06 7ratesofchange

# Rates of change

.

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

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.

.

### Example 1

The pressure P (Pa), of a given gass kept at constant temperature, and its volume V (m3)
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 $V = 10$ m3?

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

… … … $PV = 500$

… … … $P=\dfrac{500}{V}$

… … … $P = 500V^{-1}$
.

… … … $\dfrac{dP}{dV}=-500V^{-2}$

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

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

.

b) .. What is the rate of change when $V = 10$ m3?

… … … When $V=10$

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

… … … $\dfrac{dP}{dV}=-5$ Pa/m3
.

… … … The rate of change of the pressure when $V = 10$ m3

… … … is $-5$ Pa/m3 … (a decreasing rate of 5 Pa/m3).

.

NOTE: The units of $\dfrac{dP}{dV}$ come from the units of P (Pa) divided by the units of V (m3)

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

.

### 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?

.

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

.

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

.

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
.

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

.

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

.

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
.

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

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

… … … $\dfrac{dQ}{dt}=-2000(20-t)$

… … … $-2000 (20 - t) = -20000$

… … … $20 - t = 10$

… … … $t = 10$ minutes

.

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
.

… … … The average rate of water flow in the first 5 minutes is $–35,000$ litres/min
… … … … (a decrease of $35,000$ litre per minute).

.