October 28, 2024 · 6:51 am

Related Rates Question

A boat is moving towards the beach line at 3 metres/minute. On the boat is a rotating light, revolving at 4 revolutions/minute clockwise, as observed from the beach. There is a long straight wall on the beach line, as the boat approaches the beach, the light moves along the wall. Let x equal the displacement of the light from the point O on the wall, which faces the boat directly. See the diagram below.
Determine the velocity, in metres/minute, of the light when x=5 metres, and the distance of the boat from the beach D is 12 metres.

Mathematics Specialist Semester 2 Exam 2018


The light is rotating at 4 revolutions/minute, which means

    \begin{equation*}\frac{d\theta}{dt}=4\times\pi\end{equation}

We want to find \frac{dx}{dt} and we know \frac{d\theta}{dt} and \frac{dD}{dt}.

We need to find an equation connecting x, \theta, and D.

    \begin{equation*}tan (\theta)=\frac{x}{D}\end{equation}

Differentiate (implicitly) with respect to time.

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

Now we know x=5, and D=12, using pythagoras we can calulate the hypotenuse.


h=\sqrt{12^2+3^2}=13
sec(\theta)=\frac{13}{12}

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

    \begin{equation*}(\frac{13}{12})^2\times 4\pi=\frac{\frac{dx}{dt}12-3\times 5}{12^2}\end{equation}

    \begin{equation*}169\times4\pi=12\frac{dx}{dt}-15\end{equation}

    \begin{equation*}676\pi+15=12\frac{dx}{dt}\end{equation}

    \begin{equation*}\frac{dx}{dt}=178.2\end{equation}

The velocity of the light is 178.2 m/minute.

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October 22, 2024 · 6:55 am

Year 12 Mathematics Applications – Finance Question

What happens with an investment or a loan when the compounding periods are not the same as the payment periods?

For example,

Gerald invests $450 000 at a rate of 6.4% p.a. compounding monthly. At the end of every quarter he receives $35 000 from his investment.

(a) Write a recursive rule that will enable you to find the balance of the account, T_n after n payments.

(b) Find the balance of the account after 3 years.

(c) How long will it take for the balance in the account to reach zero?

(d) What will be the amount of Gerald’s last payment?

The interest is compounding at a different rate to the payments.
T_{n+1}=T_n(1+\frac{6.4}{100\times12})^{12\div4}-35 000, T_0=450 000

We need to raise the multiplier (1+\frac{6.4}{100\times12}) by the number of compound periods per payment, i.e. 12\div4

(a) T_{n+1}=T_n(1.0161)^3-35 000, T_0=450 000

(b) 3 years is 12 payments, \therefore n=12

T_12=245548.28

(c)

21 years

(d) The final repayment is 35 000-6535.03=28 464.97

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October 18, 2024 · 7:06 am

Fibonacci Sequence – Finding the Closed Form

I have been reading An Imaginary Tale – The Story of \sqrt{-1} by Paul J Nahin, which is fabulous. There was a bit in chapter 4 where he found the closed form of the generalised Fibonacci sequence. I thought it would be a good exercise to find the closed from of the Fibonacci sequence.

Just to remind you, the Fibonacci sequence is

1, 1, 2, 3, 5, 8, 13, 21, ...

and it is defined recursively

    \begin{equation*}T_{n+2}=T_{n+1}+T_n, T_0=1, T_1=1\end{equation}

That is, the next term is the sum of the two previous terms, i.e.

    \begin{equation*}T_3=T_2+T_1=1+1=2\end{equation}

Now the starting off point is slightly dodgy as it involves and educated guess as Paul Nahin writes,

How do I know that works? Because I have seen it before, that’s how! […] There is nothing dishonourable about guessing correct solutions – indeed, great mathematicians and scientists, are invariable great guessers – just as long as eventually the guess is verified to work. The next time you encounter a recurrence formula, you can guess the answer too because then you will have already seen how it works.

