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\textbf{Vector Functions and Curves}

\textit{\textbf{One variable functions}}
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\textbf{Question}

A particle is moving around a circle at constant speed. Given that the
equation of the circle is $x^2+y^2=25$ and that the particle makes one
revolution every two seconds, find its acceleration at the point
$(3,4)$.


\textbf{Answer}

The position of the particle is given by
$$\underline{r}=5\cos(\omega t)\underline{i} +5 \sin(\omega
t)\underline{j},$$
where $\omega=pi$ ensures that $\underline{r}$ has period
$2\pi/\omega =2s$.

Thus
$$\underline{a}=\frac{d^2\underline{r}}{dt^2} = -\omega^2\underline{r}
=-\pi^2\underline{r}.$$

The acceleration at $(3,4)$ is $-3\pi^2\underline{i}-4\pi^2\underline{j}$.

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