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=5t^0.8 -3t + 20 , with t>=0, mass in grams and t in seconds
A) at what time is the water mass greatest,and B) what is the greatest mass? In kilogram per minute what is the rate of mass change at t=3 s, and t=5 s.

m = 5 t^.8 - 3 t + 20

dm/dt = (5*.8) t^-.2 - 3
so at max or min
0 = 4/T^.2 - 3
T^.2 = 4/3 = 1.333
.2 log T = log 1.333
Log T = .6246
T = 10^.6246 = 4.21 seconds

find m at t = 4.21
and find dm/dt at 3 and 5 (signs better be different)

To find the time at which the water mass is greatest, we need to find the maximum of the given function.

Step 1: Take the derivative of the function with respect to time (t):

First, let's rewrite the function for simplicity:
m = 5t^0.8 - 3t + 20

Taking the derivative:
dm/dt = 4t^(-0.2) - 3

Step 2: Set the derivative equal to zero and solve for t to find the critical points:

4t^(-0.2) - 3 = 0
4t^(-0.2) = 3
t^(-0.2) = 3/4

Raise both sides to the power of -5 (reciprocal of -0.2):
(t^(-0.2))^(-5) = (3/4)^(-5)
t = (4/3)^(5)
t ≈ 1.8636

We can disregard any negative solutions since t must be greater than or equal to zero.

Therefore, the water mass is greatest at approximately t = 1.8636 seconds.

To find the greatest mass, we can substitute this value of t back into the original equation:

m = 5(1.8636)^0.8 - 3(1.8636) + 20
m ≈ 21.85 grams

So, the greatest mass is approximately 21.85 grams.

Now, let's calculate the rate of mass change at t = 3 seconds and t = 5 seconds in kilograms per minute.

For t = 3 seconds:
m = 5(3)^0.8 - 3(3) + 20
m ≈ 32.789 grams

The rate of mass change at t = 3 seconds can be found by taking the derivative:

dm/dt = 4t^(-0.2) - 3

Substituting t = 3 in the derivative equation:
dm/dt = 4(3)^(-0.2) - 3
dm/dt ≈ -1.157 grams/second

Now, to convert grams/second to kilograms/minute:
1 kilogram = 1000 grams
1 minute = 60 seconds

Therefore, the rate of mass change at t = 3 seconds is approximately -0.001157 kilograms/minute.

Similarly, for t = 5 seconds:
m = 5(5)^0.8 - 3(5) + 20
m ≈ 50.727 grams

dm/dt = 4t^(-0.2) - 3
dm/dt = 4(5)^(-0.2) - 3
dm/dt ≈ -0.685 grams/second

Converting to kilograms/minute:
dm/dt ≈ -0.000685 kilograms/minute.

So, the rate of mass change at t = 5 seconds is approximately -0.000685 kilograms/minute.

To find the time at which the water mass is greatest, you need to find the maximum point of the function. The maximum point of a function can be found by taking its derivative and setting it equal to zero.

Let's differentiate the given function to find the derivative:

m = 5t^0.8 - 3t + 20

Differentiating both sides with respect to t:

dm/dt = d(5t^0.8)/dt - d(3t)/dt + d(20)/dt

Now, let's take the derivatives of each term:

dm/dt = 4t^-0.2 - 3 + 0

Simplifying the expression:

dm/dt = 4t^-0.2 - 3

To find the time when the mass is greatest, we set the derivative equal to zero:

4t^-0.2 - 3 = 0

Adding 3 to both sides:

4t^-0.2 = 3

Dividing both sides by 4:

t^-0.2 = 3/4

Taking the reciprocal on both sides:

t^0.2 = 4/3

Raising both sides to the power of 5 (to eliminate the fractional exponent):

(t^0.2)^5 = (4/3)^5

t = (4/3)^5

Calculating t using a calculator:

t ≈ 2.49 seconds

So, the water mass is greatest at approximately t = 2.49 seconds.

To find the greatest mass, substitute the value of t = 2.49 into the original equation:

m = 5(2.49)^0.8 - 3(2.49) + 20

Calculating the value:

m ≈ 24.16 grams

Therefore, the greatest mass of water is approximately 24.16 grams.

To find the rate of mass change at t=3 s and t=5 s, we need to find the derivative at those points.

First, let's find the derivative:

dm/dt = 4t^-0.2 - 3

For t = 3 s:

dm/dt = 4(3)^-0.2 - 3

Calculating the value:

dm/dt ≈ -0.73 grams per second

So, at t = 3 seconds, the rate of mass change is approximately -0.73 grams per second.

For t = 5 s:

dm/dt = 4(5)^-0.2 - 3

Calculating the value:

dm/dt ≈ -0.49 grams per second

So, at t = 5 seconds, the rate of mass change is approximately -0.49 grams per second.

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

For t = 3 s:

Rate of mass change = (-0.73 grams per second) * (1 kilogram / 1000 grams) * (60 seconds / 1 minute)

Calculating the value:

Rate of mass change ≈ -0.0438 kilograms per minute

So, at t = 3 seconds, the rate of mass change is approximately -0.0438 kilograms per minute.

For t = 5 s:

Rate of mass change = (-0.49 grams per second) * (1 kilogram / 1000 grams) * (60 seconds / 1 minute)

Calculating the value:

Rate of mass change ≈ -0.0294 kilograms per minute

So, at t = 5 seconds, the rate of mass change is approximately -0.0294 kilograms per minute.