Unit outcomes

By the end of this unit you will be able to:

• Solve practical problems involving rates of change.

What you should know

Before you start this unit, make sure you can:

• Find the derivative by using the rules of differentiation. To revise the rules of differentiation refer to level 3 subject outcome 2.5 and unit 2 of this subject outcome.

Introduction

What makes calculus so useful is the application of the derivative to determine how fast (the rate) things are changing in relation to each other. We have seen that the derivative may be thought of as the instantaneous rate of change of a function.

From graphs we can see the rate at which the function values change as the independent (input) variable changes. This rate of change is described by the gradient of the graph, and is found by calculating the derivative.

We have learnt how to find the average gradient of a curve and how to determine the gradient of a curve at a given point. These concepts are also referred to as the average rate of change and the instantaneous rate of change.

[latex]\scriptsize \begin{align*}\text{Average rate of change}&=\displaystyle \frac{{f(x+h)-f(x)}}{h}\\\text{Instantaneous rate of change}&=\underset{{h\to 0}}{\mathop{{\lim }}}\,\displaystyle \frac{{f(x+h)-f(x)}}{h}\end{align*}[/latex]

When we mention rate of change, the instantaneous rate of change (the derivative) is implied. When average rate of change is required, it will be specifically referred to as the average rate of change.

Applications of rate of change

Velocity is the rate of change of distance over a corresponding change in time. If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. Velocity is one of the most common forms of rate of change.

If the function [latex]\scriptsize s(t)[/latex] gives the position of an object at time [latex]\scriptsize t[/latex] then:

1. The velocity of the object at time [latex]\scriptsize t[/latex] is given by [latex]\scriptsize v(t)=\displaystyle \frac{{ds}}{{dt}}={s}'(t)[/latex].
2. The acceleration of the object at [latex]\scriptsize t[/latex] is given by [latex]\scriptsize a(t)={v}'(t)={{s}'}'(t)[/latex].

Example 4.1

A ball is hit up into the air. Its height (in metres) [latex]\scriptsize t[/latex] seconds later is given by [latex]\scriptsize s(t)=-5{{t}^{2}}+20t[/latex]. Determine:

1. The average velocity of the ball during the first two seconds.
2. The velocity of the ball after [latex]\scriptsize 1.5\text{ }[/latex]seconds.
3. The time at which the velocity is zero.
4. The velocity with which the ball hits the ground.
5. The acceleration of the ball.

Solutions

1. .
   [latex]\scriptsize \begin{align*}{{v}_{{\text{ave}}}}&=\displaystyle \frac{{s(2)-s(0)}}{{2-0}}\\&=\displaystyle \frac{{[-5{{{(2)}}^{2}}+20(2)]-[-5{{{(0)}}^{2}}+20(0)]}}{2}\\&=\displaystyle \frac{{20}}{2}\\&=10\text{ m}\text{.}{{\text{s}}^{{-1}}}\end{align*}[/latex]
2. Calculate the instantaneous velocity.
   [latex]\scriptsize \begin{align*}v(t)&={s}'(t)\\&=-10t+20\\{s}'(1.5)&=-10(1.5)+20\\&=5\text{ m}\text{.}{{\text{s}}^{{-1}}}\end{align*}[/latex]
3. .
   [latex]\scriptsize \begin{align*}v(t)&=0\\-10t+20&=0\\t&=2\text{ }\end{align*}[/latex]
   Therefore, the velocity is zero after [latex]\scriptsize 2\text{ }[/latex] seconds.
4. The ball hits the ground when [latex]\scriptsize s(t)=0[/latex].
   [latex]\scriptsize \begin{align*}s(t)&=0\\-5{{t}^{2}}+20t&=0\\-5t(t-4)&=0\\t&=0\text{ or }t=4\end{align*}[/latex]
   The balls hits the ground after [latex]\scriptsize 4\text{ }[/latex] seconds. The velocity after [latex]\scriptsize 4\text{ }[/latex] seconds will be:
   .
   [latex]\scriptsize \begin{align*}v(4)&={s}'(4)\\&=-10(4)+20\\&=-20\text{ m}\text{.}{{\text{s}}^{{-1}}}\end{align*}[/latex]
   The ball hits the ground at a speed of [latex]\scriptsize -20\text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]. The sign of the velocity is negative, which means that the ball is moving downward (a positive velocity is used for upwards motion).
5. Acceleration is the derivative of velocity.
   [latex]\scriptsize \begin{align*}a(t)&={v}'(t)={{s}'}'(t)\\&=-10\text{ m}\text{.}{{\text{s}}^{{-2}}}\end{align*}[/latex]

