another question:)

Hello, turkeyy!

Quote:

As a cable is pulled from a large spool, it unwinds in layers of constant radius.

If the cable is pulled at a steady 6 ft/sec, use the equation $s\:=\:r\theta$
to find how fast (radians/sec) the spool is turning when the radius is 1.2 ft.

Code:

* * * * * * * * * * O * * * * * | * * | θ * * 1.2 | * * * | o Q * | * * o * → → → → → → P 6 ft/s

The cable is pulled horizontally at 6 ft/sec.
Then point $P$ is moving around the circle at the same rate.

We have: . $r = 1.2$

$\text{Let }\,s \:=\:\text{arc(PQ)}\text{, then: }\,\frac{ds}{dt} \,=\,6$

From $s \:=\:r\theta$ , we have: . $s \:=\:1.2\,\theta$

Differentiate with respect to time: . $\frac{ds}{dt} \:=\:1.2\,\frac{d\theta}{dt} \quad\Rightarrow\quad \frac{d\theta}{dt} \:=\:\frac{1}{1.2}\,\frac{ds}{dt}$

Since $\frac{ds}{dt} = 6$ , we have: . $\frac{d\theta}{dt} \:=\:\frac{1}{1.2}(6) \:=\:5\text{ rad/sec}$