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# Given x(t)=3sin(t) - 3, y(t)=t-1 for 0 is less than or equal to t is less than or equal to 2pi How do you find the velocity of the particle at t=3?

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mason m Share
Jun 23, 2018

The particle's velocity is $\sqrt{9 {\cos}^{2} \left(3\right) + 1} \approx 3.1338$

#### Explanation:

We can find the $x$- and $y$-velocities by taking the derivatives of each position function.

$\left\{\begin{matrix}x ' \left(t\right) = 3 \cos \left(t\right) \\ y ' \left(t\right) = 1\end{matrix}\right.$

So at $t = 3$, we have

$\left\{\begin{matrix}x ' \left(3\right) = 3 \cos \left(3\right) \\ y ' \left(3\right) = 1\end{matrix}\right.$

The magnitude of the velocity is then the hypotenuse of the right triangle whose legs are the $x$ and $y$ components of the velocity:

$v \left(3\right) = \sqrt{{\left(x ' \left(3\right)\right)}^{2} + {\left(y ' \left(3\right)\right)}^{2}} = \sqrt{9 {\cos}^{2} \left(3\right) + 1} \approx 3.1338$

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