Example 4.2

A particle moves along a coordinate axis in the positive direction to the right. Its position at time [latex]\scriptsize t[/latex] is given by [latex]\scriptsize s(t)={{t}^{3}}-4t+2[/latex]. Find [latex]\scriptsize v(1)[/latex] and [latex]\scriptsize a(1)[/latex], and use these values to answer the following questions:

1. Is the particle moving from left to right or from right to left at time [latex]\scriptsize t=1[/latex]?
2. Is the particle speeding up or slowing down at time [latex]\scriptsize t=1[/latex]?

Solutions

Begin by finding [latex]\scriptsize v(t)[/latex] and [latex]\scriptsize a(t)[/latex].

[latex]\scriptsize \begin{align*}v(t)&={s}'(t)\\&=3{{t}^{2}}-4\end{align*}[/latex]

[latex]\scriptsize \begin{align*}a(t)&={v}'(t)={{s}'}'(t)\\&=6t\end{align*}[/latex]

Calculate [latex]\scriptsize v(1)[/latex] and [latex]\scriptsize a(1)[/latex].

[latex]\scriptsize \begin{align*}v(1)&=3{{(1)}^{2}}-4\\&=-1\end{align*}[/latex] and [latex]\scriptsize \begin{align*}a(1)&=6(1)\\&=6\end{align*}[/latex]

1. Since [latex]\scriptsize v(1) \lt 0[/latex], the particle is moving from right to left.
2. Since [latex]\scriptsize a(1) \gt 0[/latex] and [latex]\scriptsize v(1) \lt 0[/latex], velocity and acceleration are acting in opposite directions. In other words, the particle is decelerating or slowing down.

In addition to velocity, speed, acceleration and position, we can use derivatives to analyse various types of populations, including bacteria colonies and cities. We can use a current population, together with a growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population and can therefore be represented by the derivative of the size of the population.

Example 4.3

In a small town the rate of change of the town’s population (measured in thousands of people) can be modelled by the function [latex]\scriptsize \displaystyle P\left( t \right)=-\displaystyle \frac{1}{3}{{t}^{3}}+64t+3000[/latex], where [latex]\scriptsize t[/latex] is measured in years.

1. Find the rate of change of population function.
2. Find [latex]\scriptsize {P}'(1),\text{ }{P}'(2),\text{ }{P}'(3)\text{ }[/latex]and [latex]\scriptsize {P}'(4)\text{ }[/latex]. Interpret what the results mean for the town’s population.

Solutions

1. [latex]\scriptsize \displaystyle {P}'\left( t \right)=-{{t}^{2}}+64[/latex]
2. .
   [latex]\scriptsize \begin{align*}{P}'(1)&=63\\{P}'(2)&=60\\{P}'(3)&=55\\{P}'(4)&=48\end{align*}[/latex]
   The town’s population is decreasing.

