { r }\) metres, at time \(t\) seconds, is given by
$$\mathbf { r } = \mathrm { e } ^ { t } \cos t \mathbf { i } + \mathrm { e } ^ { t } \sin t \mathbf { j }$$
Show that the magnitude of the acceleration of the particle, \(a \mathrm {~ms} ^ { - 2 }\), is given by
$$a = 2 \mathrm { e } ^ { t }$$
Fully justify your answer.
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-27_2490_1728_217_141}
# Question 17:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiates with evidence of correct product rule | M1 | AO 3.4. Condone sign errors |
| Finds $\mathbf{v}$ or $\frac{d\mathbf{r}}{dt}$ with either $\mathbf{i}$ or $\mathbf{j}$ component fully correct | M1 | AO 1.1a |
| $\mathbf{v}=\frac{d\mathbf{r}}{dt}=(e^t\cos t - e^t\sin t)\mathbf{i}+(e^t\sin t+e^t\cos t)\mathbf{j}$ | A1 | AO 1.1b. Condone poor brackets if fully correct acceleration seen |
| Differentiates $\mathbf{v}$ with evidence of correct product rule to find $\mathbf{a}$ with at least one component correct | M1 | AO 3.4. Condone sign errors |
| $\mathbf{a}=\frac{d\mathbf{v}}{dt}=-2e^t\sin t\,\mathbf{i}+2e^t\cos t\,\mathbf{j}$ | A1 | AO 1.1b. May be unsimplified |
| Obtains expression for $|\mathbf{a}|$: $|\mathbf{a}|=\sqrt{(-2e^t\sin t)^2+(2e^t\cos t)^2}$ | M1 | AO 1.1a. Provided $\mathbf{a}$ has non-zero $\mathbf{i}$ and $\mathbf{j}$ components |
| Completes argument: $=\sqrt{4e^{2t}(\sin^2 t+\cos^2 t)}$, $\therefore |\mathbf{a}|=2e^t$ **AG** | R1 | AO 2.1. Must see factor of $(\sin^2 t+\cos^2 t)$, e.g. $\sqrt{4e^{2t}(\sin^2 t+\cos^2 t)}$ |
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17 A particle is moving such that its position vector, $\mathbf { r }$ metres, at time $t$ seconds, is given by
$$\mathbf { r } = \mathrm { e } ^ { t } \cos t \mathbf { i } + \mathrm { e } ^ { t } \sin t \mathbf { j }$$
Show that the magnitude of the acceleration of the particle, $a \mathrm {~ms} ^ { - 2 }$, is given by
$$a = 2 \mathrm { e } ^ { t }$$
Fully justify your answer.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-27_2490_1728_217_141}
\hfill \mbox{\textit{AQA Paper 2 2022 Q17 [7]}}