I am re-posting this question because i wrote the incorrect function last time.
An oil tank is being drained for cleaning. After t minutes there are v litres of oil left in the tank, where v(t)=35(25-t)^2, 0<=t<=25
a) determine the average rate of change of volume during the first 15 minutes
I got -1225 as my answer, is that correct?
b) determine the rate of change of volume at the time t=15 minutes
I'm not sure how to answer this.
v(0) = 35(25)^2 = 21875
v(15) = 35(10)^2 = 3500
avg rate of change for those 15 minutes
= (2500 - 21875)/(15-0)
= -18375/15 L/min
= -1225 L/min
You are correct
at t=15, you are looking for the instantaneous rate of change, that is,
the derivative of v(t)
v ' (t) = 70(25-t) (-1)
v ' (15) = -70(25-15) = -700 L/min
Hmm, let me try to help you out while also bringing some humor into the mix!
a) To find the average rate of change of volume during the first 15 minutes, we need to calculate the change in volume divided by the change in time. So, let's go ahead and do that calculation.
Using the given function v(t) = 35(25 - t)^2, we can plug in the values for t = 15 minutes and t = 0 minutes to find the respective volumes. So, v(15) would be 35(25 - 15)^2, and v(0) would be 35(25 - 0)^2.
Now, we can calculate the change in volume by subtracting v(0) from v(15). And, of course, we need to divide this change in volume by the change in time, which is 15 minutes.
Okay, are you ready for the answer? Drumroll, please...
The average rate of change of volume during the first 15 minutes is -1225 litres per minute.
Wait, what? That's the same answer you got, so you're correct! But I must say, it's no surprise that the tank is in a hurry to get rid of all that oil. It's draining at quite a speedy rate, isn't it?
b) Now, let's move on to finding the rate of change of volume at the time t = 15 minutes. To do this, we need to find the derivative of the volume function v(t).
The derivative of v(t) = 35(25 - t)^2 with respect to t is -70(25 - t). So, to find the rate of change of volume at t = 15 minutes, we simply need to plug in t = 15 minutes into the derivative.
Wait for it... wait for it...
The rate of change of volume at t = 15 minutes is 0 litres per minute.
Yep, you heard that right! It's zero, zilch, nada. The tank has hit a standstill at that moment. It's a shame, but at least you won't have to worry about any sudden oil spills at t = 15 minutes!
Keep up the great work, and remember that even though oil tanks may not have a sense of humor, we're here to add a little fun to your math problems!
a) To find the average rate of change of volume during the first 15 minutes, we need to calculate the change in volume and divide it by the change in time.
Given the volume function v(t) = 35(25 - t)^2, we can find the change in volume from t = 0 to t = 15 by subtracting the initial volume from the volume at t = 15.
v(0) = 35(25 - 0)^2 = 35(25^2) = 35(625) = 21875
v(15) = 35(25 - 15)^2 = 35(10^2) = 35(100) = 3500
Change in volume = v(15) - v(0) = 3500 - 21875 = -18375
The average rate of change of volume during the first 15 minutes is then:
Average rate of change = Change in volume / Change in time = -18375 / 15 = -1225
So, your answer of -1225 for part (a) is correct.
b) To determine the rate of change of volume at the time t = 15 minutes, we can find the derivative of the volume function with respect to time.
v(t) = 35(25 - t)^2
Differentiating v(t) with respect to t:
v'(t) = 35 * 2(25 - t)(-1) * (-1) = 70(25 - t)
To find v'(15), substitute t = 15 into the derivative:
v'(15) = 70(25 - 15) = 700
The rate of change of volume at t = 15 minutes is 700.
Therefore, the rate of change of volume at the time t = 15 minutes is 700.
To determine the average rate of change of volume during the first 15 minutes, we need to find the change in volume over that time period and divide it by the change in time.
a) Average rate of change of volume during the first 15 minutes:
To find the change in volume, we need to evaluate v(15) - v(0), where v(15) represents the volume at 15 minutes and v(0) represents the volume at 0 minutes.
v(15) = 35(25 - 15)^2 = 35(10)^2 = 3500
v(0) = 35(25 - 0)^2 = 35(25)^2 = 35(625) = 21875
Change in volume = v(15) - v(0) = 3500 - 21875 = -18375
Next, we need to find the change in time, which is 15 - 0 = 15.
Average rate of change of volume = (Change in volume) / (Change in time) = -18375 / 15 = -1225
So, yes, your answer of -1225 is correct for the average rate of change of volume during the first 15 minutes.
b) To determine the rate of change of volume at the time t = 15 minutes, we need to take the derivative of the volume function v(t) and evaluate it at t = 15.
The derivative of v(t) with respect to t can be found using the chain rule:
v'(t) = 2 * 35(25 - t) * (-1)
Now, substitute t = 15 into the derivative function:
v'(15) = 2 * 35(25 - 15) * (-1) = 2 * 35(10) * (-1) = -700
The rate of change of volume at t = 15 minutes is -700 liters per minute.