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 = 4.8t^0.8 - 2.9t + 21, 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? In kilograms per minute, what is the rate of mass change at (c) t = 1.7 s and (d) t = 5.2 s?

To find the time at which the water mass is greatest, we need to find the maximum point of the function. We can do this by taking the derivative of the function and setting it equal to zero.

Step 1: Take the derivative of the function m = 4.8t^0.8 - 2.9t + 21
dm/dt = 4.8*0.8*t^(-0.2) - 2.9

Step 2: Set the derivative equal to zero and solve for t
4.8*0.8*t^(-0.2) - 2.9 = 0
3.84*t^(-0.2) = 2.9
t^(-0.2) = 2.9/3.84
t^(-0.2) = 0.7552
Taking the reciprocal of both sides,
t^(0.2) = 1/0.7552
t^(0.2) = 1.3221
Taking both sides to the power of 5,
t = 1.3221^5
t ≈ 4.445 seconds

So, at approximately t = 4.445 seconds, the water mass is greatest.

To find the greatest mass, we can substitute this time value back into the function.

Step 3: Substitute t = 4.445 into m = 4.8t^0.8 - 2.9t + 21
m = 4.8*(4.445)^0.8 - 2.9*(4.445) + 21
m ≈ 37.53 grams

Therefore, at t = 4.445 seconds, the greatest mass of the water is approximately 37.53 grams.

To find the rate of mass change at different time points, we need to take the derivative of the function and evaluate it at those given times.

Step 4: Take the derivative of the function m = 4.8t^0.8 - 2.9t + 21
dm/dt = 4.8*0.8*t^(-0.2) - 2.9

Step 5: Substitute the given time values and calculate the rate of mass change
(a) At t = 1.7 seconds:
dm/dt = 4.8*0.8*(1.7)^(-0.2) - 2.9
dm/dt ≈ -1.529 grams per second

(b) At t = 5.2 seconds:
dm/dt = 4.8*0.8*(5.2)^(-0.2) - 2.9
dm/dt ≈ -0.697 grams per second

To convert the rate of mass change from grams per second to kilograms per minute, we divide by 1000 to convert from grams to kilograms and multiply by 60 to convert from seconds to minutes.

(c) At t = 1.7 seconds:
Rate of mass change = (-1.529 grams per second) * (1 kilogram / 1000 grams) * (60 minutes / 1 minute)
Rate of mass change ≈ -0.0917 kilograms per minute

(d) At t = 5.2 seconds:
Rate of mass change = (-0.697 grams per second) * (1 kilogram / 1000 grams) * (60 minutes / 1 minute)
Rate of mass change ≈ -0.0418 kilograms per minute

Therefore, at t = 1.7 seconds, the rate of mass change is approximately -0.0917 kilograms per minute and at t = 5.2 seconds, the rate of mass change is approximately -0.0418 kilograms per minute.

To find the time when the water mass is greatest, we need to find the maximum point on the function. To do this, we can take the derivative of the function and set it equal to zero.

(a) To find the time when the water mass is greatest:
Step 1: Take the derivative of the function m(t) with respect to t.
dm/dt = 4.8 * 0.8 * t^(0.8-1) - 2.9
Simplifying, we get dm/dt = 3.84t^(-0.2) - 2.9.

Step 2: Set dm/dt equal to zero and solve for t.
3.84t^(-0.2) - 2.9 = 0
3.84t^(-0.2) = 2.9
t^(-0.2) = 2.9 / 3.84
t^(-0.2) = 0.754167

Step 3: Take both sides of the equation to the power of -5 (reciprocal of -0.2).
(t^(-0.2))^(-5) = (0.754167)^(-5)
t = 0.754167^(5)
t ≈ 1.706 seconds

So, at approximately t = 1.706 seconds, the water mass is greatest.

(b) To find the greatest mass when the water mass is at its peak, substitute the value of t into the formula m(t).
m(t) = 4.8t^0.8 - 2.9t + 21
m(1.706) = 4.8(1.706)^0.8 - 2.9(1.706) + 21
m(1.706) ≈ 22.95 grams

So, the greatest mass of water is approximately 22.95 grams.

(c) To find the rate of mass change at t = 1.7 seconds, we need to take the derivative of m(t) with respect to t and convert it to kilograms per minute.

Step 1: Take the derivative of m(t) with respect to t.
dm/dt = 4.8 * 0.8 * t^(0.8-1) - 2.9
Simplifying, we get dm/dt = 3.84t^(-0.2) - 2.9.

Step 2: Substitute t = 1.7 into the derivative to find the rate of mass change at t = 1.7 seconds.
dm/dt = 3.84(1.7)^(-0.2) - 2.9
dm/dt ≈ 0.4105 grams per second

To convert this to kilograms per minute, we divide by 1000 to convert grams to kilograms and then multiply by 60 to convert seconds to minutes.
dm/dt = 0.4105 grams per second
dm/dt ≈ 0.4105/1000 kilograms per second
dm/dt ≈ 0.0004105 kilograms per second
dm/dt ≈ 0.0004105 * 60 kilograms per minute
dm/dt ≈ 0.02463 kilograms per minute

So, at t = 1.7 seconds, the rate of mass change is approximately 0.02463 kilograms per minute.

(d) To find the rate of mass change at t = 5.2 seconds, follow the same steps:

Step 1: Take the derivative of m(t) with respect to t.
dm/dt = 3.84t^(-0.2) - 2.9

Step 2: Substitute t = 5.2 into the derivative to find the rate of mass change at t = 5.2 seconds.
dm/dt = 3.84(5.2)^(-0.2) - 2.9
dm/dt ≈ -0.39 grams per second

To convert this to kilograms per minute:
dm/dt = -0.39 grams per second
dm/dt ≈ -0.39/1000 kilograms per second
dm/dt ≈ -0.00039 kilograms per second
dm/dt ≈ -0.00039 * 60 kilograms per minute
dm/dt ≈ -0.0234 kilograms per minute

So, at t = 5.2 seconds, the rate of mass change is approximately -0.0234 kilograms per minute (negative because the mass is decreasing).