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STPM PBS Assignment

STPM 2015 Mathematics (T) Term 2 Assignment

By KK LEE

February 10, 2015

Sample question of STPM 2015 Mathematics (T) term 2 coursework assignment question. Click the question button below to download the complete question.

Many school teachers said that my answers are wrong. Please use the answer wisely. 😛

Question

A hyperboloid container of height 14 m is generated by the hyperbola $25x^2-16(y-9)^2=225, 0\leq y\leq 14$ as shown in the diagram below.

Question 1
Initially the container is empty. Water is pumped into the container from the top at a constant rate of $180\ \text{m}^3$ per hour.

Part (a)
Calculate the time taken to fill up the container.

Part (b)
Find an expression for the volume of water in terms of the height of the water level in the container.

Part (c)
Determine the rate of increase of the height of water level in the container when the height is 9 m.

Question 2
In order to avoid overflow, a mechanism is now installed. This mechanism controls the water being pumped into the container at a rate of $k(14-y)^n\ \text{m}^3$ per hour, where k and n are positive constants.

Part (a) (i)
If k=20 and n=1, state, with reasons(s), whether the time taken to fill up the container is longer or shorter than that in part 1(a),

Part (a) (ii)
calculate the time taken to fill up the container.

Part (b)
If k=15 and n=0.5, determine the minimum and the maximum rate of increase of the height of water level in the container.

Below is the sample answer and solution for the STPM 2015 Term 2 Coursework assignment. I will try to discuss the questions and answers in the class.

Answer for Question 1c and 2b are updated. -10th March 2015

Solution 1 (a)

Solution 1 (b)

Solution 1 (c)

Solution 2 (a) (i)

Old Version

Old Version (wrong)

Updated Version

Thanks to Shen Sing and the students who to help me and tell me the problem. The time taken to fill the container should be longer.

Table to show average of the rate of water

Reason: Average of the rate is 140 which is lower than 180.

Solution 2 (a) (ii)

Updated Version

Old version (wrong because ln 0 was counted as 1. =.=!!)

Updated Version

The container cannot be 100% full.

Solution 2 (b)

Download the Term 2 PBS Full solution here. Available for Ai Tuition and Teras Usaha students only. Join my class to get the full solution. Full solution will be printed for my students.

Feel free to leave comment below about your school teacher special requirement to help others. 🙂

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Tags: Differential-Equations Differentiation Integration MT Term 2 STPM 2015 STPM MT

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KK LEE

KK LEE has been a STPM Mathematics tuition teacher since June 2006, but his love of Maths dates back to at least 1999 when he was Form 4. KK LEE started teaching in 2006 at Pusat Tuisyen Kasturi. He was known as "LK" when he was teaching in PTK. After teaching STPM Mathematics for 8 years in PTK, he joined Ai Tuition Centre in 2014. Over the years he has taught Mathematics (T), Mathematics (S), Mathematics (M), Additional Mathematics.

113 Comments

Rishaliney a/p Selva Rajoo on February 24, 2015 at 9:16 pm

Sir, hw to do intro, methodology and conclusion?
Reply
- KK LEE on February 24, 2015 at 10:26 pm
  
  Hi. I do the mathematical parts. U share the intro, methodology, and conclusion parts. Fair? 😀
  
  I will help students on mathematical parts only.
- shen sing on March 10, 2015 at 10:11 pm
  
  sir,for 2 a)i)after y=5, the rate of volume of water pumped is dicreasing already.hence,is it the time will still shorter?
- KK LEE on March 11, 2015 at 2:26 am
  
  The volume of the water in the container for y=5 is more than half of the container. Volume of water when y=5 is 587.06. You can obtain this value from part 1b. Btw when the y approaches 14, the rate is very low compared to 180. I have to think again. 😀
sherrynn on March 10, 2015 at 6:49 pm

sir,where the tuition places that you teaching maths (T)for STPM?
Reply
- KK LEE on March 10, 2015 at 7:16 pm
  
