1. ## another question:)

As television cable is pulled from a large spool to be strung from the telephone poles along a street,it unwinds from the spool in layers of constant radius.If the truck pulling the cable moves at a steady 6ft/sec (a touch over 4 mph),use the equation s=rθ to find how fast (radians per second )the spool is turning when the layer of radius 1.2ft is being unwound.

2. Hello, turkeyy!

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