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Water Flow: Net Change and Accumulation

Created by @nathanedwards
 at November 3rd 2023, 11:47:27 am.
AP_Calculus_AB-Questions
#definite-integral
#net-change
#accumulation

Net Change and Accumulation

Question:

The rate at which water is flowing into a tank is modeled by the function f(t)=t2−4t+3f(t) = t^2 - 4t + 3, where tt is the time in minutes and f(t)f(t) is the rate of flow in gallons per minute.

  1. Determine the net change of water in the tank over the time interval 0≤t≤50 \leq t \leq 5.

  2. Find the accumulation of water in the tank after 5 minutes.

Answer:

  1. To find the net change of water in the tank over the time interval 0≤t≤50 \leq t \leq 5, we need to evaluate the definite integral of the rate function f(t)f(t) over that time period.

The net change of water in the tank can be found using the formula:

Net Change=∫05f(t)dt \text{Net Change} = \int_{0}^{5} f(t) dt

Let's evaluate the definite integral:

Net Change=∫05(t2−4t+3)dt=[13t3−2t2+3t]05=(13(5)3−2(5)2+3(5))−(13(0)3−2(0)2+3(0))=(13(125)−2(25)+15)−0=1253−50+15=1253−1503+453=−203 gallons \begin{align*} \text{Net Change} &= \int_{0}^{5} (t^2 - 4t + 3) dt \\ &= \left[\frac{1}{3}t^3 - 2t^2 + 3t\right]_{0}^{5} \\ &= \left(\frac{1}{3}(5)^3 - 2(5)^2 + 3(5)\right) - \left(\frac{1}{3}(0)^3 - 2(0)^2 + 3(0)\right) \\ &= \left(\frac{1}{3}(125) - 2(25) + 15\right) - 0 \\ &= \frac{125}{3} - 50 + 15 \\ &= \frac{125}{3} - \frac{150}{3} + \frac{45}{3} \\ &= -\frac{20}{3} \text{ gallons} \end{align*}

Therefore, the net change of water in the tank over the time interval 0≤t≤50 \leq t \leq 5 is −203-\frac{20}{3} gallons.

  1. The accumulation of water in the tank after 5 minutes can be found by evaluating the definite integral of the rate function f(t)f(t) from 0 to 5:
Accumulation=∫05f(t)dt \text{Accumulation} = \int_{0}^{5} f(t) dt

Let's calculate the accumulation:

Accumulation=∫05(t2−4t+3)dt=13t3−2t2+3t∣05=13(5)3−2(5)2+3(5)−(13(0)3−2(0)2+3(0))=1253−50+15=1253−1503+453=−203 gallons \begin{align*} \text{Accumulation} &= \int_{0}^{5} (t^2 - 4t + 3) dt \\ &= \frac{1}{3}t^3 - 2t^2 + 3t \bigg|_0^5 \\ &= \frac{1}{3}(5)^3 - 2(5)^2 + 3(5) - \left(\frac{1}{3}(0)^3 - 2(0)^2 + 3(0)\right) \\ &= \frac{125}{3} - 50 + 15 \\ &= \frac{125}{3} - \frac{150}{3} + \frac{45}{3} \\ &= -\frac{20}{3} \text{ gallons} \end{align*}

Therefore, the accumulation of water in the tank after 5 minutes is also −203-\frac{20}{3} gallons, which confirms the same value as the net change.

Note: The negative sign implies that the water is being drained from the tank, rather than being accumulated.

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