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) ( see in the diagram )
,
By the Pythagoras theorem,
![d^2 + (\frac{w}{2})^2 = 9^2](https://tex.z-dn.net/?f=d%5E2%20%2B%20%28%5Cfrac%7Bw%7D%7B2%7D%29%5E2%20%3D%209%5E2)
![(9-\frac{t}{6})^2 +\frac{w^2}{4} = 81](https://tex.z-dn.net/?f=%289-%5Cfrac%7Bt%7D%7B6%7D%29%5E2%20%2B%5Cfrac%7Bw%5E2%7D%7B4%7D%20%3D%2081)
![\frac{t^2}{36}-\frac{18t}{6} + \frac{w^2}{4}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bt%5E2%7D%7B36%7D-%5Cfrac%7B18t%7D%7B6%7D%20%2B%20%5Cfrac%7Bw%5E2%7D%7B4%7D%3D0)
![\frac{t^2 - 108t + 9w^2}{36}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bt%5E2%20-%20108t%20%2B%209w%5E2%7D%7B36%7D%3D0)
![t^2 - 108t + 9w^2 =0](https://tex.z-dn.net/?f=t%5E2%20-%20108t%20%2B%209w%5E2%20%3D0)
![9w^2 = 108t - t^2](https://tex.z-dn.net/?f=9w%5E2%20%3D%20108t%20-%20t%5E2)
![w = \frac{1}{3}\sqrt{108t - t^2}](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Csqrt%7B108t%20-%20t%5E2%7D)
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,
![6 =\frac{1}{3}\sqrt{108t - t^2}](https://tex.z-dn.net/?f=6%20%3D%5Cfrac%7B1%7D%7B3%7D%5Csqrt%7B108t%20-%20t%5E2%7D)
![18 =\sqrt{108t-t^2}](https://tex.z-dn.net/?f=18%20%3D%5Csqrt%7B108t-t%5E2%7D)
![324 = 108t - t^2](https://tex.z-dn.net/?f=324%20%3D%20108t%20-%20t%5E2)
![\implies t^2 - 108t+ 324=0](https://tex.z-dn.net/?f=%5Cimplies%20t%5E2%20-%20108t%2B%20324%3D0)
By using quadratic formula,
![\implies t = 3.088\text{ or }t = 104.912](https://tex.z-dn.net/?f=%5Cimplies%20t%20%3D%203.088%5Ctext%7B%20or%20%7Dt%20%3D%20104.912)
Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.
(c) ![w = \frac{1}{3}\sqrt{108t- t^2}](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Csqrt%7B108t-%20t%5E2%7D)
![\implies 3w = \sqrt{108t- t^2}](https://tex.z-dn.net/?f=%5Cimplies%203w%20%3D%20%5Csqrt%7B108t-%20t%5E2%7D)
![9w^2 = 108t - t^2](https://tex.z-dn.net/?f=9w%5E2%20%3D%20108t%20-%20t%5E2)
![-9w^2 = -108t + t^2](https://tex.z-dn.net/?f=-9w%5E2%20%3D%20-108t%20%2B%20t%5E2)
![-9w^2 + 2916 = 2916 - 108t + t^2](https://tex.z-dn.net/?f=-9w%5E2%20%2B%202916%20%3D%202916%20-%20108t%20%2B%20t%5E2)
![2916 - 9w^2 = (t - 108)^2](https://tex.z-dn.net/?f=2916%20-%209w%5E2%20%3D%20%28t%20-%20108%29%5E2)
![(t-108) = \sqrt{2916 - 9w^2}](https://tex.z-dn.net/?f=%28t-108%29%20%3D%20%5Csqrt%7B2916%20-%209w%5E2%7D)
![t = \sqrt{2916 - 9w^2} + 108](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B2916%20-%209w%5E2%7D%20%2B%20108)
For 0 ≤ w ≤ 18,
0 ≤ t ≤ 108,
So, Domain = [0, 18]
Range = [0, 108]