The radius of the puddle is zero when time is zero but expands at a rate of one-quarter unit per second instead of one-half unit per second. By what factor does the denominator of the resultant composite function A(r(t)) change from that in the opening scenario?

Given that the radius of the puddle expands, originally, at a rate of 1/4 unit/second instead of 1/2 unit/second.

The current rate of expension of the radius,

\frac {r(t)}{t}=frac 1 4 unit/second ...(i)

The resultant composite function is area, A(r(t)).

The area of a circular region having radius r is,

If the radius is the function of time, then the composite function is

To determine the rate of change of area with respect to time, differentiate the composite function A(r(t)) once with respect to time, we have

So, the composite function is changing twice of the rate of change of the radius.

By using equation (i), if r(i) is changing with R= 1/4 unit / second.

Then, A(r(t)) will change 2 x 1/4 = 1/2 square unit/second.

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this