Using Parameterization to find Speed

**iheartthemusic29** · November 13th 2009, 07:38 PM

A particle moves with its position given by x=cos6t and y=sint , where positions are given in feet from the origin and time t is in seconds.
Find the speed of the particle.

I took the derivative of each equation and then plugged it into the formula velocity=sqrt((dx/dt)^2+(dy/dt)^2).
I got ((-6sin6t)^2+cost^2)^(1/2)ft/s for the answer, but that is incorrect.

Am I going about this the wrong way?

**redsoxfan325** · November 13th 2009, 08:01 PM

Originally Posted by iheartthemusic29

A particle moves with its position given by x=cos6t and y=sint , where positions are given in feet from the origin and time t is in seconds.
Find the speed of the particle.

I took the derivative of each equation and then plugged it into the formula velocity=sqrt((dx/dt)^2+(dy/dt)^2).
I got ((-6sin6t)^2+cost^2)^(1/2)ft/s for the answer, but that is incorrect.

Am I going about this the wrong way?

Plugging the derivatives into the formula gives:

$\sqrt{36\sin^26t+\cos^2t}$

Either the answer is some variation of that, or there is something else wrong with the problem. But the math is fine.

**novice** · November 13th 2009, 08:55 PM

Originally Posted by iheartthemusic29

A particle moves with its position given by x=cos6t and y=sint , where positions are given in feet from the origin and time t is in seconds.
Find the speed of the particle.

I took the derivative of each equation and then plugged it into the formula velocity=sqrt((dx/dt)^2+(dy/dt)^2).
I got ((-6sin6t)^2+cost^2)^(1/2)ft/s for the answer, but that is incorrect.

Am I going about this the wrong way?

You need to do trigonometric work by expressing $x=\cos 6t$ in terms of $\sin t$ ; then substitute $y$ for all the $\sin t$ 's.

Your $x$ in the end should look like this: $x=1-4y-4y^3$ and work from there.

**novice** · November 14th 2009, 08:00 AM

Originally Posted by iheartthemusic29

A particle moves with its position given by x=cos6t and y=sint , where positions are given in feet from the origin and time t is in seconds.
Find the speed of the particle.

I took the derivative of each equation and then plugged it into the formula velocity=sqrt((dx/dt)^2+(dy/dt)^2).
I got ((-6sin6t)^2+cost^2)^(1/2)ft/s for the answer, but that is incorrect.

Am I going about this the wrong way?

There is nothing wrong with your answer. It's all depending on what form of expression you want.

You can see the velocity vector very clearly: $v=\frac{dr}{dt}= -6\sin 6t \overrightarrow i + \cos t \overrightarrow j$

The restriction on displacement on either components are $-1 \leq x \leq 1$ and $-1 \leq y \leq 1$

Velocities of each component are $0 \leq v_x \leq 6$ and $0 \leq v_y \leq 1$

The object moves in harmonic motion following an elliptic path, where $v_x = -6 \sin (2 \pi f t) + \phi$ and $v_y = \cos (2\pi f t)+ \phi$ . $f$ and $\phi$ are frequency and phase angle, respectively.

If you are interested in find time in relation to $x$ and $y$ , you have a lot of work to do. You must find the relation between $x$ and $y$ expressed in $\sin t$ , where $\sin{6t} = 4 \sin t \cos^2 {t}$ , $t = \frac{dy}{dx}$ which will involve ODE.

That's all, I have nothing more to give. Good luck!

**iheartthemusic29** · November 14th 2009, 05:01 PM

Thanks!
How could I use this to find the first positive time when the particle comes to a stop?

**novice** · November 15th 2009, 12:16 PM

Originally Posted by iheartthemusic29

Thanks!
How could I use this to find the first positive time when the particle comes to a stop?

When the instantaneous velocity = 0, $(-6\sin {6t})^2+cos^2{t}=0$ .

If you do the algebra, it should give you the first stop at $1.5 < t_1 < 1.7$ ; the subsequent $t$ , will be $t=t_1+nT$ , $n = 1, 2, 3, ..., N$ , $T = \frac{1}{f}$

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