This lesson centers around solving Word Problems pertaining to Average Rate of Change and Instanteneous Rate of Change. Here are the formulas for the respective Rates of Change:
For additional information and practice with these formulas, review Average Rate of Change and Instantaneous Rate of Change!!
It's important to identify what Rate of Change you're calculating to determine which formula to use. Additionally, we will set \(h = 0.01\) when solving questions involving Instantaneous Rate of Change.
When calculating Rates of Change, it's good to use variables to represent different quantities. Also, remember to include units in your final answer!
Water is draining from a large tank. After \( t \) min, there are \(V(t) = 150000 - 7500t + t^2 \) liters of water in the tank.
i. First, we can evaluate the amount of water in the tank at \(10\) minutes by substituting \(10\) for \(t\) in the original function:
\(V(10) = 1500000 - 7500(10) + (10)^2\)
\(V(10) = 1500000 - 75000 + 100\)
\(\textcolor{red}{V(10) = 75100}\)
First, we can evaluate the amount of water in the tank at \(5\) minutes by substituting \(5\) for \(t\) in the original function:
\(V(5) = 1500000 - 7500(5) + (5)^2\)
\(V(5) = 1500000 - 37500 + 25\)
\(\textcolor{blue}{V(5) = 112525}\)
Finally, we can substitute all pertinent values into the Average Rate of Change formula to determine the average rate the water empties from the tank within this interval:
Therefore, we can determine the water empties at an average rate of \(\boldsymbol{-7485 \left[\cfrac{\text{L}}{\text{minute}}\right]}\) between \(5\) and \(10\) minutes.
ii. First, we can determine from the first step that the tank contains \(75100\;[\text{L}]\) of water at \(10\) minutes.
Next, we can evaluate the amount of water in the tank at \(9\) minutes by substituting \(9\) for \(t\) in the original function:
\(V(9) = 1500000 - 7500(9) + (9)^2\)
\(V(9) = 1500000 - 67500 + 81\)
\(\textcolor{blue}{V(10) = 82581}\)
Finally, we can substitute all pertinent values into the Average Rate of Change formula to determine the average rate the water empties from the tank within this interval:
Therefore, we can determine the water empties at an average rate of \(\boldsymbol{-7481 \left[\cfrac{\text{L}}{\text{minute}}\right]}\) between \(9\) and \(10\) minutes.
iii. First, we can determine from the first step that the tank contains \(75100\;[\text{L}]\) of water at \(10\) minutes.
Next, we can evaluate \(f(a+h)\) by substituting \(10.01\) for \(t\):
\(V(10.01) = 1500000 - 7500(10.01) + (10.01)^2\)
\(V(10.01) = 1500000 - 75075 + 100.2001\)
\(\textcolor{red}{V(10.01) = 75025.2001}\)
Finally, we can substitute all pertinent values into the Instantaneous Rate of Change formula to determine its slope at these points:
Therefore, we can determine the water empties at a rate of \(\boldsymbol{-7479.99 \left[\cfrac{\text{L}}{\text{minute}}\right]}\) at \(10\) minutes.
The population, \(P\), of a small town is modelled by the function, \(P(t) = -2t^3 + 55t^2 + 15t + 22000\), where \(t=0\) represents the beginning of this year.
i. We can determine the initial population by setting the time, \(t = 0\):
\(P(0) = -2(0)^3 + 55(0)^2 + 15(0) + 22000\)
\(P(0) = -2(0) + 55(0) + 22000\)
\(P(0) = 22000\)
Therefore, we can determine that the initial population is \(\boldsymbol{22000}\).
ii. We can determine the population at the end of \(10\) years by setting the time, \(t = 10\):
\(P(10) = -2(10)^3 + 55(10)^2 + 15(10) + 22000\)
\(P(10) = -2(1000) + 55(100) + 150 + 22000\)
\(P(10) = -2000 + 5500 + 22150\)
\(P(10) = 25650\)
Therefore, we can determine that the population at the end of \(10\) years is \(\boldsymbol{25650}\).
iii. First, we can determine from the previous steps that the populations at \(0\) and \(10\) years are \(22000\) and \(25650\) respectively.
Next, we can substitute all pertinent values into the Average Rate of Change formula to determine the average rate the water empties from the tank within this interval:
Therefore, we can determine that the population increases at an average rate of \(\boldsymbol{365 \left[\cfrac{\text{people}}{\text{year}}\right]}\) between \(0\) and \(10\) years.
iv. First, we can determine from the previous step that the town has a population of \(25650\) people after \(10\) years.
Next, we can evaluate \(f(a+h)\) by substituting \(10.01\) for \(t\):
\(P(10.01) = -2(10.01)^3 + 55(10.01)^2 + 15(10.01) + 22000\)
\(P(10.01) = -2(1003.003001) + 55(100.2001) + 150.15 + 22000\)
\(P(10.01) = -2006.006002 + 5511.0055 + 22150.15\)
Finally, we can substitute all pertinent values into the Instantaneous Rate of Change formula to determine its slope at these points:
Therefore, we can determine that the population grows at a rate of \(\boldsymbol{514.9498\; \left[\cfrac{\text{people}}{\text{year}}\right]}\) by the end of the \(10\)th year.
The height of a rocket can be modelled by the function \(d(t) = -10t^2 + 55t\) where \(d\) represents the height in meters and \(t\) represents the time in seconds.
i. First, we can determine the height of the rocket at \(5\;[\text{s}]\):
\(d(5) = -10(5)^2 + 55(5)\)
\(d(5) = -10(25) + 275\)
\(d(5) = -250 + 275\)
\(\textcolor{red}{d(5) = 25}\)
Next, we can determine the height of the rocket at \(0\;[\text{s}]\):
\(d(0) = -10(0)^2 + 55(0)\)
\(\textcolor{blue}{d(0) = 0}\)
Finally, we can substitute all pertinent values into the Average Rate of Change formula to determine the average rate the water empties from the tank within this interval:
Therefore, we can determine that the average velocity of the rocket on the interval \([0, 5]\) is \(\boldsymbol{5\;\left[\cfrac{\text{m}}{\text{s}}\right]}\).
ii. First, we can evaluate \(d(a+h)\) by substituting \(4.01\) for \(t\):
\(d(4.01) = -10(4.01)^2 + 55(4.01)\)
\(d(4.01) = -10(16.0801) + 220.55\)
\(d(4.01) = -160.801 + 220.55\)
\(\textcolor{red}{(4.01) = 59.749}\)
Next, we can evaluate \(d(a)\) by substituting \(4\) for \(t\):
\(d(4) = -10(4)^2 + 55(4)\)
\(d(4) = -10(16) + 220\)
\(d(4) = -160 + 220\)
\(\textcolor{blue}{(4) = 60}\)
Finally, we can substitute all pertinent values into the Instantaneous Rate of Change formula to determine its slope at these points:
Therefore, we can determine that the instantaneous velocity of the rocket at \(t = 4\) is \(\boldsymbol{-25.1\;\left[\cfrac{\text{m}}{\text{s}}\right]}\).