A trough is 10 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 15 ft3/min, how fast is the water level rising when the water is 8 inches deep

7.5 ft³/min

Step-by-step explanation:

Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.

Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²

So, V = Ah = 2h = 2(10 - x)

The rate of change of volume is thus

dV/dt = d[2(10 - x)]/dt = -2dx/dt

Since dV/dt = 15 ft³/min,

dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min

Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt

= -dx/dt

= -(-7.5 ft³/min)

= 7.5 ft³/min

And the height at this point when x = 8 inches = 8 in × 1 ft/12 in  = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft

my dear correct answer is option number 3