Need Help w/ Add Maths Plz

• Lawtoc

  15 Jun 09, 12:23

  (a) In an experiment, the growth rate of bacteria in a liquid at any time is proportional to N, the number of bacteria after t seconds. Given that the initial number of bacteria is A, show that N = Ae^kt , where k is a constant.

  (b) After t minutes, the radius of a circle r decreases at a rate of r/10 cm per minute and the initial radius is 8 cm.
  (i) Express r in terms of t.
  (ii) Find the time when the radius is 4 cm.

  Do take note that this question is under the topic Integration and can be found in the Panpac Additional Mathematics textbook by EPB Pan Pacific, Miscellaneous Exercise 19 Question 5 (page 421).

  Thanks a lot.

   

• wee_ws

  15 Jun 09, 12:33

  Hi,

    They are both differential equations and I doubt students at O-level can do it, since it is an A-level topic.

    In (a), dN/dt = kN, where k is the constant of proportionality.

    Then we go on to solve by variable separable, i.e.

    integral 1/N dN = integral k dt

    ln N = kt + c, where c is a constant of integration

    N = Ae^(kt), where A = e^c.

    In (b), we have the differential equation dr/dt = -r/10.

    Again by variable separable, integral 1/r dr = -1/10 integral dt

    ln r = -0.1t + c, where c is the constant of integration.

    ...

    Thanks!

  Cheers,
  Wen Shih

• Lee012lee

  15 Jun 09, 12:47

  TS, are you an "O" level Add Maths student ?

  Part (a) has no question ?

• Lawtoc

  15 Jun 09, 12:57

  Erm, part (a) is a show question =)

  Thanks Mod! I'm Secondary 4 and is learning this topic already haha.

  By the way, answers are :

  (b)(i) r = 8 e^(-t/10)
      (ii) 6 minutes 56 seconds

• Lee012lee

  15 Jun 09, 13:28

  TS, oops, I did not read the question carefully.

  This type of question will not appear in the actual "O" level add maths exam.

  This type of question will only appear in the H2 Maths exam as it involves the use

  of differential equation. 

  ------------------------------------------------------------------------------------------------------

  Question (a)

  In an experiment, the growth rate of bacteria in a liquid at any time is proportional to N, the number of bacteria after t seconds. Given that the initial number of bacteria is A, show that N = Ae^kt , where k is a constant.

  Solution (copied and paste from Mr Wee's answer)

  dN/dt = kN, where k is the constant of proportionality.

  Then we go on to solve by variable separable, i.e.

    integral 1/N dN = integral k dt

    ln N = kt + c, where c is a constant of integration

    N = Ae^(kt), where A = e^c.

  Question (b)(i)

  After t minutes, the radius of a circle r decreases at a rate of r/10 cm per minute and the initial radius is 8 cm.

   Express r in terms of t.

  Solution (copied and paste from Mr Wee's answer)

  differential equation dr/dt = -r/10.

    By variable separable, integral 1/r dr = -1/10 integral dt

    ln r = -0.1t + c, where c is the constant of integration.

    r = e ^(-0.1t + c)

    r = e^c x e^(-0.1t)

  Given that initial radius = 8 cm, so t = 0,

    8 = e^c x e^(-0.1(0))

    e^c = 8

  Hence, r = e^c x e^(-0.1t)

             = 8e^(-0.1t)

           r = 8 e^(-t/10)

  Question (b) (ii)

  Find the time when the radius is 4 cm

  Solution

  when t = 4, r = 8 e^(-t/10)

                   4 = 8 e^(-t/10)

                  0.5 = e(-0.1t)

  ln on both sides

              ln 0.5 = ln e(-0.1t)

              ln 0.5 = -0.1t

              ln 0.5 / (- 0.1) = t

              t = 6.931471806

              t = 6 min 56 sec 

   

   

• Lawtoc

  15 Jun 09, 13:52

  About part (a),

  N = Ae^(kt), where A = e^c.

  Please explain how you arrived at A = e^c . I am rather obtuse. Thanks a lot.

• Lee012lee

  15 Jun 09, 14:01

  ln N = kt + c

  ln N = (kt + c) ln e          ( ln e = 1)

  ln N = ln e ^(kt + c)

  Remove ln on both sides

  N = e ^(kt + c)

  N = e^(kt) x e^c

  Let A = e^c

  N = Ae^(kt)  (Shown)

• Lawtoc

  15 Jun 09, 14:21

  Removed by author. =D

• Lee012lee

  15 Jun 09, 15:04

  Alternatively,

  N = e^(kt) x e^c

  Given that the initial number of bacteria is A ie t = 0, N = A

  When t = 0,

  N = e^(kt) x e^c

  N = e^(k(0))x e^c

  N = e^c = A

  Hence, A = e^c

• wee_ws

  15 Jun 09, 15:22

  Hi Lee012lee,

    Thanks for taking the effort to clarify :)

  Cheers,
  Wen Shih

• Lawtoc

  15 Jun 09, 17:06

  Thanks for helping in the previous question! Here's another one =S I think this is Polynomials.

  Find the value of a and of b if the remainder when 2x³ – 4x² + ax – 2 is divided by x² + x – 2 is  5x + b .

  Answers are :

  a = -5
  b = -14

  Thanks again =D

• wee_ws

  15 Jun 09, 17:36

  Hi,

    Given the information,we have 

    2x^3 - 4x^2 + ax - 2 = (x^2 + x - 2)(2x - 6) + 5x + b.

    The factor 2x - 6 can be obtained easily by comparing coefficients of x^3 and x^2.

    Comparing constants, we have 12 + b = -2.

    Comparing coefficients of x, we obtain the value of a.

    Thanks!

  Cheers,
  Wen Shih

• Lawtoc

  15 Jun 09, 18:29

  Thanks for the solution Mod.

• wee_ws

  15 Jun 09, 21:49

  Hi Lawtoc,

    Thanks for your question!

  Cheers,
  Wen Shih

• Lawtoc

  15 Jun 09, 22:38

  More coming up lol. Here's another:

  Given that p = cos A + sin A and q = cos A - sin A, express p/q in terms of tan A.

  Answer :

  (1 + tan A) / (1 - tan A)

  Thanks!

• Mikethm

  15 Jun 09, 23:48

  Originally posted by Lawtoc:

  More coming up lol. Here's another:

  Given that p = cos A + sin A and q = cos A - sin A, express p/q in terms of tan A.

  Answer :

  (1 + tan A) / (1 - tan A)

  Thanks!

  
  p/q

  = (cos A + sin A)/(cosA -sinA) divided by (cos A/cos A)*

  = ( 1+ tan A)/ (1 - tan A)

  * Divide the numerator by cos A and divide the denominator by cos A.