Answer:
(a) Let h represents the height of water and w represents the width of the water,
Since, the depth of the water is increasing at a rate of 2 inches per hour,
So, after t hours,
The height of water, h(t) = 2t inches = t/6 ft,
( ∵ 1 foot = 12 inches ⇒ 1 inch = 1/12 ft )
Thus, the distance distance from the centre to the top of the water, d = 9 - h(t) Â ( see in the diagram )
[tex]d=9-\frac{t}{6}[/tex],
By the Pythagoras theorem,
[tex]d^2 + (\frac{w}{2})^2 = 9^2[/tex]
[tex](9-\frac{t}{6})^2 +\frac{w^2}{4} = 81[/tex]
[tex]\frac{t^2}{36}-\frac{18t}{6} + \frac{w^2}{4}=0[/tex]
[tex]\frac{t^2 - 108t + 9w^2}{36}=0[/tex]
[tex]t^2 - 108t + 9w^2 =0[/tex]
[tex]9w^2 = 108t - t^2[/tex]
[tex]w = \frac{1}{3}\sqrt{108t - t^2}[/tex]
Since, diameter of the semicircular cross section is 18 ft,
So, 0 ≤ w ≤ 18,
i.e Range = [0, 18]
Also, w will be defined if 108t - t² ≥ 0
⇒ (108 - t)t ≥ 0,
⇒ 0 ≤ t ≤ 108
i.e Domain = [0, 108]
(b) If w = 6,
[tex]6 =\frac{1}{3}\sqrt{108t - t^2}[/tex]
[tex]18 =\sqrt{108t-t^2}[/tex]
[tex]324 = 108t - t^2[/tex]
[tex]\implies t^2 - 108t+ 324=0[/tex]
By using quadratic formula,
[tex]\implies t = 3.088\text{ or }t = 104.912[/tex]
Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.
(c) [tex]w = \frac{1}{3}\sqrt{108t- t^2}[/tex]
[tex]\implies 3w = \sqrt{108t- t^2}[/tex]
[tex]9w^2 = 108t - t^2[/tex]
[tex]-9w^2 = -108t + t^2[/tex]
[tex]-9w^2 + 2916 = 2916 - 108t + t^2[/tex]
[tex]2916 - 9w^2 = (t - 108)^2[/tex]
[tex](t-108) = \sqrt{2916 - 9w^2}[/tex]
[tex]t = \sqrt{2916 - 9w^2} + 108[/tex]
For 0 ≤ w ≤ 18,
0 ≤ t ≤ 108,
So, Domain = [0, 18]
Range = [0, 108]