Exercise 4.1

1. A soccer ball is kicked vertically into the air and its motion is represented by the equation [latex]\scriptsize \text{D}(t)=1+18t-3{{t}^{2}}[/latex], where [latex]\scriptsize \text{D}[/latex] is the distance in metres and [latex]\scriptsize t[/latex] is the time elapsed in seconds.
   1. Determine the initial height of the ball at the moment it is being kicked.
   2. Find the initial velocity of the ball.
   3. Determine the velocity of the ball after [latex]\scriptsize \displaystyle 1.5\text{ s}[/latex].
   4. Calculate the maximum height of the ball.
   5. Determine the acceleration of the ball after 1 second and explain the meaning of the answer.
2. If the displacement [latex]\scriptsize \displaystyle s[/latex] (in metres) of a particle at time[latex]\scriptsize \displaystyle ~t[/latex] (in seconds) is given by the equation [latex]\scriptsize s(t)=\displaystyle \frac{1}{2}{{t}^{3}}-2t[/latex], find its acceleration after two seconds.
3. During an experiment the temperature [latex]\scriptsize \displaystyle T[/latex] (in degrees Celsius) varies with time [latex]\scriptsize \displaystyle ~t[/latex] (in hours) according to the formula [latex]\scriptsize T(t)=30+4t-\displaystyle \frac{1}{2}{{t}^{2}},\text{ }t\in [1;10][/latex].
   1. Determine an expression for the rate of change of temperature with time.
   2. During which time interval was the temperature dropping?

The full solutions are at the end of the unit.

Summary

In this unit you have learnt the following:

• How to find the rate of change of different functions.

Unit 4: Assessment

Suggested time to complete: 45 minutes

1. A rocket is fired vertically upward from the ground. The distance [latex]\scriptsize \displaystyle ~s[/latex] in metres that the rocket travels from the ground after [latex]\scriptsize t[/latex] seconds is given by [latex]\scriptsize \displaystyle s\left( t \right)=-16{{t}^{2}}+560t[/latex].
   1. Find the velocity of the rocket 3 seconds after being fired.
   2. Find the acceleration of the rocket 3 seconds after being fired.
2. A particle moves along a coordinate axis in such a way that its position at time [latex]\scriptsize t[/latex] is given by for [latex]\scriptsize \displaystyle 0\le \text{ }t\text{ }\le 360{}^\circ[/latex]. At what times is the particle at rest?
3. A water reservoir has both an inlet and an outlet pipe to regulate the depth of the water in the reservoir. The depth is given by the function [latex]\scriptsize \text{D}(h)=3+\displaystyle \frac{1}{2}h-\displaystyle \frac{1}{4}{{h}^{3}}[/latex], where [latex]\scriptsize \text{D}[/latex] is the depth in metres and [latex]\scriptsize h[/latex] is the hours after [latex]\scriptsize 06\text{h}00[/latex].
   1. Determine the rate at which the depth of the water is changing at [latex]\scriptsize \text{10h}00[/latex].
   2. Is the depth of the water increasing or decreasing?
   3. At what time will the inflow of water be the same as the outflow?

The full solutions are at the end of the unit.

