Question 1185007
<pre>
The diameter is 10 ft., so the radius is 5 ft.
Let the arc at the top be "s".  We will need the formula 
{{{s}}}{{{""=""}}}{{{r*theta}}}, 
{{{s}}}{{{""=""}}}{{{5*theta}}},

and its derivative with respect to time t:

{{{ds/dt}}}{{{""=""}}}{{{5*expr(d(theta)/dt)}}}

Let the length of the chord be x. Since it is decreasing in length 1 foot per
minute, taken negative since it is decreasing.

{{{dx/dt}}}{{{""=""}}}{{{-1}}}

{{{drawing(300,300,-1.1,1.1,-1.1,1.1, circle(0,0,1),
locate(0,1,s),
line(0,0,-.9,.4358899),line(0,0,.9,.4358899),locate(.33,.3,5),locate(-.4,.3,5),
locate(-.04,.14,theta),locate(-.03,.57,x), red(arc(0,0,2,-2,60,120)),
line(.9,.4358899,-.9,.4358899) )}}}

Next we have to find an equation for x in terms of the chord and &theta;.  So we draw
this green line perpendicular to the chord, which bisects the chord and &theta;
and gives us two congruent right triangles.

{{{drawing(300,300,-1.1,1.1,-1.1,1.1, circle(0,0,1),locate(0,1,s),
line(0,0,-.9,.4358899),line(0,0,.9,.4358899),locate(.33,.3,5),locate(-.4,.3,5),
locate(-.09,.3,theta/2),locate(-.4,.7,x/2), red(arc(0,0,2,-2,60,120)),
line(.9,.4358899,-.9,.4358899), green(line(0,0,0,.4358899)) )}}}

From the right triangle,

{{{sin(theta/2)}}}{{{""=""}}}{{{opposite/hypotenuse}}}{{{""=""}}}{{{(x/2)/5^""}}}{{{""=""}}}{{{expr(x/2)*expr(1/5)}}}{{{""=""}}}{{{x/10}}}

{{{sin(theta/2)}}}{{{""=""}}}{{{x/10}}}

{{{x}}}{{{""=""}}}{{{10*sin(theta/2)}}}

Take the derivative with respect to time t:

{{{dx/dt}}}{{{""=""}}}{{{10*cos(theta/2)*(1/2)expr(d(theta)/dt)}}}

{{{dx/dt}}}{{{""=""}}}{{{5*cos(theta/2)*expr(d(theta)/dt)}}}

Since {{{dx/dt}}}{{{""=""}}}{{{-1}}}

{{{-1}}}{{{""=""}}}{{{5*cos(theta/2)*expr(d(theta)/dt)}}}

We redraw the figure at the instant when the chord x = 8, which
makes the left half of it 4.  And since it's a 3-4-5 right triangle,
the green line is 3.

{{{drawing(300,300,-1.1,1.1,-1.1,1.1, circle(0,0,1),locate(0,1,s),
line(0,0,-.9,.4358899),line(0,0,.9,.4358899),locate(.33,.3,5),locate(-.4,.3,5),
locate(-.09,.3,theta/2),
locate(.05,.25,3),
locate(-.4,.57,4), red(arc(0,0,2,-2,60,120)),
line(.9,.4358899,-.9,.4358899), green(line(0,0,0,.4358899)) )}}}

{{{-1}}}{{{""=""}}}{{{5*cos(theta/2)*expr(d(theta)/dt)}}}

Since {{{cos(theta/2)}}}{{{""=""}}}{{{adjacent/hypotenuse}}}{{{""=""}}}{{{3/5}}}

{{{-1}}}{{{""=""}}}{{{5*expr(3/5)*expr(d(theta)/dt)}}}

{{{-1}}}{{{""=""}}}{{{3*expr(d(theta)/dt)}}}

{{{-1/3}}}{{{""=""}}}{{{d(theta)/dt}}}

Finally we substitute in

{{{ds/dt}}}{{{""=""}}}{{{5*expr(d(theta)/dt)}}}

{{{ds/dt}}}{{{""=""}}}{{{5*(-1/3)}}}

{{{ds/dt}}}{{{""=""}}}{{{-5/3}}}{{{feet/minute}}}

{{{ds/dt}}}{{{""=""}}}{{{-1&2/3}}}{{{feet/minute}}}

The negative sign indicates that the arc is decreasing

Edwin</pre>