Part 3: Applications of Differentiation | Beginner’s Guide to Year 11 Ext 1

Are you having a hard time understanding inflection points and rates of change? Don’t worry, in this article, we’re going to breakdown Year 11 Extension 1 Maths Applications of differentiation into clear and accessible chunks with step-by-step process to ensure your marks see positive related rates of change!

In this article we discuss:

• The NESA Syllabus Outcomes
• Assumed knowledge
• Maximum and Minimum Problems
• The Second Derivative
• Related Rates of Change
• Concept Check Questions

Year 11 Extension 1 Mathematics: Applications of Differentiation

In this article, we discuss the various applications of differentiation. Being able to successfully apply calculus and to solve harder problems is an essential skill in mathematics. By doing so, students are not only able to solidify their knowledge on calculus, but they are also able to solve real-life problems including maximising profit, etc.

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NESA Syllabus Outcomes

NESA expects students to be proficient in the following syllabus outcomes:

• C3.1: The first and second derivatives
  • Define and interpret the concept of the second derivative as the rate of change of the first derivative function in a variety of contexts, for example recognise acceleration as the second derivative of displacement with respect to time (ACMMM108, ACMMM109) AAM
  • Use the second derivative to determine concavity and the nature of stationary points
  • Understand that when the second derivative is equal to \(0\) this does not necessarily represent a point of inflection
• C3.2: Applications of the derivative
  • Solve optimisation problems for any of the functions covered in the scope of this syllabus, in a wide variety of contexts including displacement, velocity, acceleration, area, volume, business, finance and growth and decay AAM
  • Use calculus to establish the location of local and global maxima and minima, including checking endpoints of an interval if required
• C1.3: Related rates of change
  • Solve problems involving related rates of change as instances of the chain rule (ACMSM129) AAM
  • Develop models of contexts where a rate of change of a function can be expressed as a rate of change of a composition of two functions, and to which the chain rule can be applied

Assumed Knowledge

Students should already be familiar with introductory calculus, including finding the first derivative and identifying stationary points.

This content can be found in our Year 11 Maths Advanced Guides should students want to solidify their understanding:

• Year 11 Maths Advanced Calculus 

Maximum & Minimum Problems

One of the primary uses of differentiation revolves around being able to solve for maximum and minimum quantities. The practical applications of maximisation and minimisation are vast and include:

• Maximising volume
• Maximising profit
• Minimising use of materials such as fuel, metal, etc.

When solving maximum and minimum problems, there are some key steps which are always followed:

Example 1:

A right-angled triangle has base \(30cm\) and height \(40cm\). A rectangle is inscribed in the triangle so that one of its sides lies along the base of the triangle. Find the dimensions of the rectangle which maximises its length.

Solution 1:

Step 1: Introduce the required variables. In this case, we have to introduce three variables:

Let \(A\) = area of the rectangle (quantity being maximised)
Let \(y\) = width of the rectangle (variable quantity)
Let \(x\) = length of the rectangle (variable quantity)

Step 2:

We must form a function with only two variables, one of them being \(A\). We know:

\(A = xy\). Therefore, we must express \(y\) in terms of \(x\) (or vice versa) and then use substitution. To relate \(y\) and \(x\), we can use similar triangles:

\begin{align*}
∴\frac{30}{40}&=\frac{(30-y)}{x}\\
1200-40y&=30x\\
40y&=1200-30x\\
∴y&=\frac{1}{40}(1200-30x)\\
∴A&=\frac{x}{40}(1200-30x)\\
\end{align*}

Step 3:

Now we solve our function to find the maximum:

\begin{align*}
A&=30x-\frac{(3x^{2})}{4}\\
∴A^{‘}&=30-\frac{3x}{2}\\
When A^{‘}=0,\\
30-\frac{3x}{2}&=0\\
60-3x&=0\\
∴x&=20\\
\end{align*}

Step 4:

As length cannot be \(0\), in this case, the local maximum is equal to the global maximum.

Step 5:

We have one dimension \(x\), however we still need to find y:

\begin{align*}
y&=\frac{1}{40}(1200-30x)\\
∴y&=\frac{1}{40}(1200-30(20))\\
∴y&=15\\
∴\text{The required dimensions are} \ 20 \ cm \ \text{by} 15 \ cm\\
\end{align*}

The Second Derivative

The second derivative is obtained by differentiating the first derivative. There are two main applications for the second derivative:

• To find the nature of stationary points
• To find points of inflexion

Second Derivative Test

With regards to the nature of stationary points, the second derivative test offers a quicker and easier alternative to the ‘table method’. The test abides by the following rules:

• If \(f”(a) > 0\): the curve is concave up. Therefore, there is a local minimum at \(a\).
• If \(f”(a) < 0\): the curve is concave down. Therefore, there is a local maximum at \(a\).
• If \(f”(a) = 0\): there is most likely a horizontal point of inflexion at a. However, the ‘table method’ must be performed on the second derivative to confirm.

