# Rates of change

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At the beginning of chapter five, we covered the theory related to **average rates of change** and **instantaneous rates of change**.

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Recall that:

**Average rate of change**between two points is equal to the gradient of a straight line between those points

**Instantaneous rate of change**at a point is equal to the gradient of the tangent at that point- This is therefore, the derivative of the function at that point.

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#### Example 1

The pressure **P** (Pa), of a given gass kept at constant temperature, and its volume **V** (m^{3})

are connected by the formula $PV = 500$ (joules).

… **a)** .. Find the rate of change of the pressure with respect to the volume.

… **b)** .. What is the rate of change when $V = 10$ m^{3}?

**Solution:**

… **a)** .. Find the rate of change of the pressure with respect to the volume.

… … … We want rate of change of pressure, so make **P** the subject

… … … $PV = 500$

… … … $P=\dfrac{500}{V}$

… … … $P = 500V^{-1}$

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… … … $\dfrac{dP}{dV}=-500V^{-2}$

… … … $\dfrac{dP}{dV}=-\dfrac{500}{V^2}$

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… … … The rate of change is negative since the pressure is decreasing as the volume increases.

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… **b)** .. What is the rate of change when $V = 10$ m^{3}?

… … … When $V=10$

… … … $\dfrac{dP}{dV}=-\dfrac{500}{10^2}$

… … … $\dfrac{dP}{dV}=-5$ Pa/m^{3}

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… … … The rate of change of the pressure when $V = 10$ m^{3}

… … … is $-5$ Pa/m^{3} … (a decreasing rate of 5 Pa/m^{3}).

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**NOTE:** The units of $\dfrac{dP}{dV}$ come from the units of **P** (Pa) divided by the units of **V** (m^{3})

An understanding of chemistry etc is not required to take the units given in the question and put them into your answer.

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#### Example 2

A water tank is being emptied and the quantity of water, **Q** litres, remaining in the tank **t** minutes after it starts to empty is given by

… … $Q(t)=1000(20-t)^2, \qquad t \geqslant 0$

… **a)** .. At what rate is the tank being emptied at any time **t**?

… **b)** .. How long does it take to empty the tank?

… **c)** .. At what time is the water flowing out at the rate of $20,000$ litres per minute?

… **d)** .. What is the average rate at which the water flows out in the first 5 minutes?

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**Note:** In modelling questions, time will always be greater than or equal to zero. We do not consider negative time.

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**Note:** If asked to graph this, only graph the first quadrant. Neither **t** nor **Q** can become negative so reflect that in your graph.

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**Solution:**

… **a)** .. At what rate is the tank being emptied at any time **t**?

… … … $Q(t)=1000(20-t)^2$

… … … $\dfrac{dQ}{dt}=-2000(20-t)$ litres/minute

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**Note:** The units for $\dfrac{dQ}{dt}$ come from the units of **Q** (litres) divided by the units for **t** (minutes).

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… **b)** .. How long does it take to empty the tank?

… … … Tank is empty when $Q = 0$

… … … $Q(t) = 0$

… … … $1000 (20 – t)^2 = 0$

… … … $t = 20$ minutes

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… **c)** .. At what time is the water flowing out at the rate of $20,000$ litres per minute?

… … … We want rate at a specific time, so **instantaneous rate of change**

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… … … Find **t** when $\dfrac{dQ}{dt}=-20000$

… … … the rate is negative since water is flowing out

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… … … $\dfrac{dQ}{dt}=-2000(20-t)$

… … … $-2000 (20 - t) = -20000$

… … … $20 - t = 10$

… … … $t = 10$ minutes

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… **d)** .. What is the average rate at which the water flows out in the first 5 minutes?

… … … Average Rate of Change $= \dfrac{ Q(t_2) - Q(t_1) }{ t_2 - t_1}$

… … … In the first five minutes so between $t = 0$ and $t = 5$

… … … **Average Rate of Change**

… … … …$=\dfrac{Q(5)-Q(0)}{5}$

… … … … $=\dfrac{1000(20-5)^2-1000(20-0)^2}{5}$

… … … … $=-35000$ litres per minute

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… … … The average rate of water flow in the first **5** minutes is $–35,000$ litres/min

… … … … (a decrease of $35,000$ litre per minute).

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