Water is draining from a large tank. After t min, there are 150000-7500t+t^2 litres of water in the tank.?
a) Determine the average rate at which the water empties from the tank in the interval between
5 and 10 minutes.
b) Determine the average rate at which the water empties from the tank in the interval between
9 and 10 min.
c) Estimate the rate at which the water runs out after exactly 10 min.
d) How much water is in the tank at exactly 10 min?
a) The average rate at which the water empties from the tank in the interval between 5 and 10 minutes can be found by taking the difference in the amounts of water at those two times and dividing it by the difference in time. So, the average rate is (150000 - 7500(10) + 10^2 - (150000 - 7500(5) + 5^2)) / (10 - 5).
b) The average rate at which the water empties from the tank in the interval between 9 and 10 minutes can be found in a similar way. We take the difference in the amounts of water at the two times and divide it by the difference in time. So, the average rate is (150000 - 7500(10) + 10^2 - (150000 - 7500(9) + 9^2)) / (10 - 9).
c) To estimate the rate at which the water runs out after exactly 10 minutes, we can calculate the rate of change at 10 minutes by finding the derivative of the function that represents the amount of water in the tank at time t. This will give us the instantaneous rate at 10 minutes.
d) To find the amount of water in the tank at exactly 10 minutes, we can substitute t=10 into the function that represents the amount of water in the tank at time t. This will give us the value of the function at t=10.
To answer these questions, we need to find the derivative of the function that describes the amount of water in the tank at any given time. The function is given as:
V(t) = 150000 - 7500t + t^2
a) To find the average rate at which the water empties from the tank in the interval between 5 and 10 minutes, we need to find the average rate of change of the function V(t) within that interval.
Average Rate = (V(10) - V(5)) / (10 - 5)
= (150000 - 7500(10) + 10^2) - (150000 - 7500(5) + 5^2) / 5
= (150000 - 75000 + 100) - (150000 - 37500 + 25) / 5
= (75000 + 100) - (112500 - 37500 + 25) / 5
= 75100 - (112525 - 37525) / 5
= 75100 - 75000 / 5
= 75100 - 15000
= 60100 / 5
= 12020
Therefore, the average rate at which the water empties from the tank in the interval between 5 and 10 minutes is 12020 liters per minute.
b) To find the average rate at which the water empties from the tank in the interval between 9 and 10 minutes, we can use the same approach.
Average Rate = (V(10) - V(9)) / (10 - 9)
= (150000 - 7500(10) + 10^2) - (150000 - 7500(9) + 9^2) / 1
= (150000 - 75000 + 100) - (150000 - 67500 + 81) / 1
= 75100 - (150000 - 67500 + 81)
= 75100 - 82481
= -7381
Therefore, the average rate at which the water empties from the tank in the interval between 9 and 10 minutes is -7381 liters per minute.
c) To estimate the rate at which the water runs out after exactly 10 minutes, we need to find the value of the derivative of V(t) at t = 10.
V'(t) = -7500 + 2t
Rate at t = 10 = V'(10) = -7500 + 2(10)
= -7500 + 20
= -7480
Therefore, the rate at which the water runs out after exactly 10 minutes is estimated to be -7480 liters per minute.
d) To find the amount of water in the tank at exactly 10 minutes, we can substitute t = 10 into the function V(t).
V(10) = 150000 - 7500(10) + 10^2
= 150000 - 75000 + 100
= 75100
Therefore, the amount of water in the tank at exactly 10 minutes is 75100 liters.
To solve these questions, we will differentiate the given function, which represents the amount of water in the tank at any given time.
The given function is: V(t) = 150000 - 7500t + t^2
a) To determine the average rate at which the water empties from the tank in the interval between 5 and 10 minutes, we need to calculate the derivative of the function V(t) with respect to time (t).
Find the derivative of V(t):
V'(t) = dV(t)/dt = -7500 + 2t
Evaluate the derivative at t = 5 and t = 10 to find the average rate of change:
V'(5) = -7500 + 2(5) = -7500 + 10 = -7490 (litres per minute)
V'(10) = -7500 + 2(10) = -7500 + 20 = -7480 (litres per minute)
Therefore, the average rate at which the water empties from the tank in the interval between 5 and 10 minutes is approximately -7490 litres per minute.
b) To determine the average rate at which the water empties from the tank in the interval between 9 and 10 minutes, we will again use the derivative of V(t).
Evaluate the derivative at t = 9 and t = 10 to find the average rate of change:
V'(9) = -7500 + 2(9) = -7500 + 18 = -7482 (litres per minute)
V'(10) = -7500 + 2(10) = -7500 + 20 = -7480 (litres per minute)
Therefore, the average rate at which the water empties from the tank in the interval between 9 and 10 minutes is approximately -7482 litres per minute.
c) To estimate the rate at which the water runs out after exactly 10 minutes, we need to examine the derivative of V(t) at t = 10.
Evaluate the derivative at t = 10 to find the instantaneous rate of change:
V'(10) = -7500 + 2(10) = -7500 + 20 = -7480 (litres per minute)
Therefore, the rate at which the water runs out after exactly 10 minutes is estimated to be approximately -7480 litres per minute.
d) To find the amount of water in the tank exactly at 10 minutes, we substitute t = 10 into the given function V(t):
V(10) = 150000 - 7500(10) + (10)^2
V(10) = 150000 - 75000 + 100
V(10) = 150000 - 75000 + 100
V(10) = 75,100 (litres)
Therefore, there are approximately 75,100 litres of water in the tank at exactly 10 minutes.
(a) that would be the slope of the line between the two points on the graph of f(t) as given:
(f(10)-f(5))/(10-5)
plug in your t values
(b) same thing, different interval
(c) pick a small interval centered at t=10, such as [9.9,10.1]
(d) naturally, that will be f(10)