Points of Inflexion

Further, much like how the first derivative is used to find stationary points, the second derivative can be used to find points of inflexion. Points of inflexion are points on a curve where it changes from concave up to concave down (or vice versa). Mathematically, these are generally the points which satisfy the following:

Condition: Second Derivative = \(0\)

However, this is not always be the case and as such more work, in the form of a table, is needed to confirm this. If there is a change in the sign of the second derivative (i.e. change in concavity) before and after, then there is indeed a point of inflexion where \(f”(a) = 0\).

Example 1: Stationary points

Find the stationary points of the cubic \(y = 2x^3-3x^2-12x+7\) and determine their nature.

Solution 1:

\begin{align*}
y&’= 6x^2-6x-12\\
&\text{When} \ y^{‘}=0\\
6x^2-6x-12&=0\\
x^2-x-2&=0\\
(x-2)(x+1)=0\\
&∴x=2,x=-1\\
y ‘&’= 12x-6\\
&\text{Substitute} \ x=2\\
y &= 2(2)^3-3(2)^2-12(2)+7\\
y&=-13\\
y” &= 12(2) -6\\
y”&=18\\
&\text{As} \ y”>0, \ \text{there is a local minimum at} \ (2,18).\\
&\text{Substitute} \ x=-1\\
y &= 2(-1)^3-3(-1)^2-12(-1)+7\\
y&=14\\
y” &= 12(-1) -6\\
y”&=-18\\
&\text{As} \ y”<0, \ \text{there is a local maximum at} (-1,14).\\
\end{align*}

Example 2: Inflexion point(s)

Find x-coordinate for the point of inflexion(s) of the cubic \(y = 2x^3-15x^2+36x+8\).

Solution 2:

\begin{align*}
y ‘&= 6x^2-30x+36\\
y “&= 12x-30\\
&\text{When} \ y”=0\\
12x-30&=0\\
6(2x-5)&=0\\
2x-5&=0\\
∴x&=\frac{5}{2}\\
\end{align*}

Related Rates of Change

The derivative, at its core, is a rate of change. As such, by using the chain rule, we can resolve many derivative (i.e. rates) into one single derivative:

Forming related rate equations such as this allows us to find how one quantity varies with another quantity, without having an equation relating the two. Moreover, the can chain allows us to relate infinity rates, and thus we can expand to include a third rate If needed:

To solve problems revolving around related rates of change, often one rate is given, and we have to utilise formulae for volume, time, etc., to find the other rate. The following process is a general guide on how to approach these questions:

Example 1: Balloon inflation

A balloon is being inflated, with its radius increasing at a rate of \(4cm/second\). What is the rate at which the volume is changing when the radius is \(10 cm\).

Solution 1:

Step 1: Assign the required variables. In most cases, we have to introduce three variables:

Let \(V\) = volume of the balloon (quantity being maximised)
Let \(r\) = radius of the balloon (variable quantity)
Let \(t\) = time

Step 2:

Now we need to form the chain rule. We want to find how volume is changing with respect to time, therefore, \(\frac{dV}{dt}\) will be the subject of the rule. Further we are given how the radius changes with time, i.e. \(\frac{dr}{dt}\), and this rate must be incorporated:

Step 3:

From the above rule, there is only one rate which is unknown, \(\frac{dV}{dr}\). To find this rate, we use the formula for the volume of a sphere and differentiate:

\begin{align*}
V&=\frac{4}{3} πr^3\\
&\frac{dV}{dr}=4πr^2\\
\end{align*}

Step 4:

Finally, we substitute our given and found rates to find, \(dV/dt\):

\begin{align*}
\frac{dV}{dt}&=\frac{dV}{dr}×\frac{dr}{dt}\\
dV/dr&=4πr^2× 4\\
∴dV/dr&=16πr^2\\
\text{Substitute} r&=10cm:\\
\frac{dV}{dr}&=16π(10)^2\\
∴\frac{dV}{dr}&=1600π\\
∴\text{Volume is increasing at a rate of } \ 1600π/second\\
\end{align*}

Concept Check Questions

1. The sum of two positive numbers is \(36\). Find the numbers if the sum of their squares product is minimised.
2. Differentiate the following functions to find stationary points and determine their nature using the second derivative test:
   1. \(\frac{x^3}{3}-\frac{3x^2}{2}-10x+7\)
   2. \(2x^3+15x^2-144x+5\)
3. Find the points of inflexion for the following functions:
   1. \(5x^2-8x+7\)
   2. \(x^3-12x^2-36x+9 \)
4. A spherical bubble is shrinking so that its volume is decreasing at a constant rate of \(120 \ cm^3/min\). Find the rate at which its radius is decreasing when the volume is \(48π \ cm^3\).

Concept Check Solutions

1. \(18\) and \(18\)
   1. Local minimum at \((5,(\frac{-233}{3})\) and local maximum at \((-2,   55/3)\)
   2. Local minimum at \((3,-238)\) and local maximum at \((-8,1093)\)
   1. \((10,427)\)
   2. \((4,-263)\)
2. \(\frac{dr}{dt}=\frac{5}{6π} \ cm/min\)

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