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Discussion – 

113

Discussion – 

113

STPM 2015 Mathematics (T) Term 2 Assignment

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

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 2 (a) (i)

Old Version

Question 2ai

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.

Average

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

Question 2aii

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

Updated Version

Question 2aii new

The container cannot be 100% full.

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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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

  1. Rishaliney a/p Selva Rajoo

    Sir, hw to do intro, methodology and conclusion?

    Reply
    • KK LEE

      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

      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

      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. 😀

  2. sherrynn

    sir,where the tuition places that you teaching maths (T)for STPM?

    Reply
    • KK LEE

      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

      how do find the maximum
      and the minimum

    • KK LEE

      In general, use dy/dx=0

  3. Soonnee

    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

      yes. that is one of the sample of explanation only.

  4. Wendy

    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

      hi. Can i know which V u mean?

    • Wendy

      Sorry sir~I got it now~~Because I had expanded all the brackets and a bit messed up~ thanks

  5. ccm

    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

      Please try to use the quadratic formula.

  6. abc96

    Hi sir, could you please check again 1(b)? Maybe you overlook the expansion of (y-9)^3.

    Reply
    • abc96

      Opps… its’s my mistake…sry

    • KK LEE

      😀

  7. V

    Sir, how u can get the dv/dt= 15(14-y)’1/2 ? On question 2 (b)

    Reply
    • KK LEE

      Substitute k=15, n=1/2 into the rate provided. dv/dt=k(14-y)^n

    • V

      Thanks sir. How to join your tuition class?

    • KK LEE

      yes. ARe you from kl area. I am teaching at jalan Petaling KL,Subang Jaya and segambut

  8. A yun

    Sir,What can we write for title??

    Reply
    • KK LEE

      Hmm.. Investigating the rate of change of water.
      Cant help you on this. 😛

    • A yun

      Its okey..But anyway thank you,sir..

  9. jega

    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

      Sir.. I’m sooooo sorry.. 1c is correct over here.. my mistake sir sorry…

    • KK LEE

      Hi. Can share your marking scheme? 😀

  10. Tan Nee Nee

    i did not understand why minimum rate is 0

    Reply
    • KK LEE

      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. 😀

  11. sree

    I dun understand the part 2(a)(i)…can u pls help me in it.. tq

    Reply
    • KK LEE

      Since the rate is not constant. So I used average of the rate(approximation) to compare with 180

    • F. Syaf

      sorry sir, i still cannot understand part 2a(ii), how to write it down?

    • KK LEE

      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

      Thank you sir

  12. Angeline

    Sir, for 2a2, how can u get the dvdy ?

    Reply
    • KK LEE

      Differentiate v with respect to y

  13. mj

    Sir y we can use dy/dT=0 in 2a(b) to get minimum value?

    Reply
    • KK LEE

      Hi. We can use any methods to find the minimum value of functions.

  14. eyan

    sir, why do we need to use the average of water? and also how to explain that t=∞? tq

    Reply
    • KK LEE

      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

  15. wani

    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

      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

      I have no idea to get those values

  16. wani again

    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

      because i used implicit differentiation.
      you can use part 1(b)

  17. Boiboi

    Sir,why Q2b use d2y/dx2 =0? That one is for point of inflection right? Correct me pls

    Reply
    • KK LEE

      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.

  18. Boiboi

    Sir,Q2a2 why t=infinity? Coz 1875ln0?

    Reply
    • KK LEE

      yes

  19. Boiboi

    Sir,Q2a2,why substitute y=14,t=infinity?

    Reply
    • KK LEE

      Please recall that when x approaches 0, ln x approaches -infinity

  20. Shawn

    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

      Yes

    • Shawn

      so y = 14 isn’t a solution to d2y/dt2 = 0?

    • KK LEE

      ya

  21. Rachel

    Why my one fron yours? My ans for 1(b) is 308 14/75.

    Reply
    • KK LEE

      sorry. please check again

  22. unknown

    1) (c) 1/9Π x 180 = 20/Π

    Reply
    • KK LEE

      Yes

  23. shirley

    how (48ypower 2 -864y+4563)over (14-y)wil get -48y+192+(1875/(14-y))? i dont get it

    Reply
    • KK LEE

      Long division

  24. Irene SinnYue

    Hey sir, do you have Maths M pbs? TQ

    Reply
  25. AFIFAH

    sir, pbs maths t term 1 stpm 2015. tq

    Reply
    • KK LEE

      Are you kidding?

    • Kalishwaran

      Sir… Do u have solutions for term 1 assignment?? Im a private candidate.. Need to do it.

    • KK LEE

      Hi.I have no sem 1 answer

  26. Boiboi

    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

      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.

  27. Raffles

    sir soloution 1c with wrong answer
    it must be (1/9pie)x180 then the ans will be6.366m or 2.xx pie

    Reply
    • Raffles

      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

      it was written as 6.366 there.

    • KK LEE

      280+260+…+20+0 / 15

  28. kks

    Sir ,in 2(a)ii) why with a mechanism installed the water CNT fill up the container100%?

    Reply
    • KK LEE

      from the formula and calculation.

  29. kks

    Sir,in 2(a)i) why time taken will b longer?just in case teacher ask during presentation

    Reply
    • KK LEE

      Average is smaller.

  30. kks

    Oh..OK,thanks!other than the results frm calculation n formula, is thr any other reason I cn state?

    Reply
    • KK LEE

      As i mentioned earlier. I will help on Maths part only. 😀

  31. Assunta Patrick

    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

      Thanks.

  32. Alves

    Best maths T teacher ever! Clap for u! Thanks for helping us!

    Reply
    • KK LEE

      You are welcome. 😀

  33. Punitha

    sir, can u pls check my introduction tht i sent to ur email?? 🙂 tq sir

    Reply
    • KK LEE

      checked

  34. w

    sir, can i know what is that two applications of this mechanism?

    Reply
    • KK LEE

      Hi. Ask your teacher. 😛
      I will help on maths part only.

    • w

      thank you sir.

  35. che

    sir, what is the main formula to find the time taken to fill the container?

    Reply
    • KK LEE

      In short, Time=Volume divide rate

    • che

      thank you sir.

  36. Joan

    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

      long division

    • Joan

      I see!! Thank you sir!!

  37. connie

    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

      Try to explain history of Tupperware water container. 😛

  38. mumu

    hi,sir,may i noe wat is the title of this sem 2 coursework?

    Reply
    • KK LEE

      Hi. I cant help you for this title. Sorry

    • GY

      Sir,may I knw tat..why 1.225 is a maximum rate but not minimum rate?

    • KK LEE

      Hi. Please refer to other comments. 1.225 is max because it is the largest value among others

  39. cyy

    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

      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

      Ok,thank you sir but it no need to use others method to prove that is min&max value ??

    • KK LEE

      It depends on your teacher. For me, a simple explanation is enough. But i am not the teacher who give you marks.

  40. lrj

    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

      Hi. Please refer to the other comments. Thanks

  41. Peiying

    Hai Sir… How to calculate the percentage shown in the 2(a)(ii)..

    Reply
    • KK LEE

      Hi. Please ask your friends or your teacher. No more questions about term 2 assignment will be answered as it is term 3 now. 🙂

  42. Nbk

    Sir,the question 2aiii,when y=14,why t=infinity?? and the In(0) should be error or not??

    Reply
    • KK LEE

      Hi. Please ask your friends or your teacher. No more questions about term 2 assignment will be answered as it is term 3 now. 🙂

  43. WAN

    excuse me, is there any reference for assignment math t first term 2015?

    Reply
    • KK LEE

      Hi. Past Assignment solution will not be posted.

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