As water drains out of a 2000L hot tub, the amount of water remaining in the tub can be modelled by the function V=0.00002(100-t)^4 , where t is the time, in minutes (t is greater than or equal to 0 and less than or equal to 100), and V is the volume of water, in litres, remaining in the tub at time t.
Determine the average rate of cahnge of the volume of water during
- the entire 100 min
- the first 30 min
- the last 30 min
I'm really confused as to what I actually have to do. I'd really appreciate some help.
I believe what you are asked to do is something like this.
For the 100 min period--
V=0.00002(100-100)^4 = 0 L remaining.
So all 2000 L have drained; therefore, the rate of change is 2000L/100 min = 20L/min.
For the first 30 min--
V=0.00002(100-30)^4 = 480.2 L remaining or 2000-480.2 =1519.8 L drained.
rate of change is 1519.8/30 = 50.7 L/min.
Etc. Check my thinking. Check my work.
oh okay, thank you.
Oh, don't worry, I'm here to help! Let's break it down step by step.
To find the average rate of change of the volume of water, we need to calculate the change in volume over a given period of time, and then divide it by the length of that period.
First, let's find the change in volume during the entire 100-minute period. We can do this by subtracting the volume at the beginning from the volume at the end:
V(end) - V(start) = 0.00002(100-100)^4 - 0.00002(100-0)^4
Since the tub is completely drained at the end (t=100), we can simplify this to:
0 - 0.00002(100)^4 = -0.00002(100)^4
So, the change in volume during the 100-minute period is -0.00002(100)^4 liters.
Next, let's calculate the average rate of change of the volume during the first 30 minutes. We can do this with the same method:
V(30 min) - V(0 min) = 0.00002(100-30)^4 - 0.00002(100-0)^4
Simplifying this gives us:
0.00002(70)^4 - 0.00002(100)^4
So, the change in volume during the first 30 minutes is 0.00002(70)^4 liters.
Finally, to find the average rate of change during the last 30 minutes, we subtract the volume at 70 minutes from the volume at 100 minutes:
V(100 min) - V(70 min) = 0 - 0.00002(100-70)^4
Simplifying this gives us:
-0.00002(30)^4
So, the change in volume during the last 30 minutes is -0.00002(30)^4 liters.
Remember, if you want to find the average rate of change, you need to divide these values by the length of the corresponding time period. I hope this clears things up for you!
To determine the average rate of change of the volume of water during the entire 100 minutes, first, we need to find the initial volume of water and the final volume of water.
- The initial volume of water is when t = 0:
V(0) = 0.00002(100-0)^4 = 0.00002(100)^4 = 0.00002(10000) = 200 liters.
- The final volume of water is when t = 100:
V(100) = 0.00002(100-100)^4 = 0.00002(0)^4 = 0 liters.
The average rate of change of the volume of water during the entire 100 minutes is calculated by subtracting the initial volume from the final volume and dividing it by the change in time:
Average rate of change = (Final volume - Initial volume) / (Final time - Initial time)
= (0 - 200) / (100 - 0)
= -200 / 100
= -2 liters/minute.
Therefore, the average rate of change of the volume of water during the entire 100 minutes is -2 liters/minute.
To determine the average rate of change of the volume of water during the first 30 minutes, we need to find the initial volume of water and the volume of water at 30 minutes.
- The initial volume of water remains the same, which is 200 liters.
- To find the volume of water at 30 minutes:
V(30) = 0.00002(100-30)^4 = 0.00002(70)^4 ≈ 34.56 liters.
The average rate of change of the volume of water during the first 30 minutes is calculated by subtracting the initial volume from the volume at 30 minutes and dividing it by the change in time:
Average rate of change = (Volume at 30 minutes - Initial volume) / (30 - 0)
= (34.56 - 200) / 30
= -165.44 / 30
≈ -5.51 liters/minute.
Therefore, the average rate of change of the volume of water during the first 30 minutes is approximately -5.51 liters/minute.
To determine the average rate of change of the volume of water during the last 30 minutes, we need to find the volume of water at 70 minutes and the final volume of water.
- To find the volume of water at 70 minutes:
V(70) = 0.00002(100-70)^4 = 0.00002(30)^4 ≈ 2.16 liters.
- The final volume of water remains the same, which is 0 liters.
The average rate of change of the volume of water during the last 30 minutes is calculated by subtracting the volume at 70 minutes from the final volume and dividing it by the change in time:
Average rate of change = (Final volume - Volume at 70 minutes) / (100 - 70)
= (0 - 2.16) / 30
≈ -0.072 liters/minute.
Therefore, the average rate of change of the volume of water during the last 30 minutes is approximately -0.072 liters/minute.
To determine the average rate of change of the volume of water, you need to find the slope of the function over the given time intervals.
First, let's determine the average rate of change of the volume of water during the entire 100 minutes:
To find the average rate of change of the function over an interval, you need to find the difference in the function values divided by the difference in the time values.
Average rate of change over 100 minutes:
V(t) = 0.00002(100 - t)^4
We need to find the change in volume (ΔV) over the interval from t = 0 to t = 100:
ΔV = V(100) - V(0)
= [0.00002(100 - 100)^4] - [0.00002(100 - 0)^4]
= 0 - 0
= 0
Since the change in volume is 0 over the entire time interval, the average rate of change is also 0.
Next, let's determine the average rate of change of the volume of water during the first 30 minutes:
We need to find the change in volume (ΔV) over the interval from t = 0 to t = 30:
ΔV = V(30) - V(0)
= [0.00002(100 - 30)^4] - [0.00002(100 - 0)^4]
= 0.00002(70^4) - 0.00002(100^4)
≈ 55.1392 - 160
≈ -104.8608
The change in volume over the first 30 minutes is approximately -104.8608 liters.
Finally, let's determine the average rate of change of the volume of water during the last 30 minutes:
We need to find the change in volume (ΔV) over the interval from t = 70 to t = 100:
ΔV = V(100) - V(70)
= [0.00002(100 - 100)^4] - [0.00002(100 - 70)^4]
= 0 - 0.00002(30^4)
≈ 0 - 0.00002(81000)
≈ 0 - 1.62
≈ -1.62
The change in volume over the last 30 minutes is approximately -1.62 liters.
Therefore, the average rate of change of the volume of water during:
- the entire 100 minutes is 0 liters per minute.
- the first 30 minutes is approximately -104.8608 liters per minute.
- the last 30 minutes is approximately -1.62 liters per minute.