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