Question

Water is poured into a container that has a leak. The mass m of the water is given as a function of time t by m = 5.0t^0.8 - 3.3t + 23, with t ≥ 0, m in grams, and t in seconds. (a) At what time is the water mass greatest, and (b)what is that greatest mass? What is the rate of mass change at (c) t = 1.9 s and (d) t = 5.0 s?

Answer 1

Part A.

Mass will be greatest in the container when

dm/dt = 0

m = 5t^0.8 - 3.3t + 23

dm/dt = 5*0.8*t^-0.2 - 3.3

5*0.8*t^-0.2 - 3.3 = 0

t^-0.2 = 3.3/4.0

t = (3.3/4.0)^(-1/0.2) = (3.3/4.0)^(-5)

t = 2.62 sec

Part B.

Greatest mass will be

m = 5t^0.8 - 3.3t + 23

m = 5*2.62^0.8 - 3.3*2.62 + 23

m = 25.16 gm = greatest mass

m = 0.02516 kg = greatest mass

Part C.

rate of mass change is

dm/dt = 5*0.8*t^-0.2 - 3.3

dm/dt = 4*t^-0.2 - 3.3

at t = 1.9 sec

dm/dt = 4*1.9^-0.2 - 3.3 = 0.22 gm/sec

dm/dt = 0.00022 kg/sec

Part D.

at t = 5.0 sec

dm/dt = 4*5.0^-0.2 - 3.3 = -0.40 gm/sec

dm/dt = -0.00040 kg/sec

Please Upvote.

If you need answer in any different units let me know in the comment section.