A hyperboloid container of height 14 m is generated by the hyperbola 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 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 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

### Old Version

### Updated Version

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

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Sir, hw to do intro, methodology and conclusion?

Hi. I do the mathematical parts. U share the intro, methodology, and conclusion parts. Fair? 😀

I will help students on mathematical parts only.

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?

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. 😀

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

Hi. I am teaching STPM Maths tuition at Ai Tuition at Jalan Petaling Kuala Lumpur and SS19 Subang Jaya. and Teras Usaha at Segambut

how do find the maximum

and the minimum

In general, use dy/dx=0

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

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

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~

hi. Can i know which V u mean?

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

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?

Please try to use the quadratic formula.

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

Opps… its’s my mistake…sry

😀

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

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

Thanks sir. How to join your tuition class?

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

Sir,What can we write for title??

Hmm.. Investigating the rate of change of water.

Cant help you on this. 😛

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

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?

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

Hi. Can share your marking scheme? 😀

i did not understand why minimum rate is 0

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. 😀

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

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

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

Hi. It depends on your school teacher. For my sample solution, it is impossible to fill up the container to 100.000% full.

Thank you sir

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

Differentiate v with respect to y

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

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

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

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

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..

You can use any methods. As long as your school teacher accept.

i will try to check the values 0.2936 and 1.186.

I have no idea to get those values

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?

because i used implicit differentiation.

you can use part 1(b)

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

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.

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

yes

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

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

sir, when y = 14, isn’t d2y/dt2 =

144y^2 – 3552y +19629 / 2[sqr(14-y)](48y^2-864y+4563)^2 undefined?

Yes

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

ya

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

sorry. please check again

1) (c) 1/9Π x 180 = 20/Π

Yes

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

Long division

Hey sir, do you have Maths M pbs? TQ

https://kkleemaths.com/2015/03/16/stpm-2015-mathematics-m-term-2-assignment/

sir, pbs maths t term 1 stpm 2015. tq

Are you kidding?

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

Hi.I have no sem 1 answer

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?

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.

sir soloution 1c with wrong answer

it must be (1/9pie)x180 then the ans will be6.366m or 2.xx pie

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?

it was written as 6.366 there.

280+260+…+20+0 / 15

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

from the formula and calculation.

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

Average is smaller.

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

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

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]]

Thanks.

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

You are welcome. 😀

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

checked

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

Hi. Ask your teacher. 😛

I will help on maths part only.

thank you sir.

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

In short, Time=Volume divide rate

thank you sir.

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 §.

long division

I see!! Thank you sir!!

sir i just want to ask that , can you give me some idea to write the introduction ? i have no idea how to start.

Try to explain history of Tupperware water container. 😛

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

Hi. I cant help you for this title. Sorry

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

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

Sir,for question 2b,why the max value of dy/dt is 1.225 and not 0.320?Why the min value is 0?

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?

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

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

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?

Hi. Please refer to the other comments. Thanks

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

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

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

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

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

Hi. Past Assignment solution will not be posted.