VCE Mathematical Methods Units 3&4: Concise Guides
by AlphaZero (Dan)
How will this work?
Basically, you folks request a topic that you would like me to write a concise guide on! This can be any topic you like. For example: transformations of the plane with matrices, inverse functions, sketching polynomial functions, confidence intervals for the population proportion, etc. I going to aim to produce around one guide every 1-2 weeks, but keep in mind, I'm a very busy person, so I apologise in advance if you don't see anything from me for some time. If I get lots of different requests, I'll pick the most popular.
But...
Please remember that these are meant to be miniature guides for people to learn from. Do not message me with "can you write a guide on calculus?" There are entire textbooks on calculus. To write a 'concise' guide on such a topic would be futile due the amount of detail I would need to omit to make it even remotely brief. And, as most of you would know, omitting details, particularly in this subject, does not serve students well.
Contents
1. Using Inverse Functions in Integration
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1. Using Inverse Functions in Integration
9 March 2019
Prerequisite knowledge
> Finding the rule, domain and range of the inverse of a function
> The fundamental theorem of calculus
> Antidifferention of all the functions investigated in the subject excluding logarithmic functions and the tangent function
> Finding the area between curves in the plane
Introduction
Integration in VCE Mathematical Methods, aside from a few topics in probability and statistics, is mostly used in geometric applications - mainly to find areas of the plane with uncommonly shaped boundaries. Given the limited number of techniques available to us to find antiderivatives of functions by hand, we can sometimes find ourselves a bit lost trying to find the exact area of a region when given a function which we might not know how to antiderive. Further, using the geometric properties of a function's inverse, we can often simplify the amount of work required to arrive at a conclusion. In this guide, I'll present the solutions to two problems. Please attempt the questions before reading the solution.
Question 1 [Exam 1]
Let \(f:[0,\;\infty)\to\mathbb{R},\ f(x)=\log_e(x+1)\).
\(\mathbf{a.}\quad\)Find the rule of \(f^{-1}\).
Solution
\[x=\log_e\!\big(f^{-1}(x)+1\big)\implies f^{-1}(x)=e^x-1\]
\(\mathbf{b.}\quad\)Find the area of the region bounded by the graph of \(f\), the \(x\)-axis and the line \(x=2\).
Solution
Consider the following figure which shows the graph of \(f\) (in red), the graph of \(f^{-1}\) (in blue), the required region (in red), and another region (in blue).
(https://i.imgur.com/6Oanub0.png)
By symmetry about the line \(y=x\) (in black), it is clear that the required region (in red) is the same as the region bounded by the graph of \(f^{-1}\), the \(y\)-axis and the line \(y=2\) (in blue).
Thus, we have \begin{align*}A&=\int_0^{\log_e(3)}\!\Big(2-(e^x-1)\Big)dx\\
&=\int_0^{\log_e(3)}\!\Big(3-e^x\Big)dx\\
&=\Big[3x-e^x\Big]_0^{\log_e(3)}\\
&=3\log_e(3)-3-\big(0-1\big)\\
&=3\log_e(3)-2\ \ \text{units}^2 \end{align*}
For those of you that know integration by parts (which is not required in this subject), you can very easily verify the above answer.\begin{align*}A&=\int_0^2 \log_e(x+1)dx\\
&=\int_0^2\frac{d}{dx}\big[x+1\big]\log_e(x+1)dx\\
&=\Big[(x+1)\log_e(x+1)\Big]_0^2-\int_0^2 (x+1)\frac{d}{dx}\big[\log_e(x+1)\big]dx\\
&=\Big[(x+1)\log_e(x+1)\Big]_0^2-\int_0^2 dx\\
&=\Big[(x+1)\log_e(x+1)\Big]_0^2-\Big[x\Big]_0^2\\
&=3\log_e(3)-0-\big(2-0\big)\\
&=3\log_e(3)-2\ \ \text{units}^2\end{align*}
Question 2 [VCAA 2018 Exam 2 - Question 5]
Consider functions of the form \(f:\mathbb{R}\to\mathbb{R},\ f(x)=\dfrac{81x^2(a-x)}{4a^4}\) and \(h:\mathbb{R}\to\mathbb{R},\ h(x)=\dfrac{9x}{2a^2}\), where \(a\) is a positive real number.
\(\mathbf{a.}\quad\)Find the coordinates of the local maximum of \(f\) in term of \(a\).
