Answered

IDNLearn.com offers a reliable platform for finding accurate and timely answers. Find the information you need quickly and easily with our reliable and thorough Q&A platform.

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

Sagot :

Oladayoademosuidn

Answer:

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

Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.