This chapter has been dedicated to formalising mathematical methods for determining the derivative function. We have generally described this as the gradient function, but another way to think about this is as the rate of change of a function. When we substitute an $x$x value into the derivative function, this gives us the instantaneous rate of change at that point.
We will now consider various practical applications of these concepts. The rate of change of any quantity or calculation can be found by differentiation. Often, but not always, this will be in terms of $t$t. This gives us a rate of change with respect to time. For instance, if a function $V$V represents the volume of an expanding balloon then $\frac{dV}{dt}$dVdtis the rate of change of that volume over time. Substituting in a value of $t$t will give us an instantaneous rate of change of the volume.
Because we are now dealing with a practical context, we will have units. If the volume of the balloon is given in $cm^3$cm3 and the time measured in seconds, the rate of change will be in $cm^3/s$cm3/s. As you can see, the units for a rate of change will the dependent variable units "per" (or, over, as in a fraction) the independent variable units.
Practice questions
Question 1
The electrical resistance, $R$R, of a component at temperature, $t$t, is given by $R=9+\frac{t}{17}+\frac{t^2}{108}$R=9+t17+t2108.
Find $\frac{dR}{dt}$dRdt, the instantaneous rate of increase of resistance with respect to temperature.
Question 2
The effectiveness of a supplement is given by $\frac{dE}{dg}$dEdg, where $E$E is the body’s reaction to the supplement and $g$g is the number of grams of the supplement administered to the subject.
For a particular supplement, $E=\frac{g^2\left(400-g\right)}{2}$E=g2(400−g)2.
Determine the equation for the effectiveness of this particular supplement $\frac{dE}{dg}$dEdg.
Determine the effectiveness of the drug when $g=30$g=30.
After how many units of the supplement does the effectiveness of the drug start to decrease?
Give your answer to one decimal place.
Motion and differentiation
We will now look more closely at particular application of rates of change in the field of kinematics (motion) of a particle. When we talk about a particle (or sometimes called an object or body), we can be referring to anything undergoing motion, including large bodies like cars, trucks and trains. We will limit our study to motion in a straight line (rectilinear motion) and ignore friction, gravity and other forces that may influence the motion.
We can describe the motion of an object using distance, speed and acceleration. However, we will consider the displacement, velocity and acceleration of the object with respect to time. The only difference between these quantities is that displacement and velocity are defined by a magnitude (size) and direction whereas distance and speed only have magnitude. For example, if a body moves $4$4units left right, the distance moved is still $4$4 units irrespective of direction. Similarly, if the speed of a car is $50$50 km/hr, this is also independent of direction.
Displacement
Displacement $x$x is the distance of a body from a fixed point (origin) within a certain frame of reference. Normally, a positive displacement means the body has moved right of the origin and a negative displacement means that the body has moved left of the origin. A displacement of $0$0 means the body is at the origin. The initial displacement is calculated when $t=0$t=0. Displacement is usually measured using a unit of length such as metres (m) or kilometres (km).
Let's consider the following displacement vs time graph:
 |
From this graph we can see that the displacement is $0$0 at a time of $0$0 and $200$200 seconds. We can also see that the particle is not moving between points $B$B and $C$C, and $D$D and $E$E as the displacement remains constant.
When we consider displacement, we are finding the position of that object with respect to a certain point (usually the origin). If there has been a direction change during the motion for a given time period, displacement and total distance traveled will be different. For the example above, notice that after $200$200 seconds the object has travelled a distance of $300$300 metres but its displacement at this time is $0$0.
Velocity
The instantaneous velocity $v$v is the rate of change of displacement at a given point in time. Velocity is considered positive or negative based on the direction of travel. A positive velocity means the body is moving to the right (usually) and a negative displacement means that the body is moving left. A velocity of $0$0 means the body is at rest. The initial velocity is calculated when $t=0$t=0. As velocity is the rate of change of displacement, it is commonly measured in metres per second (m/s or ms-1) or kilometres per hour (km/hr).
We can find the velocity function $v(t)$v(t) by differentiating the displacement function $x(t)$x(t). We could also consider the gradient of the tangent at each point on the displacement function.