Solution
\[f'(x)=0\implies x=0\ \text{or}\ x=\frac{2a}{3}\] Since \(f\) is a 'negative cubic', the local maximum occurs at \[\left(\frac{2a}{3},\ \frac{3}{a}\right)\]
\(\mathbf{b.}\quad\)Find the \(x\)-values of all the points of intersection between the graphs of \(f\) and \(h\), in terms of \(a\) where appropriate.
Solution
\[f(x)=h(x)\implies x=0,\ \frac{a}{3},\ \frac{2a}{3}\]
\(\mathbf{c.}\quad\)Determine the total area of the regions bounded by the graphs of \(y=f(x)\) and \(y=h(x)\).
Solution
A quick sketch of the graph of \(f\) (in red) and \(h\) (in blue) for some values of \(a\) shows that \(h(x)>f(x)\) for \(0<x<\dfrac{a}{3}\) and that \(f(x)>h(x)\) for \(\dfrac{a}{3}<x<\dfrac{2a}{3}\).
(https://i.imgur.com/HoBzqKd.png)
Thus, the total area of the required regions is given by \begin{align*}A&=\int_0^{a/3}\Big[h(x)-f(x)\Big]dx+\int_{a/3}^{2a/3}\Big[f(x)-h(x)\Big]dx\\
&=\frac18\ \ \text{units}^2\end{align*} Note that we could have also evaluated integral expressions along the lines of \[2\int_0^{a/3}\Big[h(x)-f(x)\Big]dx\quad\text{or}\quad \int_0^{2a/3}\Big|f(x)-h(x)\Big|dx.\]
Consider the function \(g:\left[0,\ \dfrac{2a}{3}\right]\to\mathbb{R},\ g(x)=\dfrac{81x^2(a-x)}{4a^4}\), where \(a\) is a positive real number.
\(\mathbf{d.}\quad\)Evaluate \(\dfrac{2a}{3}\times g\!\left(\dfrac{2a}{3}\right)\).
Solution
\begin{align*}\dfrac{2a}{3}\times g\!\left(\dfrac{2a}{3}\right)&=\frac{2a}{3}\times\frac{3}{a}\\
&=2 \end{align*}
\(\mathbf{e.}\quad\)Find the area bounded by the graph of \(g^{-1}\), the \(x\)-axis and the line \(x=g\!\left(\dfrac{2a}{3}\right)\).
Solution
One should note in this question that it is not possible to find an expression for the rule of \(g^{-1}\) in terms of \(x\) and \(a\), and so we must use symmetry to answer this question.
In the following image, we are required to find the area of the region with green boundaries. I've also included red boundaries for another region, which clearly has the same area.
(https://i.imgur.com/TeJZHyz.png)
We can evaluate the required area using the technique as in Question 1 part b above, or we could utilise our result from part e, which gives the area of the rectangle with opposite vertices at the end points of the graph of \(g\).
Thus, the area of the region is given by \begin{align*}A&=2-\int_0^{2a/3}g(x)\,dx\\
&=1\ \ \text{unit}^2 \end{align*}
\(\mathbf{f.}\quad\)Find the value of \(a\) for which the graphs of \(g\) and \(g^{-1}\) have the same endpoints.
Solution
The graphs of \(g\) and \(g^{-1}\) always have an endpoint at \((0,\ 0)\), and so we just require \[\dfrac{2a}{3}=g\!\left(\dfrac{2a}{3}\right)\implies a=\frac{3\sqrt{2}}{2}.\]
\(\mathbf{g.}\quad\)Find the area enclosed by the graph of \(g\) and \(g^{-1}\) when they have the same endpoints.
Solution
Consider the following figure which shows the graphs of \(g\), \(g^{-1}\) and \(y=x\).
(https://i.imgur.com/WhaO2a2.png)
But, from part b, we know that the graph of \(h\) should intersect the graph of \(g\) at \(x=0,\ \dfrac{a}{3},\ \dfrac{2a}{3},\) and so indeed, we have \(h(x)=x\) for \(a=\dfrac{3\sqrt2}{2}\).
Hence, by symmetry, the required area is just double that of the area found in part c. \[A=\frac14\ \ \text{units}^2\]
Conclusion
Well, this guide was certainly an interesting one to write. It was a very specific topic, but I definitely enjoyed providing in-depth explanations of both questions. Thank you to Lear and undefined for suggesting this topic. If I have made any typos / errors, please let me know so I can fix them!
To request a topic, reply to this thread or send me a message!