Unit 4: Solutions

Exercise 4.1

1. .
   [latex]\scriptsize D(t)=1+18t-3{{t}^{2}}[/latex]
   1. The initial height of the ball is found when [latex]\scriptsize t=0[/latex].
      [latex]\scriptsize \begin{align*}D(0)&=1+18(0)-3(0)\\&=1\text{ m}\end{align*}[/latex]
   2. Initial velocity is found when [latex]\scriptsize t=0[/latex].
      [latex]\scriptsize \begin{align*}{D}'(t)&=18-6t\\{D}'(0)&=18\text{ m}\text{.}{{\text{s}}^{{-1}}}\end{align*}[/latex]
   3. [latex]\scriptsize {D}'(1.5)=9\text{ m}\text{.}{{\text{s}}^{{-1}}}[/latex]
   4. The maximum height of the ball is found at the turning point where [latex]\scriptsize {D}'(t)=0[/latex] .
      [latex]\scriptsize \begin{align*}18-6t&=0\\t&=3\text{ }\\D(3)&=1+18(3)-3{{(3)}^{2}}\\&=28\text{ m}\end{align*}[/latex]
   5. After 1 second:
      [latex]\scriptsize {{D}'}'(t)=-6\text{ m}\text{.}{{\text{s}}^{{-2}}}[/latex]
      This means the ball is decelerating.
2. .
   [latex]\scriptsize \begin{align*}{s}'(t)&=\displaystyle \frac{3}{2}{{t}^{2}}-2\\{{s}'}'(t)&=3t\\a(2)&=3(2)\\&=6\text{ m}\text{.}{{\text{s}}^{{-2}}}\end{align*}[/latex]
3. .
   1. [latex]\scriptsize {T}'(t)=4-t[/latex]
   2. The temperature drops when [latex]\scriptsize \displaystyle {T}'(t) \lt 0[/latex]
      [latex]\scriptsize \begin{align*}4-t & \lt 0\\-t & \lt-4\\\therefore t & \gt 4\end{align*}[/latex]
      The temperature was dropping during [latex]\scriptsize t\in (4;10][/latex].

Back to Exercise 4.1

Unit 4: Assessment

1. .
   1. .
      [latex]\scriptsize \displaystyle \begin{align*}{s}'\left( t \right)&=-32t+560\\{s}'(3)&=-32(3)+560\\&=464\text{ m}\text{.}{{\text{s}}^{{-1}}}\end{align*}[/latex]
   2. .
      [latex]\scriptsize \displaystyle \begin{align*}{{s}'}'\left( t \right)&=-32t\\{{s}'}'(3)&=-32(3)\\&=-96\text{ m}\text{.}{{\text{s}}^{{-2}}}\end{align*}[/latex]
2. The particle is at rest when [latex]\scriptsize {s}'(t)=v(t)=0[/latex].
   .
   So we must solve [latex]\scriptsize \displaystyle 2\cos t-1=0[/latex] for [latex]\scriptsize 0\le t\le 360{}^\circ[/latex] .
   [latex]\scriptsize \displaystyle \begin{align*}\cos t&=\displaystyle \frac{1}{2}\\t&=60{}^\circ \text{ and }t=300{}^\circ \end{align*}[/latex]
   Thus the particle is at rest at times [latex]\scriptsize \displaystyle t=60{}^\circ \text{ and }t=300{}^\circ[/latex].
3. .
   1. [latex]\scriptsize D(h)=3+\displaystyle \frac{1}{2}h-\displaystyle \frac{1}{4}{{h}^{3}}[/latex]
      [latex]\scriptsize \begin{align*}{D}'(h)&=\displaystyle \frac{1}{2}-\displaystyle \frac{3}{4}{{h}^{2}}\\{D}'(4)&=\displaystyle \frac{1}{2}-\displaystyle \frac{3}{4}{{(4)}^{2}}\\&=-11.5\text{ m per hour}\end{align*}[/latex]
   2. The water is decreasing as [latex]\scriptsize {D}'(h) \lt 0[/latex].
   3. .
      [latex]\scriptsize \begin{align*}{D}'(h)&=\displaystyle \frac{1}{2}-\displaystyle \frac{3}{4}{{h}^{2}}=0\\\displaystyle \frac{3}{4}{{h}^{2}}&=\displaystyle \frac{1}{2}\\{{h}^{2}}&=\displaystyle \frac{2}{3}\\h&=\sqrt{{\displaystyle \frac{2}{3}}}\end{align*}[/latex]
      There are [latex]\scriptsize 60[/latex] minutes in an hour, so find the number of minutes in order to find the time after [latex]\scriptsize 06\text{h}00[/latex] .
      [latex]\scriptsize \sqrt{{\displaystyle \frac{2}{3}}}\times 60\approx 49\min[/latex]
      Therefore, at [latex]\scriptsize 06\text{h49}[/latex] the inflow will equal the outflow.

Back to Unit 4: Assessment