0 like 0 dislike

An industrial open water tank, as shown in the picture below, has an inlet pipe and an outlet pipe. The depth of the water in the tank changes continually.

The equation $D(t)=4+0.5t^2-0.25t^3$ gives the depth (in metres) of the water, where $t$ represents the time (in hours) that has lapsed since the depth reading was taken at 09:00


  1. The depth of the water in the tank at 11:00
  2. The rate of change of the depth of the water in the tank at 12:00



in Grade 12 Technical Maths by Diamond (39.8k points) | 6 views

1 Answer

0 like 0 dislike
1. After 2 hrs the depth = D(2) =  4+ 0,5(2)^2 - 0,25(2)^3 m
                                                     = 4 m

2. D = 4 + 0,5t^2 - 0,25t
    dD/dt =  t - 0,75t
At 12:00 (3 hours later) we have
                    (3) - 0,75(3)
                     = -3,75m/hr
by Silver Status (31.3k points)
Welcome to MathsGee Skills Question and Answer Bank, a platform, where you can ask study questions and receive answers from other members of the community. Help is always 100% FREE!
MathsGee Q&A is the knowledge-sharing community where millions of students and experts put their heads together to crack their toughest homework questions.

Enter your email address:


[consumerlti id="python_coding"]