We start with T_n=kz^n

This means T_{n+2}=T_{n+1}+T_n is kz^{n+2}=kz^{n+1}+kz^n

    \begin{equation*}kz^{n+2}=kz^{n+1}+kz^n\end{equation}

    \begin{equation*}kz^n(z^2-z-1)=0\end{equation}

    \begin{equation*}z^2-z-1=0\end{equation}

    \begin{equation*}z=\frac{1\pm\sqrt{(-1)^2-4(1)(-1)}}{2}\end{equation}

\therefore z=\frac{1+\sqrt{5}}{2} or z=\frac{1-\sqrt{5}}{2}

Hence T_n=k_1(\frac{1+\sqrt{5}}{2})^n+k_2(\frac{1-\sqrt{5}}{2})^n and we can use the initial conditions T_0=1 and T_1=1 to find k_1 and k_2

When n=0, T_0=1

(1)   \begin{equation*}1=k_1+k_2\end{equation*}

When n=1, T_1=1

(2)   \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+k_2(\frac{1-\sqrt{5}}{2})\end{equation*}

From equation 1, k_2=(1-k_1), substitute into equation 2

    \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+(1-k_1)(\frac{1-\sqrt{5}}{2})\end{equation}

    \begin{equation*}1=k_1(\frac{1+\sqrt{5}}{2})+\frac{1-\sqrt{5}}{2}-k_1(\frac{1-\sqrt{5}}{2})\end{equation}

    \begin{equation*}1=k_1\sqrt{5}+\frac{1-\sqrt{5}}{2}\end{equation}

    \begin{equation*}1-(\frac{1-\sqrt{5}}{2})=\sqrt{5}k_1\end{equation}

    \begin{equation*}k_1=\frac{1}{\sqrt{5}}(\frac{1}{2}+\frac{\sqrt{5}}{2})\end{equation}

    \begin{equation*}k_1=\frac{1}{2}(\frac{1}{\sqrt{5}}+1)\end{equation}

    \begin{equation*}k_1=\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}

    \begin{equation*}k_2=1-\frac{1+\sqrt{5}}{2\sqrt{5}}\end{equation}

    \begin{equation*}k_2=-(\frac{1-\sqrt{5}}{2\sqrt{5}})\end{equation}

\therefore T_n=(\frac{1+\sqrt{5}}{2\sqrt{5}})(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2\sqrt{5}})(\frac{1-\sqrt{5}}{2})^n

T_n=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1})

Does it work?

Remember the sequence is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...

If n=5, T_n=8

    \begin{equation*}T_5=\frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^6-\frac{1-\sqrt{5}}{2})^6)\end{equation}

As you can see it works!

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October 13, 2024 · 4:59 am

Tricky Area Ratio Question

A circle of radius r, is inscribed within an isosceles triangle ABC. CA=CB=5r. Given that \angle{ACB} is acute, find the ratio of the area of the circle to that of triangle ABC.

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know AC, AB and BC are tangents to the circle. Because the triangle is isosceles, the distance from A to the circle is the same as the distance from B to the circle.

CP is perpendicular to AB (because the triangle is isosceles).

Because it is proportional, i.e. r and 5r, we can let r=1

Let h=CP

(1)   \begin{equation*}h=\sqrt{25-a^2}\end{equation*}

but h also equals

(2)   \begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}

Set equation 1 equal to equation 2

    \begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}

Square both sides

    \begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}

Expand and simplify

    \begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}

Divide by 2

    \begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}

    \begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}

Square both sides

    \begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}

    \begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}

    \begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}

I solved this using a graphics calculator

a=-0.8434, a=1.3068, a=4.5367, a=5

We can reject a=-0.8434, a=4.5367, and a=5. If a=5, there isn’t a triangle, and if a=4.5367 \angle{ACB} is not acute.

Hence the area of triangle ABC=a\times h

    \begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}

(3)   \begin{equation*}A_{\Delta}=6.3069\end{equation*}

(4)   \begin{equation*}A_{\circ}=\pi\end{equation*}

Hence, equation 4 divided by equation 3 is

    \begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}

Perhaps I approached this question in the wrong way. Is there an easier process?

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October 7, 2024 · 7:43 am

Arithmetic Sequence

I did this question with on of my year 11 students. I think the algebra and the subscripts can be a bit tricky.

