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A particle is moving at $x=3\cos\left(2t\right)$. Find the expression for velocity in terms of $x$.

I'm not sure where to start.

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    @GregorBruns Thanks for that! I'll have a play around and see if I can find the answer.2012-06-14

3 Answers 3

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All right, here's a slightly cleaner version of previous answers. We have

$x=3\cos 2t,\frac{dx}{dt}=-6\sin 2t$

Again, we use the trig identity

$\sin^2u+\cos^2u=1$

Substituting $u=2t$ we have

$\sin^2 2t+\cos^2 2t=1$

Now we multiply this equation by 9.

$9\sin^22t+9\cos^22t=9$

$9\sin^22t+x^2=9$

$9\sin^22t=9-x^2$

$3\sin2t=\pm\sqrt{9-x^2}$

$\frac{dx}{dt}=-6\sin 2t=\pm2\sqrt{9-x^2}$

The plus or minus depends on the value of $t$.

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Since the velocity is the derivative of the displacement we are looking for $\frac{dx}{dt}$ in terms of x $x=3\cos(2t)$ $\frac{dx}{dt}=-6\sin(2t)$ Solving for $t$ in the first equation shows that $t=\frac{1}{2}\arccos\left(\frac{x}{3}\right)$ Using the following trig identity $\sin^2x+\cos^2x=1$ and solving for $\sin x$ gives $\sin x=\pm\sqrt{1-\cos^2x}$ Therefore, plugging in for $t$ in the second equation gives $\frac{dx}{dt}=-6\sin\left(\arccos\left(\frac{x}{3}\right)\right)$ $\frac{dx}{dt}=\pm6\sqrt{1-\cos^2\left(\arccos\left(\frac{x}{3}\right)\right)}$ $\frac{dx}{dt}=\pm6\sqrt{1-\frac{x^2}{9}}$ $\fbox{$\frac{dx}{dt}=\pm2\sqrt{9-x^2}$}$

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You have: $\dot x(t)=-6\sin(2t)$ From the equation: $x(t)=3\cos(2t)$ you have: $t=\frac{1}{2}\arccos\left(\frac{x(t)}{3}\right)$ So: $\dot x(t)=-6\sin\left(\arccos\frac{x(t)}{3}\right)$