A trough is 8 ft long and its ends have the shape of isosceles triangles that are 3 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 8 ft3/min, how fast is the water level rising when the water is 7 inches deep

The level of the water is 4/7 feet when it is at 7inches.

Step-by-step explanation:

Given data:

Length of trough = 8ft

Shapes at the end = 3ft, and 1ft height.

Rate of filling the trough = 8ft/3min.

Solution:

When depth of the water is x feet, the base of the rough is 3x.

A = x * 3x / 2

= 1.5x^2

So the volume of water ( depth )

= 8 * 1.5x^2

= 12x^2.

The rate of change of depth

dv/dt = 24* (dx/dt)

Where dv/dt = 8ft3min.

8ft3min = 24* (dx/dt)

dx/dt = 8/24

= 8/ 24 (7/12)

= 8/14

= 4/7.

The level of the water is 4/7 feet when it is at 7inches.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23