If T_m=n and T_n=m, then prove that T_{m+n}=0. Here where T_n and T_m are terms of an arithmetic sequence.
Mathematics Methods Units 1&2 – Exercise 15B Question 19

If T_m=n then,

(1)   \begin{equation*}n=a+(m-1)d\end{equation*}


And if T_n=m then,

(2)   \begin{equation*}m=a+(n-1)d\end{equation*}


Subtract equation (2) from equation (1)

    \begin{equation*}n-m=(m-1)d-((n-1)d)\end{equation*}


    \begin{equation*}n-m=md-nd\end{equation*}


(3)   \begin{equation*}n-m=d(m-n)\end{equation*}


Therefore d must equal -1
Substitute d=-1 into equation (1)

    \begin{equation*}n=a+(m-1)(-1)\end{equation*}


(4)   \begin{equation*}n=a-m+1\end{equation*}


Therefore a=n+m-1


(5)   \begin{equation*}T_{m+n}=a+(m+n-1)d\end{equation*}


Substitute a=n+m-1 and d=-1 into equation (5)

    \begin{equation*}$T_{m+n}=n+m-1+(m+n-1)(-1)$\end{equation*}


    \begin{equation*}$T_{m+n}=n+m-1-m-n+1$\end{equation*}


(6)   \begin{equation*}$T_{m+n}=0$\end{equation*}

As you can see from equation (6), T_{m+n}=0

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October 1, 2024 · 7:31 am

Using De Moivre’s Theorem for Trigonometric Identities

We are going to use De Moivre’s theorem to prove trigonometric identities.

Remember, De Moivre’s Theorem

If z=r(cos(\theta)+isin(\theta)), then z^n=r^n(cos(n\theta)+isin(n\theta))

Or a shorter version z=rcis(\theta), then z^n=r^ncis(n\theta)

Now, let z=cos(\theta)+isin(\theta), find z+\frac{1}{z}

z+z^{-1}=cos(\theta)+isin(\theta)+cos(-\theta)+isin(-\theta)

Remember cos(\theta)=cos(\theta) and sin(-\theta)=-sin(\theta)

z+\frac{1}{z}=cos(\theta)+isin(\theta)+cos(\theta)-isin(\theta)

z+\frac{1}{z}=2cos(\theta)

It is the same for z^n+\frac{1}{z^n}

z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(-n\theta)+isin(-n\theta)

z^n+\frac{1}{z^n}=2cos(n\theta)

Prove cos(2\theta)=2cos^2(\theta)-1
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})+z\times\frac{1}{z}-z\times\frac{1}{z}
LHS=\frac{1}{2}(z^2+2z\times\frac{1}{z}+\frac{1}{z^2})-z\times\frac{1}{z}
LHS=\frac{1}{2}(z+\frac{1}{z})^2-1
LHS=\frac{1}{2}(2cos(\theta))^2-1
LHS=\frac{1}{2}(4cos^2(\theta))-1
LHS=2cos^2(\theta)-1
LHS=RHS

We can do something similar with sine.

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(\theta)-isin(\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-cos(\theta)+isin(\theta)

z-\frac{1}{z}=2isin(\theta)

Hence z^n-\frac{1}{z^n}=2isin(n\theta)

Prove sin(2\theta)=2sin(\theta)cos(\theta)
LHS=sin(2\theta)
LHS=\frac{1}{2i}(z^2-\frac{1}{z^2})
LHS=\frac{1}{2i}(z-\frac{1}{z})(z+\frac{1}{z})
LHS=\frac{1}{2i}(2isin(\theta)(2cos(\theta))
LHS=sin(\theta)2cos(\theta)
LHS=2sin(\theta)cos(\theta)
LHS=RHS

Let’s find an identity for cos(3\theta)

cos(3\theta)=\frac{1}{2}(z^3+\frac{1}{z^3})

=\frac{1}{2}(z^3+\frac{1}{z^3}+3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}-3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2})

=\frac{1}{2}((z+\frac{1}{z})^3-3z-\frac{3}{z})

=\frac{1}{2}((z+\frac{1}{z})^3-3(z+\frac{1}{z}))

=\frac{1}{2}(2cos(\theta))^3-3(2cos(\theta)))

=\frac{1}{2}(8cos^3(\theta)-6cos(\theta))

=4cos^3(\theta)-3cos(\theta)

\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)

And sin(3\theta)?

sin(3\theta)=\frac{1}{2i}(z^3-\frac{1}{z^3})

=\frac{1}{2i}(z^3-\frac{1}{z^3}-3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}+3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2}

=\frac{1}{2i}((z-\frac{1}{z})^3+3z-\frac{3}{z})

=\frac{1}{2i}(2isin(\theta))^3+3(z-\frac{1}{z}))

=\frac{1}{2i}(-8isin^3(\theta)+6isin(\theta))

=-4sin^3(\theta)+3sin(\theta)

\therefore sin(3\theta)=3sin(\theta)-4sin^3(\theta)

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September 29, 2024 · 10:18 am

Intersecting Circles

Two circles of radius L and 2L intersect as shown. What is the area of the shaded region?