  Hi. I am teaching STPM Maths tuition at Ai Tuition at Jalan Petaling Kuala Lumpur and SS19 Subang Jaya. and Teras Usaha at Segambut
- ngiam kee wei on March 31, 2015 at 7:10 pm
  
  how do find the maximum
  and the minimum
- KK LEE on April 1, 2015 at 12:46 am
  
  In general, use dy/dx=0
Soonnee on March 18, 2015 at 8:40 pm

The average 140 rate of flow is invalid as the dy/dt,the rate of increase of y, is not constant.In my opinion,the time taken to COMPLETELY fill up the container is infinite as the answer has shown above.By the way, thank you sir for sharing these solutions, thank you very much
Reply
- KK LEE on March 18, 2015 at 10:05 pm
  
  yes. that is one of the sample of explanation only.
Wendy on March 19, 2015 at 2:01 pm

Dear sir, for question 1(c),isnt it the V should be 1/75(pi)(y)? Then dv/dy using the uv formula? Thank you sir~
Reply
- KK LEE on March 19, 2015 at 7:10 pm
  
  hi. Can i know which V u mean?
- Wendy on April 8, 2015 at 12:14 am
  
  Sorry sir~I got it now~~Because I had expanded all the brackets and a bit messed up~ thanks
ccm on March 20, 2015 at 6:27 pm

Sir, I can seem to understand the second differentiation you did for 2b. How did you get y= 37/3 plus/minus 5/12 sqaure root 9?
Reply
- KK LEE on March 20, 2015 at 8:15 pm
  
  Please try to use the quadratic formula.
abc96 on March 21, 2015 at 7:33 pm

Hi sir, could you please check again 1(b)? Maybe you overlook the expansion of (y-9)^3.
Reply
- abc96 on March 21, 2015 at 7:40 pm
  
  Opps… its’s my mistake…sry
- KK LEE on March 21, 2015 at 10:35 pm
  
  😀
V on March 22, 2015 at 11:51 pm

Sir, how u can get the dv/dt= 15(14-y)’1/2 ? On question 2 (b)
Reply
- KK LEE on March 22, 2015 at 11:57 pm
  
  Substitute k=15, n=1/2 into the rate provided. dv/dt=k(14-y)^n
- V on March 23, 2015 at 6:51 pm
  
  Thanks sir. How to join your tuition class?
- KK LEE on March 23, 2015 at 7:45 pm
  
  yes. ARe you from kl area. I am teaching at jalan Petaling KL，Subang Jaya and segambut
A yun on March 24, 2015 at 8:54 pm

Sir,What can we write for title??
Reply
- KK LEE on March 24, 2015 at 10:31 pm
  
  Hmm.. Investigating the rate of change of water.
  Cant help you on this. 😛
- A yun on March 26, 2015 at 3:59 pm
  
  Its okey..But anyway thank you,sir..
jega on March 27, 2015 at 2:57 pm

Sir… question 1b and 1c the answer given by my teacher was v=Π/25[1521h-144h^2+16/3h^3
And for c) 6.37m/h… its her skema jawapan… why the answer over here is different sir?
Reply
- jega on March 27, 2015 at 3:13 pm
  
  Sir.. I’m sooooo sorry.. 1c is correct over here.. my mistake sir sorry…
- KK LEE on March 27, 2015 at 4:24 pm
  
  Hi. Can share your marking scheme? 😀
Tan Nee Nee on March 28, 2015 at 2:10 pm

i did not understand why minimum rate is 0
Reply
- KK LEE on March 28, 2015 at 5:55 pm
  
  Minimum value of the rate refers to the smallest value of the rate. And 0 is the value when y=14. Thats why the minimum value is 0. But i heard some teachers said there is no minimum value. Please confirm with your teacher. Thanks. 😀
sree on March 29, 2015 at 12:39 pm

