Comments on: Novena Under Water! http://www.exampaper.com.sg/questions/a-maths/novena-under-water 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/novena-under-water#comment-6929 Miss Loi Tue, 08 Apr 2008 16:15:05 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6929 <b>Someone:</b> Done. Your wish is Miss Loi's command :) Regarding the starting time in the question, Miss Loi actually meant it to be from the point the phone call is made (hence the word <em>initial</em> in the question). But having reread the question, she found that this can be open to different interpretation so she should just state this clearly the next time :P. In any case, it'll still overflow regardless! Someone: Done. Your wish is Miss Loi's command :)

Regarding the starting time in the question, Miss Loi actually meant it to be from the point the phone call is made (hence the word initial in the question). But having reread the question, she found that this can be open to different interpretation so she should just state this clearly the next time :P .

In any case, it'll still overflow regardless!

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6919 Someone Tue, 08 Apr 2008 13:28:09 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6919 Could you help insert my correction into my original post? It will make more sense to readers. Haha. Could you help insert my correction into my original post? It will make more sense to readers. Haha.

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6902 Someone Tue, 08 Apr 2008 08:14:35 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6902 I was under the impression it was 70cm after 10 minutes from -10 min as the start, with 50 minutes counting down from the 70cm mark. Haha... I was under the impression it was 70cm after 10 minutes from -10 min as the start, with 50 minutes counting down from the 70cm mark. Haha...

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6898 Miss Loi Tue, 08 Apr 2008 07:23:47 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6898 <em>Rate of Change</em> questions are pretty straightforward and will almost always involve the <a rel="external nofollow" href="http://en.wikipedia.org/wiki/Chain_rule" rel="nofollow">Chain Rule</a> i.e. given an expression <var>y</var> = f(<var>x</var>), <m>dy/dt = {dy/dx} * {dx/dt}</m> Most Rate of Change questions in your Ten-Year Series are quite kind to state explicitly the variables involved and their relationship i.e. <var>y</var> = f(<var>x</var>) But others, like this one, requires you to <em>read</em> the question carefully to determine the variables and their relationship. 1) So from the question, you're supposed to find the rate of change of the <em>height</em> (<var>h</var>) of the water level i.e. <m>dh/dt</m> 2) But you're given that the <em>width</em> of the water level (<var>w</var>) is lengthening <em class="highlight">constantly</em>, which means you can obtain <m>dw/dt</m> by simply <m>dw/dt = {(70-60)cm}/{10min} = 1 cm min^-1</m> <div class="attention"><strong class="highlight">Note:</strong> You can only divide this way when the rate of change is explicitly stated to be constant!</div> 3) But from the Chain Rule <m>dh/dt = {dh/dw} * {dw/dt}</m> we'll still need to form the expression for the relationship between <var>h</var> and <var>w</var> in order to obtain <m>dh/dw</m>. So ... <img src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/04/miss-loi-rate-of-change-solution-diagram.gif" alt="Miss Loi's Rate of Change Solution Diagram" /> See the two little right-angled triangles formed at the sides? Taking one of this triangles to form the following similarity ratio, can you see that: <m>x/25 = h/100</m> <m>doubleright x = h/4</m> So at any given time, the water level <var>w</var> can be expressed by: <m>w = 50 + 2x = 50 + 2(h/4) = 50 + h/2</m> ---- (1) and viola we have you relationship between <var>h</var> and <var>w</var>! So <m>dw/dh = 1/2</m> <m>doubleright dh/dt = {1/{dw/dh}} * {dw/dt} = 2 * 1 = 2 cm min^-1</m> which is exactly the same answer as <b>Someone</b>'s! Which means that the height of the water is rising at 2 cm per minute, and in 50 mins it would have risen by 50 x 2 = 100 cm. And since it's given that the initial width is 60cm, from (1) ⇒ initial height = (60-50) x 2 = 20 cm ⇒ height = (20 + 100) cm = 120 cm in 50 mins! ⇒ OVERFLOW! ⇒ It was REALLY scary driving the car through the flood! Miss Loi felt like she was driving a boat! Rate of Change questions are pretty straightforward and will almost always involve the Chain Rule i.e. given an expression y = f(x),

dy/dt = {dy/dx} * {dx/dt}

Most Rate of Change questions in your Ten-Year Series are quite kind to state explicitly the variables involved and their relationship i.e. y = f(x)

But others, like this one, requires you to read the question carefully to determine the variables and their relationship.

1) So from the question, you're supposed to find the rate of change of the height (h) of the water level i.e. dh/dt

2) But you're given that the width of the water level (w) is lengthening constantly, which means you can obtain dw/dt by simply

dw/dt = {(70-60)cm}/{10min} = 1 cm min^-1

Note: You can only divide this way when the rate of change is explicitly stated to be constant!

3) But from the Chain Rule dh/dt = {dh/dw} * {dw/dt} we'll still need to form the expression for the relationship between h and w in order to obtain dh/dw.

So ...

Miss Loi's Rate of Change Solution Diagram

See the two little right-angled triangles formed at the sides?

Taking one of this triangles to form the following similarity ratio, can you see that:

x/25 = h/100
doubleright x = h/4

So at any given time, the water level w can be expressed by:

w = 50 + 2x = 50 + 2(h/4) = 50 + h/2 ---- (1)

and viola we have you relationship between h and w!

So dw/dh = 1/2
doubleright dh/dt = {1/{dw/dh}} * {dw/dt} = 2 * 1 = 2 cm min^-1

which is exactly the same answer as Someone's!

Which means that the height of the water is rising at 2 cm per minute, and in 50 mins it would have risen by 50 x 2 = 100 cm.

And since it's given that the initial width is 60cm, from (1)
⇒ initial height = (60-50) x 2 = 20 cm
⇒ height = (20 + 100) cm = 120 cm in 50 mins!
⇒ OVERFLOW!

⇒ It was REALLY scary driving the car through the flood! Miss Loi felt like she was driving a boat!

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6836 Someone Mon, 07 Apr 2008 11:37:49 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6836 Please pardon me if I have any errors, I'm kind of tired now. Haha. Please pardon me if I have any errors, I'm kind of tired now. Haha.

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6835 Someone Mon, 07 Apr 2008 11:37:12 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6835 <img class="right" src="http://www.exampaper.com.sg/blog/wp-content/uploads/2008/04/someone-rate-of-change-solution-diagram.gif" alt="Someone's Rate of Change Solution Diagram" /> Considering the full triangle, with x as the 'chopped off' length, by similar triangles, [pmath] {100/50}={100+x}/x[/pmath] [pmath] 2={100+x}/x[/pmath] [pmath] 2x=100+x[/pmath] [pmath] x=100[/pmath] w is the width at any particular time, by similar triangles, [pmath] w/{h+x}=100/200[/pmath] [pmath] w/{h+100}=1/2[/pmath] [pmath] 2w=h+100[/pmath] [pmath] 2w-100=h[/pmath] [pmath] h=2w-100[/pmath] [pmath] dw/dt={70-60}/10=1[/pmath] [pmath] dh/dw = 2[/pmath] [pmath]dh/dt={dh/dw}*{dw/dt}=2*1=2[/pmath] In 50 minutes, the water will have risen by another <del>25cm</del> 100cm; from an initial [pmath] h=2w-100=2(70)-100=40[/pmath], To give a final height of <del>40+25=65 cm; which is less than 100</del> 40+2(50)=140>100, overflows <div class="highlight"> Once again Miss Loi has inserted a diagram for readers to better understand your solution - especially on the whereabouts of your 'chopped-off length' <var>x</var> :P For your solution Miss Loi was just about to pounce on you until she saw your next comment! The Chain Rule and similar triangles are definitely required to solve this Rate of Change question (<b>Papillion</b>'s epic shortcut nowithstanding) :P :P :P Miss Loi's similar triangles are a little different - though the final answer is the same, but once again it's great to be able to highlight that there's often more than one way to solve the same question :D Miss Loi's solution coming right up ... P.S. Thought the initial water level width was 60cm (not 70cm)? Hence initial <var>h</var> = 2(60)-100 = 20. CARELESS MISTAKE! Tsk tsk. </div> Someone's Rate of Change Solution Diagram

Considering the full triangle, with x as the 'chopped off' length, by similar triangles,

{100/50}={100+x}/x
2={100+x}/x
2x=100+x
x=100

w is the width at any particular time, by similar triangles,

w/{h+x}=100/200
w/{h+100}=1/2
2w=h+100
2w-100=h
h=2w-100
dw/dt={70-60}/10=1
dh/dw = 2
dh/dt={dh/dw}*{dw/dt}=2*1=2

In 50 minutes, the water will have risen by another 25cm 100cm; from an initial h=2w-100=2(70)-100=40,

To give a final height of 40+25=65 cm; which is less than 100

40+2(50)=140>100,

overflows

Once again Miss Loi has inserted a diagram for readers to better understand your solution - especially on the whereabouts of your 'chopped-off length' x :P For your solution Miss Loi was just about to pounce on you until she saw your next comment!

The Chain Rule and similar triangles are definitely required to solve this Rate of Change question (Papillion's epic shortcut nowithstanding) :P :P :P

Miss Loi's similar triangles are a little different - though the final answer is the same, but once again it's great to be able to highlight that there's often more than one way to solve the same question :D

Miss Loi's solution coming right up ...

P.S. Thought the initial water level width was 60cm (not 70cm)? Hence initial h = 2(60)-100 = 20. CARELESS MISTAKE! Tsk tsk.

]]> By: ignorantsoup http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6821 ignorantsoup Mon, 07 Apr 2008 05:33:13 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6821 hmm i assumed that the amount of water that drops inside is constant..and hence, the amount of area occupied per unit time should be the same? lol..niwae just did that for fun whilst doing my C++.. hmm i assumed that the amount of water that drops inside is constant..and hence, the amount of area occupied per unit time should be the same? lol..niwae just did that for fun whilst doing my C++..

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6808 Miss Loi Mon, 07 Apr 2008 03:39:01 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6808 Haha <b>Papillion</b>, My bad my bad ... in her neverending quest for the dramatic, Miss Loi was too complacent to see this obvious shortcut coming from the blindside :) Guess you already know that this is a question on <em>Rate of Change</em>, so she should better rephrase to the following: <div class="attention">Find the rate at which the height of the water in the drain is rising, and <em>hence</em> determine if the height of the water in the drain will overflow within 50 mins.</div> Oh and with the weather these days, please don't even predict how the rain will behave! Bring on your <strong class="highlight">Chains</strong>! Haha Papillion,

My bad my bad ... in her neverending quest for the dramatic, Miss Loi was too complacent to see this obvious shortcut coming from the blindside :)

Guess you already know that this is a question on Rate of Change, so she should better rephrase to the following:

Find the rate at which the height of the water in the drain is rising, and hence determine if the height of the water in the drain will overflow within 50 mins.

Oh and with the weather these days, please don't even predict how the rain will behave!

Bring on your Chains!

]]> By: La Papillion http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6806 La Papillion Mon, 07 Apr 2008 03:21:25 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6806 Hmm, I have two methods for this method, I'll just show the simpler one, though frankly, i'm also not too sure :) Here it goes: Let x be the width of the water level, with 60cm as the starting and 100cm being the max level. Assume that the change of x with respect to time is constant (though it's weird...cos it means that the rain must be getting progressively heavier and heavier, not exactly what I can imagine when the meteorological station says that the rain will stop in 50 mins time...oh well) 60 to 70 cm takes 10 min 70 to 80 cm takes 10 min 80 to 90 cm takes 10 min 90 to 100 cm takes 10 min In total, it'll take 40 mins for x to reach 100 cm. Another longer than that, the water will over flow out. My spidey sense is tingling...so Miss Loi pls point out any wrong assumptions I've made, thks! Hmm, I have two methods for this method, I'll just show the simpler one, though frankly, i'm also not too sure :)

Here it goes:

Let x be the width of the water level, with 60cm as the starting and 100cm being the max level. Assume that the change of x with respect to time is constant (though it's weird...cos it means that the rain must be getting progressively heavier and heavier, not exactly what I can imagine when the meteorological station says that the rain will stop in 50 mins time...oh well)

60 to 70 cm takes 10 min
70 to 80 cm takes 10 min
80 to 90 cm takes 10 min
90 to 100 cm takes 10 min

In total, it'll take 40 mins for x to reach 100 cm. Another longer than that, the water will over flow out.

My spidey sense is tingling...so Miss Loi pls point out any wrong assumptions I've made, thks!

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6794 Miss Loi Mon, 07 Apr 2008 00:35:17 +0000 http://www.exampaper.com.sg/questions/a-maths/novena-under-water#comment-6794 Wah <b>Soup</b> ... Miss Loi only mentioned that the <em>width</em> of the water level (let's call this <var>x</var>) is lengthening (ok let's be more specific here)/changing at a constant rate i.e. <m>{dx}/{dt}</m> is constant. But by simply dividing the change in <em>area</em> (<var>A</var>) (which is in another dimension) by 10 mins you are implying that <m>{dA}/{dt}</m> is also constant. Are you absolutely sure? Wah Soup ... Miss Loi only mentioned that the width of the water level (let's call this x) is lengthening (ok let's be more specific here)/changing at a constant rate i.e. {dx}/{dt} is constant.

But by simply dividing the change in area (A) (which is in another dimension) by 10 mins you are implying that {dA}/{dt} is also constant.

Are you absolutely sure?

]]>