From Professor Povey’s Perplexing Problems

My plan is to find the sum of the area of the two segments (see below).

Construct triangles

The diagonal of the square (the pink line above) has length

=\sqrt{(2L)^2+(2L)^2}=\sqrt{8L^2}=2\sqrt{2}L

From the pink triangle in the above diagram, I am going to find the angles using the cosine rule.

cos\theta=\frac{L^2+(2\sqrt{2}L)^2-(2L)^2}{2(L)(2\sqrt{2}L)}

cos\theta=\frac{5L^2}{4\sqrt{2}L^2}

cos\theta=\frac{5}{4\sqrt{2}}

\theta=0.487

cos\alpha=\frac{(2L)^2+(2\sqrt{2}L)^2-L^2}{(2\sqrt{2}L)(2L)}

cos\alpha=\frac{11L^2}{8\sqrt{2}L^2}

cos\alpha=\frac{11}{8\sqrt{2}}

\alpha=0.236

The green quadrilateral is a kite, which means the diagonals are perpendicular.

This means the segment angles are 2\theta and 2\alpha (because the triangles are isosceles and the diagonal is perpendicular to the base of the triangles).

Area of green segment

A=\frac{1}{2}r^2(\theta-sin\theta)

A=\frac{1}{2}L^2(0.9734-sin(0.9734))=0.0733L^2

Area of yellow segment

A=\frac{1}{2}4L^2(0.4721-sin(0.4721))=0.0035

Total Area =0.0733L^2+0.0035L^2=0.108L^2

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September 24, 2024 · 3:30 am

Optimisation

An optimisation question from the 2019 ATAR Mathematics Methods exam.

I always like optimisation questions. There is a nice process to follow:

  • Find the function to optimise (in terms of one variable).
  • Find the stationary points.
  • Find the nature of the stationary points.
  • Find the maximum or minimum.
(a) Volume of the cylinder V=\pi r^2h
42=2r+h
h=42-2r
\therefore V_C=\pi r^2(42-2r)
Volume of spherical decorations V_S=\frac{4}{3}\pi( r_s)^3 where r_s=\frac{r}{3}
V_S=\frac{4\pi r^3}{81}
Volume unused space V=\pi r^2(42-2r)-20(\frac{4\pi r^3}{81})
V=2\pi (21r^2-r^3-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{81r^3}{81}-\frac{40r^3}{81})
V=2\pi (21r^2-\frac{121r^3}{81})

(b) V=2\pi (21r^2-\frac{121r^3}{81})
\frac{dV}{dr}=2\pi (42r-\frac{121r^2}{27})
\frac{dV}{dr}=0
0=42r-\frac{121r^2}{27}
0=r(42-\frac{121r}{27})
r=0 or r=\frac{1134}{121}=9.372

\frac{d^2V}{dr^2}=2\pi (42-\frac{242r}{27})
(\frac{d^2V}{dr^2})_{|r=9.372}=-42
\therefore r=9.372 is a maximum.

Dimensions of the vase, internal diameter=18.7cm internal height=23.3cm

(c) Maximum volume of empty space =2\pi (21r^2-\frac{121r^3}{81})=3863.08cm^3
Volume of one sphere =\frac{4}{3}\pi r^3=3448.03cm^3

There is enough unused space for one extra decoration, but it would depend on how they are packed.