I dun understand the part 2(a)(i)…can u pls help me in it.. tq
Reply
- KK LEE on March 29, 2015 at 6:31 pm
  
  Since the rate is not constant. So I used average of the rate(approximation) to compare with 180
- F. Syaf on May 10, 2015 at 12:15 am
  
  sorry sir, i still cannot understand part 2a(ii), how to write it down?
- KK LEE on May 10, 2015 at 1:39 am
  
  Hi. It depends on your school teacher. For my sample solution, it is impossible to fill up the container to 100.000% full.
- F. Syaf on May 13, 2015 at 8:34 pm
  
  Thank you sir
Angeline on March 29, 2015 at 1:58 pm

Sir, for 2a2, how can u get the dvdy ?
Reply
- KK LEE on March 29, 2015 at 6:32 pm
  
  Differentiate v with respect to y
mj on March 30, 2015 at 10:03 pm

Sir y we can use dy/dT=0 in 2a(b) to get minimum value?
Reply
- KK LEE on March 31, 2015 at 1:09 am
  
  Hi. We can use any methods to find the minimum value of functions.
eyan on March 31, 2015 at 3:02 pm

sir, why do we need to use the average of water? and also how to explain that t=∞? tq
Reply
- KK LEE on March 31, 2015 at 3:06 pm
  
  Since the rate is not constant. So I used average of the rate(approximation) to compare with 180.
  Time taken to fill up the container(y=14.0) is infinity. So you can try to explain from y=13.5, y=13.9, and others
wani on April 2, 2015 at 5:54 pm

hi sir, my teacher’s answer for minimum rate is 0.2936 and maximum is 1.186. why is that? and can we use the hyperbola equation to find the answer for question 2(b) by finding the centre point/axis symetry etc. because this assignment is regarding differentiation and integration..
Reply
- KK LEE on April 2, 2015 at 9:05 pm
  
  You can use any methods. As long as your school teacher accept.
  i will try to check the values 0.2936 and 1.186.
- KK LEE on April 5, 2015 at 8:26 pm
  
  I have no idea to get those values
wani again on April 2, 2015 at 5:57 pm

sir, why got dy/dt in the second differentiation in question 2(b)? we need to find rate for height right why dont we use the volume from 1(b) in term of h?
Reply
- KK LEE on April 2, 2015 at 9:04 pm
  
  because i used implicit differentiation.
  you can use part 1(b)
Boiboi on April 3, 2015 at 10:54 pm

Sir,why Q2b use d2y/dx2 =0? That one is for point of inflection right? Correct me pls
Reply
- KK LEE on April 3, 2015 at 11:05 pm
  
  You can use any methods to find maximum and minimum.
  
  d2y/dx2=0 is used to find point of inflexion of the the curve y=f(x).
  
  At the same time, d2y/dx2=0 can be used to find max/min of dy/dx.
Boiboi on April 4, 2015 at 2:16 pm

Sir,Q2a2 why t=infinity? Coz 1875ln0?
Reply
- KK LEE on April 5, 2015 at 10:56 pm
  
  yes
Boiboi on April 4, 2015 at 2:18 pm

Sir,Q2a2,why substitute y=14,t=infinity?
Reply
- KK LEE on April 5, 2015 at 12:12 am
  
  Please recall that when x approaches 0, ln x approaches -infinity
Shawn on April 5, 2015 at 1:00 pm

sir, when y = 14, isn’t d2y/dt2 =
144y^2 – 3552y +19629 / 2[sqr(14-y)](48y^2-864y+4563)^2 undefined?
Reply
- KK LEE on April 5, 2015 at 3:59 pm
  
  Yes
- Shawn on April 5, 2015 at 5:24 pm
  
  so y = 14 isn’t a solution to d2y/dt2 = 0?
- KK LEE on April 5, 2015 at 6:33 pm
  
  ya
Rachel on April 5, 2015 at 2:03 pm

Why my one fron yours? My ans for 1(b) is 308 14/75.
Reply
- KK LEE on April 5, 2015 at 4:01 pm
  
  sorry. please check again
unknown on April 5, 2015 at 3:37 pm

1) (c) 1/9Π x 180 = 20/Π
Reply
- KK LEE on April 5, 2015 at 3:59 pm
  
  Yes
shirley on April 5, 2015 at 7:26 pm

how (48ypower 2 -864y+4563)over (14-y)wil get -48y+192+(1875/(14-y))? i dont get it
Reply
- KK LEE on April 5, 2015 at 8:24 pm
  
  Long division
Irene SinnYue on April 6, 2015 at 11:01 pm

Hey sir, do you have Maths M pbs? TQ
Reply
- KK LEE on April 7, 2015 at 12:34 am
  
  https://kkleemaths.com/2015/03/16/stpm-2015-mathematics-m-term-2-assignment/
AFIFAH on April 7, 2015 at 9:04 pm

sir, pbs maths t term 1 stpm 2015. tq
Reply
- KK LEE on April 7, 2015 at 11:31 pm
  
  Are you kidding?
- Kalishwaran on May 20, 2015 at 8:06 pm
  
  Sir… Do u have solutions for term 1 assignment?? Im a private candidate.. Need to do it.
- KK LEE on May 21, 2015 at 1:22 am
  
  Hi.I have no sem 1 answer
Boiboi on April 7, 2015 at 10:25 pm

Sir,Q2b, can I substitute when y=0,14 to dy/dt that founded? The 2 answer will equal to the max&min rate?am I right?
Reply
- KK LEE on April 7, 2015 at 11:32 pm
  
  You will get the rates. But are they max or min, u have to compare with other values of y range from 0 to 14.
Raffles on April 9, 2015 at 2:15 am

sir soloution 1c with wrong answer
it must be (1/9pie)x180 then the ans will be6.366m or 2.xx pie
Reply
- Raffles on April 9, 2015 at 2:27 am
  
  sir may i ans 1 more question? why is the average rate in 1)a)1) is (280+260 +….+20)divide by 13 only but not 14?
- KK LEE on April 9, 2015 at 9:27 am
  
  it was written as 6.366 there.
- KK LEE on April 9, 2015 at 9:27 am
  
  280+260+…+20+0 / 15
kks on April 12, 2015 at 7:53 pm

Sir ,in 2(a)ii) why with a mechanism installed the water CNT fill up the container100%?
Reply
- KK LEE on April 12, 2015 at 7:55 pm
  
  from the formula and calculation.
kks on April 12, 2015 at 7:55 pm

Sir,in 2(a)i) why time taken will b longer?just in case teacher ask during presentation
Reply
- KK LEE on April 12, 2015 at 7:58 pm
  
  Average is smaller.
kks on April 12, 2015 at 7:58 pm

Oh..OK,thanks!other than the results frm calculation n formula, is thr any other reason I cn state?
Reply
- KK LEE on April 12, 2015 at 8:00 pm
  
  As i mentioned earlier. I will help on Maths part only. 😀
Assunta Patrick on April 14, 2015 at 8:43 pm

Sir, the answers given by my school teacher:

1a) 308.19 -pie-
b) V=-pie- (9h+16/75 (h-9)^3 + 3888/25
[ 0《h《14]
c) 20/-pie-

2 a)i) shorter
ii) t= -pie-/ 500 [64y -8y^2 + 625 ln [14/ 14-y]]
Reply
- KK LEE on April 15, 2015 at 10:25 am
  
  Thanks.
Alves on April 16, 2015 at 10:44 pm

Best maths T teacher ever! Clap for u! Thanks for helping us!
Reply
- KK LEE on April 16, 2015 at 10:48 pm
  
  You are welcome. 😀
Punitha on April 18, 2015 at 5:43 pm

sir, can u pls check my introduction tht i sent to ur email?? 🙂 tq sir
Reply
- KK LEE on April 18, 2015 at 6:01 pm
  
  checked
w on April 25, 2015 at 11:29 pm

sir, can i know what is that two applications of this mechanism?
Reply
- KK LEE on April 26, 2015 at 12:29 am
  
