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2009

Tue

Jan

Miss Loi’s Wheel of Misfortune

(9)

Tuition given in the topic of A-Maths Tuition Questions from the desk of Miss Loi at 10:37 pm (Singapore time)

Updated on 12 September 2016

Now that the euphoria has settled somewhat, it’s time for Miss Loi to catch up on her new year shopping with just a week left to Chinese New Year.

Comprehensively out-muscled and out-jostled by the Ip Man-trained aunties in a packed Chinatown the year before, she decided to do her shopping at somewhere less competitive this time round.

Scenes from a pasar malam – courtesy of hot blogger Molemole

And so she found herself shopping tonight at a fun fair cum pasar malam, set in a vast field in the middle of a HDB estate somewhere in the northern part of Singapore – complete with smoky food stalls, tikam-tikam stalls, fun rides, ~~pirated DVDs,~~ and disturbing new year songs that went on for eternity.

Laden with boxes of bak kwa, candy and other new year goodies an hour later, she sat down with a cup of sugar cane juice for a much-needed breather, and to reflect on her tragic failure to win that pinky fluffy soft toy at the tikam-tikam stall.

It was then she spotted the entrance to the huge ferris wheel of the fair, with a large poster touting it as “The Largest Ferris Wheel in the North!”, “Taller Than A HDB Flat!”, and “Visitors will be treated to spectacular views of the northern skyline – including key landmarks like the Causeway and JB‘s magnificent cityscape!”, among other things.

No, not this one

Intrigued by the poster (and the possibility of checking if the Causeway was jammed), Miss Loi quickly gulped down her remaining sugar cane juice, bought a ticket and hopped into an empty carriage.

Surrendering herself to the subliminal effects of the endless new year songs, an oh-so-breathtaking sight unfolded as she soared high into the sky. Rising above the glitzy lights of the HDB neighbourhood and watching the aunties below receding into little dots like those in a dot diagram, she was swept away by stunning views of majestic landmarks like BLK 107, BLK 359, and the multi-storey carpark where she had parked her car.

Just when her carriage reached the top of the wheel, however, the ferris wheel suddenly jerked to a stop and (miracle of miracles) the new year songs actually stopped. She looked down to see the wheel’s operator instructing her to remain in her carriage while he looked for help.

And an obedient Miss Loi did as she was told, and it was then she realized to her dismay that she had left her boxes of bak kwa on the ground before she came on board …

[6 hours later] …

Cold and hungry from the night winds, Miss Loi was awakened by someone calling out to her from a loud hailer:

HELLO! I’M FROM THE FERRIS WHEEL RESCUE TEAM! THERE HAS BEEN A TECHNICAL GLITCH!

He continued …

SORRY WE TOOK SO LONG TO COME AS WE WERE BUSY WITH ANOTHER PROBLEMATIC WHEEL IN THE SOUTH! WE TRIED TO REACH YOU BY A LADDER BUT IT TURNED OUT TO BE TOO SHORT!

-_-

THE ONLY WAY IS TO USE A ROPE TO LOWER YOU TO THE GROUND BUT UNFORTUNATELY, I FAILED MY O-LEVEL MATHS LAST TIME AND NOW I DUNNO HOW TO CALCULATE THE LENGTH OF ROPE REQUIRED!
OH WAIT! YOU’RE THAT FAMOUS MATHS TUTOR FROM THE SUNDAY TIMES! SO YOU MUST HELP ME TO SAVE YOU!

Mustering what was left of her mathematical powers, a weakened Miss Loi could only “see” the following:

Using the ground as the x-axis, the circular ferris wheel is centered at (0, 6). A ladder left lying on the wheel can be described using the equation x + y = 10 and is tangent to the wheel.

If Miss Loi’s carriage is right at the top of the wheel, find the length of rope needed to lower her to the ground.

As she’s beginning to feel the effects of that massive gulp of sugar cane juice, can you please help Miss Loi before she has to do the unthinkable?

*Blogged to you live from Miss Loi’s mobile phone, in a cold, lonely carriage x meters high in the sky … *

NOTE: Unfortunately Miss Loi is too weak this time to give more tips or draw a diagram to illustrate further. So it’s up to you to first visualize the situation and brush up on both your basic and AMaths Geometry.

頑張って!!!

Miss Loi

circles, coordinate geometry, geometry, quadratic equations

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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9 Comments

Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
12:55am
1
*Dawn has broken ... a weak voice could be heard coming from that dilapidated carriage swinging gently to the breeze ...*
Miss Loi: In a geometrical problem like this when the person who set it was too lazy (or just being sadistic) to include a diagram, it's always a good practice to draw a quick sketch in order to get your orientation right and know exactly what you're looking for (especially if you're one of those 'visual' type of students):
From the diagram above, it's pretty clear that the total rope length required = 6+r units, since Miss Loi is right at the top of the wheel. So the only unknown is r the radius of the wheel.
P.S. Just a quick sketch will do - please don't waste time drawing Miss Loi's face during your exam!
Miss Loi is really, really glad to see your approach using the following properties from the little Intersection of Line & Curve Leading to a Quadratic Equation section (though it wasn't what Miss Loi had in mind when she set this question - see comment #5 ...) :
Our scenario is a classic problem (with one unknown, r in our case) for this oft-missed portion that's tucked inside your Quadratic Equations chapter, as some students have the impression that discriminants are only used in cases regarding intersections on the x-axis.
(1) $\begin{align*} x+y&=10 \\ y&=10-x \end{align*}$
the equation of the circle is: $(x-0)^2+(y-6)^2=r^2$
(2) $\begin{align*}x^2+(y-6)^2&=r^2\end{align*}$
When solving the 2 simultaneous equations, it's more convenient to sub y=10-x into the circle equation (2) as you've done ( vs equating (1)=(2) ), as the circle equation contains a y².
Sub y=10-x into (2):
$\begin{align*} x^2+(10-x-6)^2&=r^2 \\ x^2+(4-x)^2&=r^2 \\ x^2+4^2-2(4)(x)+x^2&=r^2 \\ x^2+16-8x+x^2&=r^2 \\ 2{x^2}+16-8x&=r^2 \\ 2{x^2}+16-8x-r^2&=0 \\ 2{x^2}-8x+16-r^2&=0 \end{align*}$
since x+y=10 was a tangent, the ${b^2}-4ac=0$
$\begin{align*} {(-8)^2}-4(2)(16-r^2)&=0 \\ 64-8(16-r^2)&=0 \\ 64-(128-8{r^2})&=0 \\ 64-128+8{r^2}&=0 \\ -64+8{r^2}&=0 \\ 8{r^2}-64&=0 \\ r^2-8&=0 \\ r^2&=8 \\ r&=\sqrt{8} \end{align*}$
the ferris wheel is a bit small... but I graphed the equations and it seems valid.
Miss Loi didn't specify the scale of the axes, so each unit could be in meter, feet, km etc 😛 And you used GC!
~~length of rope is 2r, the diameter of the wheel to reach the top of the wheel,~~
$\cancelto{r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}}{2r=2{\sqrt{8}}\approx{5.656854}\approx{5.66}}$ units
Correction: should be r+6 (see comment #2 below) - that's why sketching a quick diagram will likely to help one to visualize the problem better 🙂
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Reply Someone
Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
1:07am
2
whoops...
the length should be $r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}$
please help me to correct my original post. =)
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Reply Someone
Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
1:12pm
3
Why you cancel my working? =(
Miss Loi: ~~Oh please don't be angry at Miss Loi as she is in the process of marking and commenting on your workings so expect some amendment/re-amendment here and there in the meantime.~~
~~Is it ok with you if Miss Loi uses (1), (2) etc. to label the equations in your workings - instead of the ( ) which caused her some confusion in her carriage last night?~~
Okie done.
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Reply Someone
Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
1:35pm
4
oh... that's ok. 🙂
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Reply Someone
Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
水
曜
日
2009
Jan
21
Wed
9:11pm
5
Night has fallen again ... and yet Miss Loi remained in that lonely carriage which was still swinging eeriely to the breeze ...