The gradient of the displacement vs time curve in our previous example above tells us:
| Point | Velocity |
|---|---|
| $A-B$A−B | Constant positive velocity |
| $B-C$B−C | Zero velocity, the body is at rest |
| $C-D$C−D | Positive increasing velocity |
| $D-E$D−E | Zero velocity, the body is at rest |
| $E-F$E−F | Negative velocity, that is, the body is moving back towards the origin |
Note that between $t=0$t=0 and $t=200$t=200 secs the average speed is given by:
| Average Speed | $=$= | $\frac{\text{Distance}}{\text{Time}}$DistanceTime |
| $=$= | $\frac{300}{200}$300200 | |
| $=$= | $1.5$1.5 ms-1 |
Acceleration
The instantaneous acceleration $a$a is the rate of change of velocity at a given point in time. This corresponds to the gradient of the tangent at each point on the velocity curve. A positive acceleration is usually considered acting to the right and a negative acceleration acting to the left. However, it is necessary to consider acceleration in relation to velocity. If acceleration and velocity are in the same direction the body is accelerating (speeding up). If acceleration and velocity are in the opposite direction the body will be decelerating (slowing down). The initial acceleration is calculated when $t=0$t=0. Acceleration is the rate of change of velocity and is commonly measured in metres per second per second (ms-2).
An acceleration of $0$0 means the body has a constant velocity. We can find the acceleration function $a(t)$a(t) by differentiating the velocity function $v(t)$v(t). We could also consider the gradient of the tangent at each point on the velocity function.
The diagram below shows the velocity vs time of an object. The gradient at each point of the velocity vs time curve gives us acceleration. We can see that there are three distinct sections to the graph as follows:
- Increasing velocity, positive gradient which means positive acceleration
- Constant velocity, gradient is $0$0 which means acceleration is $0$0
- Decreasing velocity, negative acceleration
The relationships between displacement, velocity, acceleration and their derivatives can be summarised as follows:
For a displacement as a function of time $x(t)$x(t) the instantaneous velocity as a function of time $v(t)$v(t) is the derivative of the displacement function with respect to time $t$t:
$v(t)=x'(t)=\frac{dx}{dt}$v(t)=x′(t)=dxdt
The instantaneous acceleration as a function of time $a(t)$a(t) is the derivative of the velocity function with respect to time $t$t:
$a(t)=v'(t)=\frac{dv}{dt}$a(t)=v′(t)=dvdt
The worked example below shows us how we will use these formulae and our understanding of the derivative in the context of the motion of a body.
Worked example
Example 1
The displacement of a particle moving in a straight line is given by $x(t)=2t^4-4t^2+4$x(t)=2t4−4t2+4, where$x$x is the displacement in metres from the origin and t is the time in seconds.
a) Calculate the initial displacement of the particle.
When $t=0$t=0,
| $x(t)$x(t) | $=$= | $2(0)^4-4(0)^2+4$2(0)4−4(0)2+4 |
| $=$= | $4$4 |
b) Determine the function $v(t)$v(t) for the velocity of the particle
| $x(t)$x(t) | $=$= | $2t^4-4t^2+3t+4$2t4−4t2+3t+4 |
| $\frac{dx}{dt}$dxdt | $=$= | $8t^3-8t+3$8t3−8t+3 |
| $v(t)$v(t) | $=$= | $8t^3-8t+3$8t3−8t+3 |
c) Calculate the initial velocity of the particle
When $t=0$t=0
| $v(t)$v(t) | $=$= | $8t^3-8t+3$8t3−8t+3 |
| $v(0)$v(0) | $=$= | $8(0)^3-8(0)+3$8(0)3−8(0)+3 |
| $v(0)$v(0) | $=$= | $3$3 |
Practice questions
Question 3
The position (in metres) of an object along a straight line after $t$t seconds is modelled by $x\left(t\right)=3t^3-4t^2$x(t)=3t3−4t2.
State the velocity $v\left(t\right)$v(t) of the particle at time $t$t.
Which of the following represents the velocity of the particle after $4$4 seconds?
$v\left(4\right)$v(4)
A$v'\left(4\right)$v′(4)
B$x'\left(4\right)$x′(4)
C$x\left(4\right)$x(4)
DHence find the velocity of the particle after $2$2 seconds.
Question 4
The velocity, in m/s, of a body moving in rectilinear motion is given by$v\left(t\right)=t^2-11t+24$v(t)=t2−11t+24 where $t$t is time in seconds.
Solve for the time(s), $t$t, when the body is instantaneously at rest.
If the acceleration of a particle is the rate of change of the velocity, determine the function $a\left(t\right)$a(t) for the acceleration of the body.
Calculate the acceleration at time $t=2$t=2.