Atrough has a semicircular cross section with a radius of 9 feet. water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. (a) give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. make sure to specify the domain and compute the range too.

(b) after how many hours will the surface of the water have width of 6 feet?

(c) give a function t = f −^1 (w) relating the time to the width of the surface of the water. make sure to specify the domain and compute the range too.

(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,

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,

By using quadratic formula,

Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.

(c)

For 0 ≤ w ≤ 18,

0 ≤ t ≤ 108,

So, Domain = [0, 18]

Range = [0, 108]

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.

answer: i'm great at algebra

~coco

step-by-step explanation: