Comments on: Miss Loi’s Wheel of Misfortune http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20977 Miss Loi Mon, 26 Jan 2009 11:52:22 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20977 And so the rescue team finally understood the workings and reached the carriage with <strong>6+√8 units</strong> of rope, at the very moment <a rel="external nofollow" href="http://www.singaporeflyer.com/en/latest-announcements/singapore-flyer-to-fly-high-once-again.html" rel="nofollow">another wheel began turning again</a>. To their surprise the carriage was empty when they opened it, save a note left on the seat that said <blockquote> 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, <strong class="highlight big">新年快乐!</strong> </blockquote> 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!

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20771 Someone Thu, 22 Jan 2009 09:54:59 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20771 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... 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...

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20727 Someone Wed, 21 Jan 2009 14:14:46 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20727 <div class="highlight"> Dawn brings forth another day ... and a faint whisper was heard again from that carriage ... Yes, we can also get the same answer using <a href="/tag/coordinate-geometry" rel="nofollow">Coordinate Geometry</a>, which most of us would've learnt in Sec 3. <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/01/ferris-wheel-coordinate-geometry-solution-diagram.gif" alt="Solution Diagram using Coordinate Geometry" /> From the above diagram, and from your basic <a href="/tag/geometric-properties-of-circles" rel="nofollow">Geometrical Properties of Circles</a>, we see that the radius is perpendicular to the tangent line <var>x</var> + <var>y</var> = 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 (<var>x</var><sub>1</sub>, <var>y</var><sub>1</sub>) and the circle's center (0, 6)) via the <em class="highlight">Length of Line Segment</em> formula from your E-Maths <a href="/tag/coordinate-geometry" rel="nofollow">Coordinate Geometry</a>: [tex]r=\sqrt{{(x_1-0)^2}+{(y_1-6)^2}}[/tex] </div> [tex]x+y=10[/tex] [tex]y=10-x[/tex] <span class="highlight"> ---------- (1) This is the equation of the tangent</span> A line passing through (0, 6) will intersect the line [tex]x+y=10[/tex] at right angles; perpendicular tangent to a circle. <span class="highlight">Yup yup - tangent ⊥ radius AMEN</span> hence [tex]{m_1}{m_2}=-1[/tex] [tex]m{\times}(-1)=-1[/tex] [tex]m=1[/tex] a line of gradient 1 passing through (0, 6) is: [tex]y=(1)(x)+6[/tex] [tex]y=x+6[/tex] <div class="highlight"> Yes since: <ol> <li>The radius ⊥ tangent ⇒ gradient of radius = 1 (∵<var>m</var><sub>1</sub><var>m</var><sub>2</sub>=-1)</li> <li>The <var>y</var>-intercept = 6 since the circle is centered at (0, 6) </ol> ∴ equation of the radius: <var>y</var> = <var>x</var> + 6 --------- (2) </div> finding the point of intersection between [tex]x+y=10[/tex] and [tex]y=x+6[/tex] [tex]10-x=x+6[/tex] <span class="highlight">Equating (1) and (2) to obtain (<var>x</var><sub>1</sub>, <var>y</var><sub>1</sub>)</span> [tex]10=x+6+x[/tex] [tex]10=2x+6[/tex] [tex]10-6=2x[/tex] [tex]4=2x[/tex] [tex]2x=4[/tex] [tex]x=\frac{4}{2}=2[/tex] [tex]y=2+6=8[/tex] The point of intersection is ( 2, 8 ) [tex]L=\sqrt{{(x_2-x_1)^2}+{(y_2-y_1)^2}}[/tex] find r, the distance from ( 0, 6 ) to ( 2, 8 ) [tex]r=\sqrt{{(2-0)^2}+{(8-6)^2}}=\sqrt{{2^2}+{2^2}}=\sqrt{4+4}=\sqrt{8}[/tex] [tex]r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}[/tex] <span class="highlight">Yes, and you don't even need to know the equation of the circle to get the answer!</span> I was thinking of using implicit differentiation at first too... Haha. <span class="highlight">Don't stress our poor Sec Three rescuer (with A-Level stuff) already <i>lah</i> :P Miss Loi hasn't eaten in two days!</span>

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.

Solution Diagram using Coordinate Geometry

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}}

x+y=10
y=10-x ---------- (1) 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
{m_1}{m_2}=-1
m{\times}(-1)=-1
m=1

a line of gradient 1 passing through (0, 6) is:
y=(1)(x)+6
y=x+6

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: y = x + 6 --------- (2)

finding the point of intersection between x+y=10 and y=x+6
10-x=x+6 Equating (1) and (2) to obtain (x1, y1)
10=x+6+x
10=2x+6
10-6=2x
4=2x
2x=4
x=\frac{4}{2}=2
y=2+6=8
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 )
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}
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 :P Miss Loi hasn't eaten in two days!

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20724 Someone Wed, 21 Jan 2009 13:12:59 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20724 Yay! I'm correct. Eh... I never used GC... I just graphed my original equation substituted back with [tex]r^2=8[/tex] and [tex]x+y=10[/tex] to make sure. 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.

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20723 Miss Loi Wed, 21 Jan 2009 13:11:52 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20723 Night has fallen again ... and yet Miss Loi remained in that lonely carriage which was still swinging eeriely to the breeze ... The <del>KBKP</del> voice from the loud hailer brought an abrupt end to the silence. <blockquote> THANK YOU SO MUCH <b>SOMEONE</b> 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 <strong>UNDERSTANDING</strong> THE WORKINGS! SO DO YOU KNOW ANOTHER APPROACH THAT A SEC 3 DROPOUT LIKE ME WILL UNDERSTAND?! </blockquote> 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?!

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20705 Someone Wed, 21 Jan 2009 05:35:13 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20705 oh... that's ok. :) oh... that's ok. :)

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20703 Someone Wed, 21 Jan 2009 05:12:57 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20703 Why you cancel my working? =( <div class="highlight"> Miss Loi: <del>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.</del> <del>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?</del> Okie done. </div> 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.

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20673 Someone Tue, 20 Jan 2009 17:07:44 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20673 whoops... the length should be [tex]r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}[/tex] please help me to correct my original post. =) whoops...
the length should be r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}
please help me to correct my original post. =)

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20672 Someone Tue, 20 Jan 2009 16:55:56 +0000 http://www.exampaper.com.sg/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20672 <div class="highlight"> *Dawn has broken ... a weak voice could be heard coming from that dilapidated carriage swinging gently to the breeze ...* <strong>Miss Loi:</strong> 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 <em>exactly</em> what you're looking for (especially if you're one of those 'visual' type of students): <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/01/ferris-wheel-diagram.gif" alt="Ferris Wheel Question Diagram" /> From the diagram above, it's pretty clear that the total rope length required = 6+<var>r</var> units, since Miss Loi is right at the top of the wheel. So the only unknown is <var>r</var> 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 <em class="highlight">Intersection of Line & Curve Leading to a Quadratic Equation</em> section (though it wasn't what Miss Loi had in mind when she set this question - see <a href="/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20723" rel="nofollow">comment #5</a> ...) : <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2009/01/line-curve-intersection-qudratic-equation.gif" alt="Discriminants For Intersection of A Line & A Curve" /> Our scenario is a classic problem (with one unknown, <var>r</var> in our case) for this oft-missed portion that's tucked inside your <a href="/tag/quadratic-equations" rel="nofollow">Quadratic Equations</a> chapter, as some students have the impression that discriminants are only used in <a href="/questions/a-maths/quadratic-equations-so-easy-my-grandma-cant-do-it" rel="nofollow">cases regarding intersections on the <var>x</var>-axis</a>. </div> [tex]x+y=10[/tex] [tex]y=10-x[/tex] ---------- (1) the equation of the circle is: [tex](x-0)^2+(y-6)^2=r^2[/tex] [tex]x^2+(y-6)^2=r^2[/tex] ---------- (2) <span class="highlight">When solving the 2 simultaneous equations, it's more convenient to sub <var>y</var>=10-<var>x</var> into the circle equation (2) as you've done ( vs equating (1)=(2) ), as the circle equation contains a <var>y</var><sup>2</sup>.</span> Sub <var>y</var>=10-<var>x</var> into (2): [tex]x^2+(10-x-6)^2=r^2[/tex] [tex]x^2+(4-x)^2=r^2[/tex] [tex]x^2+4^2-2(4)(x)+x^2=r^2[/tex] [tex]x^2+16-8x+x^2=r^2[/tex] [tex]2{x^2}+16-8x=r^2[/tex] [tex]2{x^2}+16-8x-r^2=0[/tex] [tex]2{x^2}-8x+16-r^2=0[/tex] since x+y=10 was a tangent, the [tex]{b^2}-4ac=0[/tex] [tex]{(-8)^2}-4(2)(16-r^2)=0[/tex] [tex]64-8(16-r^2)=0[/tex] [tex]64-(128-8{r^2})=0[/tex] [tex]64-128+8{r^2}=0[/tex] [tex]-64+8{r^2}=0[/tex] [tex]8{r^2}-64=0[/tex] [tex]r^2-8=0[/tex] [tex]r^2=8[/tex] [tex]r=sqrt{8}[/tex] the ferris wheel is a bit small... but I graphed the equations and it seems valid. <span class="highlight">Miss Loi didn't specify the scale of the axes, so each unit could be in meter, feet, km etc :P And you used <acronym title="Graphical calculator">GC</acronym>!</span> <del class="highlight">length of rope is 2r, the diameter of the wheel to reach the top of the wheel,</del> <del class="highlight">[tex]2r=2{sqrt{8}}\approx{5.656854}\approx{5.66}[/tex] .</del> <span class="highlight">Correction: should be <var>r</var>+6 (see <a href="/questions/a-maths/miss-lois-wheel-of-misfortune#comment-20673" rel="nofollow">comment #2</a> below) - that's why sketching a quick diagram will likely to help one to visualize the problem better :)</span> [tex]r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83}[/tex] <strong class="highlight">units</strong>

*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):

Ferris Wheel Question Diagram

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 ...) :

Discriminants For Intersection of A Line & A Curve

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.

x+y=10
y=10-x ---------- (1)

the equation of the circle is: (x-0)^2+(y-6)^2=r^2

x^2+(y-6)^2=r^2 ---------- (2)

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):

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

since x+y=10 was a tangent, the {b^2}-4ac=0

{(-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} 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 :P And you used GC!

length of rope is 2r, the diameter of the wheel to reach the top of the wheel,
2r=2{sqrt{8}}\approx{5.656854}\approx{5.66} .

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 :)
r+6=\sqrt{8}+6\approx{8.8284271}\approx{8.83} units

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