The ~~KBKP~~ voice from the loud hailer brought an abrupt end to the silence.
THANK YOU SO MUCH SOMEONE FOR YOUR HELP!
BUT I FAILED MY AMATHS AND DROPPED OUT OF SCHOOL BEFORE I REACHED SEC 4! I'VE NOT DONE EQUATION OF CIRCLES BEFORE SO I DON'T FOLLOW YOUR WORKINGS!
AS MY CHER ONCE TOLD ME, CANNOT ANYHOW TRUST ANSWERS WITHOUT UNDERSTANDING THE WORKINGS! SO DO YOU KNOW ANOTHER APPROACH THAT A SEC 3 DROPOUT LIKE ME WILL UNDERSTAND?!
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Reply Miss Loi
Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
9:12pm
6
Yay! I'm correct. Eh... I never used GC... I just graphed my original equation substituted back with $r^2=8$ and $x+y=10$ to make sure.
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Reply Someone
Someone commented in tuition class
水
曜
日
2009
Jan
21
Wed
10:14pm
7
Dawn brings forth another day ... and a faint whisper was heard again from that carriage ...
Yes, we can also get the same answer using Coordinate Geometry, which most of us would've learnt in Sec 3.
From the above diagram, and from your basic Geometrical Properties of Circles, we see that the radius is perpendicular to the tangent line x + y = 10.
So if we can obtain the coordinate of the intersection point between the tangent and the radius, we can then calculate the length of the radius (i.e. between the intersection point (x₁, y₁) and the circle's center (0, 6)) via the Length of Line Segment formula from your E-Maths Coordinate Geometry:
$r=\sqrt{{(x_1-0)^2}+{(y_1-6)^2}}$
(1) $\begin{align*} x+y=10 \\ y=10-x \end{align*}$
This is the equation of the tangent
A line passing through (0, 6) will intersect the line $x+y=10$ at right angles; perpendicular tangent to a circle. Yup yup - tangent ⊥ radius AMEN
hence
$\begin{align*} {m_1}{m_2}&=-1 \\ m{\times}(-1)&=-1 \\ m&=1 \end{align*}$
a line of gradient 1 passing through (0, 6) is:
$\begin{align*} y&=(1)(x)+6 \\ y&=x+6 \end{align*}$
Yes since:
1. The radius ⊥ tangent ⇒ gradient of radius = 1 (∵m₁m₂=-1)
2. The y-intercept = 6 since the circle is centered at (0, 6)
∴ equation of the radius:
(2) $\begin{equation*}y = x + 6\end{equation*}$
finding the point of intersection between $x+y=10$ and $y=x+6$
Equating (1) and (2) to obtain (x₁, y₁)
$\begin{align*} 10-x&=x+6 \\ 10&=x+6+x \\ 10&=2x+6 \\ 10-6&=2x \\ 4&=2x \\ 2x&=4 \\ x&=\frac{4}{2}=2 \\ y&=2+6=8 \end{align*}$
The point of intersection is ( 2, 8 )
$L=\sqrt{{(x_2-x_1)^2}+{(y_2-y_1)^2}}$
find r, the distance from ( 0, 6 ) to ( 2, 8 )
$\begin{align*} r&=\sqrt{{(2-0)^2}+{(8-6)^2}}\\&=\sqrt{{2^2}+{2^2}}=\sqrt{4+4}\\&=\sqrt{8} \\ r+6&=\sqrt{8}+6\approx{8.8284271}\approx{8.83} \end{align*}$
Yes, and you don't even need to know the equation of the circle to get the answer!
I was thinking of using implicit differentiation at first too... Haha.
Don't stress our poor Sec Three rescuer (with A-Level stuff) already lah 😛 Miss Loi hasn't eaten in two days!
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Reply Someone
Someone commented in tuition class
木
曜
日
2009
Jan
22
Thu
5:54pm
8
Haha... It just seemed obvious to me to use the equation of the circle; it was the first way to form a relationship of r with x+y=10 , although it is simpler to use the geometrical property of tangents. Sorry about my strange ( ) notation, its just something I use to track the balancing of my equations. Lol...
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Reply Someone
Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class
月
曜
日
2009
Jan
26
Mon
7:52pm
9
And so the rescue team finally understood the workings and reached the carriage with 6+√8 units of rope, at the very moment another wheel began turning again.
To their surprise the carriage was empty when they opened it, save a note left on the seat that said
Now that you've understood the many concepts behind this problem (i.e. intersection of line and curve, gradients of perpendicular lines etc.), do note the circumstances in which they were used and try to look out for opportunities to apply them again in your future geometrical questions, where appropriate.
Meanwhile, 新年快乐!
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