Question

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Tutorial Exercise 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 12 ft3/min, how fast is the water level rising when the water is 8 inches deep?

Answer 1

Answer:

0.75 ft/min

Step-by-step explanation:

Given;

Length of trough l = 8 ft

base length of trough b = 3 ft

height of trough h = 1 ft

Change in volume dV/dt = 12 ft^3/min

At height hi = 8 in = 2/3 ft

Ratio of base to height;

b/h = 3/1

b = 3h .....1

The volume of the trough;

V = area of triangle × length

V = 1/2 × b×h × l

V = 1/2(bhl) .....2

Substituting l = 8ft and equation 1 (b = 3h) into equation 2;

V = 1/2(3h × h × 8)

V = 12h^2

differentiating V, change in volume per time is;

dV/dt = 12 × 2h .dh/dt

dV/dt = 24 hi .dh/dt

Substituting the values of dV/dt and hi;

12 = 24(2/3) .dh/dt

dh/dt = 12 ÷ 24(2/3) = 12/16

dh/dt = 3/4 ft/min = 0.75 ft/min