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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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Do you know that velocity is the derivative of $x(t)$? – Gregor Bruns Jun 14 '12 at 10:37
@GregorBruns If I differentiate the function, wouldn't that give me the velocity but in terms of t instead of x? – Jallah Jun 14 '12 at 10:40
Yes, it does. But there is the relation $\sin(s)^2+\cos(s)^2=1$, which applies also to $s=2t$, that is $\sin(2t)^2+\cos(2t)^2=1$. You have to play around a little bit with the numbers, though. In this way you can express $v(t)=\frac{dx}{dt}$ in terms of $x(t)$. – Gregor Bruns Jun 14 '12 at 10:42
@GregorBruns Thanks for that! I'll have a play around and see if I can find the answer. – Jallah Jun 14 '12 at 10:43
up vote 2 down vote accepted

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


Substituting $u=2t$ we have

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

Now we multiply this equation by 9.





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

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: $$\dot x(t)=-6\sin\left(\arccos\frac{x(t)}{3}\right)$$

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