A Brief Introduction to Calculus: Part 1.4 - Using the Rate of Change

Math
Calculus
Further Math

Published on Wed Aug 03 2022. Views

An introduction to basic concepts in Calculus; mostly based on the IGCSE Further Math textbook.

Introduction

This is the fourth part of first series — Derivatives, focused on applying the rate of change in relatively simple scenarios and questions.

Note: the content introduced in this series is mostly based on the IGCSE Further Math textbook.

Quick Access

Problems Involving Displacement, Velocity, and Acceleration

Displacement is a vector quantity represented by $s$ .
Velocity is a vector quantity represented by $v$ .
Acceleration is a vector quantity represented by $a$ .

If you've learned a little bit of physics, you would know that $v = s/t$ where t is time, and $a = v/t$ .

Generally, the expressions' variable would be time, and we would have to differentiate the functions with respect to $t$ (time) to work out the rate of something.

Now, if you get a question that, for example, gives you a polynomial of some degree indicating the relationship between two of these quantities stated above (e.g. $s = 3t^2 + 7t - 13$ ), you'll have to calculate the derivative of the function to obtain some result, for example, velocity.

Key Point

$v = \frac{ds}{dt}$
$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Example 1

A particle $P$ is moving along a straight line that passes through the fixed point $O$ . The displacement, $s$ meters, of $P$ from $O$ at time $t$ seconds is given by:

\begin{align*} s = t^3 - 6t^2 + 5t - 4 \end{align*}

Find the value of t for which the acceleration of P is $3m/s^2$ .

\begin{align*} a &= \frac{d^2s}{dt^2} = 6t -12 \\ \\ 3 &= 6t - 12 \\ 6t &= 15 \\ t &= \frac{5}{2} \space seconds \end{align*}

Example 2

The velocity, $v$ m/s, of a particle after t seconds is given by $v = 160 - 32t$ . Find the acceleration.

\begin{align*} a &= \frac{dv}{dt} \\ &= -32 m/s^2 \\ \\ \end{align*}

Example 3

The displacement, $s$ meters, of a particle after $t$ seconds is given by $s = t^3 - 2t^2 + 3t + 1$

Find an expression for $v$ .

\begin{align*} v &= \frac{ds}{dt} \\ \\ v &= 3t^2 - 4t + 3 \end{align*}

Find an expression for $a$ .

\begin{align*} a &= \frac{dv}{dt} \\ \\ a &= 6t - 4 \end{align*}

Work out the velocity and acceleration of the particle after two seconds, in the correct units.

\begin{align*} v &= 3 \cdot 2^2 - 4 \cdot 2 + 3 \\ &= 12 - 8 + 3 \\ &= 7 m/s \\ \\ a &= 6t - 4 \\ &= 12 - 4 \\ &= 8m/s^2 \end{align*}

Maximization/Minimization Problems

Oftentimes, you will have to construct a function that relates one quantity to another, say $x$ is the independent variable and $y$ is the dependent variable, and you will have to find the value of $x$ for which $y$ is a maximum/minimum.

Once you have a function, calculate its derivative and set its value to 0, thus providing an equation whose solutions would be the coordinates of the turning points. Refer back to the "Stationary Points" section for more explanation of the method.

Example 1

Prerequisite:

The minor arc length of a sector is $l = r\theta$ , where $\theta$ is the angle in radians (180 degrees = $\pi$ radians).

$OMT$ is a minor sector of a circle with center $O$ and radius $r$ cm. The perimeter of the sector is 200cm.

Find a function for the area of the sector, A, in $cm^2$ .

\begin{align*} P &= 200 \\ 200 &= 2r + r\theta \\ \frac{200}{r} &= 2 + \theta \\ \theta &= \frac{200}{r} - 2 \\ \\ A &= \pi r^2 \frac{\theta}{2\pi} \\ A &= r^2 \frac{\theta}{2} \\ \\ \because \theta &= \frac{200}{r} - 2 \\ A &= r^2 \cdot \frac{1}{2}\left[\frac{200}{r} - 2\right] \\ &= r^2 \left[\frac{100}{r} - 1\right] \\ &= 100r - r^2 \end{align*}

Find the value of $r$ for which $A$ is a maximum.

\begin{align*} \frac{dA}{dr} &= 100 - 2r \\ \\ 0 &= 100 - 2r \\ r &= 50 \end{align*}

Prove that the area of a sector with the radius calculated above is indeed a maximum.

\begin{align*} \frac{d^2A}{dr^2} &= -2 < 0 \end{align*}

Find the maximum area of the sector $OMN$ .

\begin{align*} r &= 50 \\ A &= 100 \cdot 50 - 50^2 \\ &= 5000 - 2500 \\ &= 2500 cm^2 \end{align*}

Example 2

A large tank in the shape of a cuboid is to be made from $54m^2$ of sheet metal. There is no top, i.e. only 5 faces. The height and width of the tank are both $x$ meters.

Find a function for the volume of the cuboid.

\textnormal{Let } y \textnormal{ be the length of the cuboid} \\ \begin{align*} \\ 54 &= 2x^2 + 3xy \\ y &= \frac{54 - 2x^2}{3x} \end{align*}

\begin{align*} V &= x^2y \\ &= x^2\left[\frac{54 - 2x^2}{3x}\right] \\ &= 18x - \frac{2}{3}x^3 \end{align*}

Find the maximum stationary point of $V$ and justify that it's a maximum.

\begin{align*} \frac{dV}{dx} &= 18 - 2x^2 \\ 0 &= 18 - 2x^2 \\ x &= {3} \\ \\ \frac{d^2V}{dx^2} &= -4x \\ \end{align*}

\begin{align*} &\textnormal{Thus a coordinate with a positive x value} \\ &\textnormal{would be a maximum stationary point.} \\ \end{align*}

\begin{align*} x &= 3 \\ V &= 18\cdot 3 - \frac{2}{3}\cdot 3^3 \\ &= 54 - 18 \\ &= 36 m^3 \end{align*}

Example 3

A community is planning the greenery for a land of area $1000m^2$ . One part would be planted with flowers and the other part with grass.

Let the area of grass be $x$ $m^2$ ; the cost of the grass, $y_1$ , is given by $y_1 = 30x$ . The cost of the flowers, $y_2$ is given by $y_2 = -0.01x^2 - 20x + 30000$ .

Let $W$ be the total cost of greenery for the whole area, calculate the maximum value of $W$ ; and thus find the area of grass when $W$ is maximum.

\begin{align*} W &= y_1 + y_2 \\ W &= -\frac{1}{100}x^2 + 10x + 30000 \\ \\ \frac{dW}{dx} &= - \frac{1}{50}x + 10 \\ 0 &= - \frac{1}{50}x + 10 \\ 10 &= \frac{1}{50}x \\ x &= 500 \\ \\ W_{max} &= -\frac{1}{100}\cdot 500^2 + 10 \cdot 500 + 30000 \\ &= 32500 \end{align*}

Conclusion

This is the end of the first section in the series — Derivatives. Hopefully the content was intuitive and helpful.

Thanks for reading.