  Hi. Ask your teacher. 😛
  I will help on maths part only.
- w on April 26, 2015 at 1:00 pm
  
  thank you sir.
che on April 29, 2015 at 7:40 am

sir, what is the main formula to find the time taken to fill the container?
Reply
- KK LEE on April 29, 2015 at 9:31 am
  
  In short, Time=Volume divide rate
- che on April 29, 2015 at 9:47 am
  
  thank you sir.
Joan on May 4, 2015 at 9:37 pm

Hi sir, I wanna ask about the Question 2a(ii). To find the dy/dt,there is a step §(48y²-864y+4563)/14-y dy = §1500/μ
After that d step becomes §-48y+192y+1875/14-y dy… how do sir do this step? I’m a bit confuse.. hope sir can solve my problem sorry sir that I can’t find an integration sign,n I replace it with §.
Reply
- KK LEE on May 4, 2015 at 11:44 pm
  
  long division
- Joan on May 5, 2015 at 12:12 am
  
  I see!! Thank you sir!!
connie on May 5, 2015 at 6:04 pm

sir i just want to ask that , can you give me some idea to write the introduction ? i have no idea how to start.
Reply
- KK LEE on May 5, 2015 at 11:15 pm
  
  Try to explain history of Tupperware water container. 😛
mumu on May 22, 2015 at 6:12 pm

hi,sir,may i noe wat is the title of this sem 2 coursework?
Reply
- KK LEE on May 22, 2015 at 7:09 pm
  
  Hi. I cant help you for this title. Sorry
- GY on May 26, 2015 at 11:32 pm
  
  Sir,may I knw tat..why 1.225 is a maximum rate but not minimum rate?
- KK LEE on May 28, 2015 at 10:38 am
  
  Hi. Please refer to other comments. 1.225 is max because it is the largest value among others
cyy on May 24, 2015 at 10:56 am

Sir,for question 2b,why the max value of dy/dt is 1.225 and not 0.320?Why the min value is 0?
Reply
- KK LEE on May 24, 2015 at 9:24 pm
  
  Why max is 1.225 and not 0.320????
  
  Max means largest. You can ask any kindergarten kids, which is larger among 1 and 0?
- cyy on May 27, 2015 at 10:05 pm
  
  Ok,thank you sir but it no need to use others method to prove that is min&max value ??
- KK LEE on May 28, 2015 at 10:37 am
  
  It depends on your teacher. For me, a simple explanation is enough. But i am not the teacher who give you marks.
lrj on June 9, 2015 at 3:11 pm

sir…i dont understand part b) …d2y/dx2 is used to find point of inflexion while dy/dx is used to find maximum and minimum point right? but y in this case we need to find inflexion instead of using dy/dx?
Reply
- KK LEE on June 10, 2015 at 12:47 am
  
  Hi. Please refer to the other comments. Thanks
Peiying on June 10, 2015 at 7:09 pm

Hai Sir… How to calculate the percentage shown in the 2(a)(ii)..
Reply
- KK LEE on June 10, 2015 at 10:50 pm
  
  Hi. Please ask your friends or your teacher. No more questions about term 2 assignment will be answered as it is term 3 now. 🙂
Nbk on June 10, 2015 at 10:21 pm

Sir,the question 2aiii,when y=14,why t=infinity?? and the In(0) should be error or not??
Reply
- KK LEE on June 10, 2015 at 10:50 pm
  
  Hi. Please ask your friends or your teacher. No more questions about term 2 assignment will be answered as it is term 3 now. 🙂
WAN on July 24, 2015 at 4:58 pm

excuse me, is there any reference for assignment math t first term 2015?
Reply
- KK LEE on July 29, 2015 at 10:13 am
  
  Hi. Past Assignment solution will not be posted.