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September 23, 2024 · 3:11 am

Cats and Dogs

In my town 10% of the dogs think they are cats and 10% of the cats think they are dogs. All the other cats and dogs are perfectly normal. When all the cats and dogs in my town were rounded up and subjected to a rigorous test, 20% of them thought they were cats. What percentage of them really were cats?
Hamilton Olympiad 2003 B4 – The Ultimate Mathematical Challenge

Let x be the number of cats and y be the number of dogs.
Then 0.9x+0.1y think they are cats.
But we also know 20% of the total think they are cats.
0.2(x+y)
Therefore, 0.9x+0.1y=0.2(x+y)
0.9x+0.1y=0.2x+0.2y
0.7x=0.1y
7x=y
Percentage of cats is \frac{x}{x+y}\times100
Substitute 7x for y
\frac{x}{x+7x}\times100=\frac{x}{8x}\times100=\frac{1}{8}\times100=12.5%
\therefore 12.5% of the animals are cats

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September 19, 2024 · 9:43 am

Solving Cubic Functions

I have been thinking about cubics a bit lately because some of my students are solving and then sketching cubics. Plus I am reading An Imaginary Tale by Paul Nahin, which talks about solving cubics and complex numbers.

Cubics must have at least one real root. If one of the roots is a rational number, then we can use the Factor and Remainder Theorem.

For example,

Solve 2x^3-3x^2-3x+2=0

We know the root(s) must be a factor of 2\times3=6.
I always start with 1 or -1
2(1)^3-3(1)^2-3(1)+2=2-3-3+2=-2 \therefore x\neq=1
Try -1
2(-1)^3-3(-1)^2-3(-1)+2=-2-3+3+2=0 \therefore x=-1 and (x+1) is a factor.
Then we can do polynomial long division.

Now we know that 2x^3-3x^2-3x+2=(x+1)(2x^2-5x+2)
And we can factorise the quadratic (or using the quadratic equation formula)
2x^2-5x+2=2x^2-4x-x+2
=2x(x-2)-1(x-2)
=(2x-1)(x-2)
\therefore x=-1, \frac{1}{2}, 2

But what if it is not factorisable?

For example,

Solve 2x^3+5x^2-2x+4=0

How many roots does this equation have?

We could find the derivative and find out how many stationary points the function has.

f'(x)=6x^2+10x-2

This is a quadratic function. Find the discriminant to determine the number of roots.

\Delta=b^2-4ac=100-4(6)(-2)=148

As \Delta>0, there are two stationary points, which means we could have 1, 2 (one root is repeated) or three roots, depending on if the function crosses the x-axis between stationary points. So not much use.

We could try the discriminant of a cubic.

\Delta=18abcd-4b^3d+b^2c^2-4ac^2-27a^2d^2

\Delta=18(2)(5)(-2)(4)-4(5^3)(4)+(5^2)(-2)^2-4(2)(-2)^2-27(2)^2(4)^2=-5100

The discriminant is negative so there is one real root.

From my reading, we need to turn the cubic into a depressed cubic (cubics of the form x^3+px+q=0).

We can do this by using a change of variable.

Let x=t-\frac{5}{6}
2(x^3+\frac{5}{2}x^2-x+2)=0
\therefore x^3+\frac{5}{2}x^2-x+2=0
Substitute t-\frac{5}{6} into the cubic.
(t-\frac{5}{6})^3+\frac{5}{2}(t-\frac{5}{6})^2-(t-\frac{5}{6})+2
(t^3-3(\frac{5}{6})t^2+3(\frac{5}{6})^2t-(\frac{5}{6})^3+\frac{5}{2}(t^2-2(\frac{5}{6})t+(\frac{5}{6})^2)-t+\frac{5}{6}+2
t^3-\frac{5}{2}t^2+\frac{25}{12}t-\frac{125}{216}+\frac{5}{2}t^2-\frac{25}{6}t+\frac{125}{72}-t+\frac{5}{6}+2
t^3-\frac{37}{12}t+\frac{431}{108}

We can then use Cardano’s formula

x=\sqrt[3]{-\frac{q}{2}+\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}+\sqrt[3]{-\frac{q}{2}-\sqrt{(-\frac{q}{2})^2+(\frac{p}{3})^3}}

p=-\frac{37}{12} and q=\frac{431}{108}
\frac{p}{3}=\frac{-37}{36}
\frac{q}{2}=\frac{431}{216}
(\frac{431}{216})^2+(\frac{-37}{36})^3=\frac{139}{48}
t=\sqrt[3]{\frac{-431}{216}+\sqrt{\frac{139}{48}}}+\sqrt[3]{\frac{-431}{216}-\sqrt{\frac{139}{48}}}
t=-2.21095...
\therefore x=-2.21095...-\frac{5}{6}=-3.044

We can see from the sketch below that there is only one solution and it is about -3.

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