A.Maths Qn

• wishboy

  21 Apr 08, 19:53

  The variables x and y are related in such a way that, when ln y is plotted against ln x, a straight line is obtained which passes through the points (1, 1) and (4, 3).

  Express y in terms of x.

  need help with this qn

• ohnoez!

  21 Apr 08, 20:03

  Y=mX + c

  Y= ln y, X = ln x

   

  ln y = m ln x + c

   

  sub (1,1)

   

  1 = m(1) + c

   

  m = (3-1)/(4-1) = 2/3

   

  c = 1/3

   

  ln y = 2/3ln x + 1/3

   

  y = e^(2/3 ln x +1/3)???

   

• weewee

  21 Apr 08, 20:10

  Originally posted by wishboy:

  need help with this qn

  
  A straight line is in the form Y=mX+c

   

  here gradient m = (3-1)/(4-1)=2/3

   

  Hence Y=(2/3)X + c

  Sub in X=1, Y=1, we get 1=2/3+c so c=1/3

   

  Therefore, Y=(2/3)X + 1/3

  Now Y is actually ln y and X is actually ln x

  ln y = (2/3) ln x +1/3

  y = exp(2/3 ln x +1/3)

  y = x exp(2/3)exp(1/3)

  y=x exp(1)

   

  Not 100% sure.

   

• wishboy

  21 Apr 08, 20:17

  Now Y is actually ln y and X is actually ln x

  ok i understand liao

  previously i subbed the coords into the x/y in ln x/ln y instead of the whole ln x/ln y

  tyvm!

• wishboy

  21 Apr 08, 22:55

  another qn

  solve the equation

  2 sin² x = 3 cos x , for 0 <= x <= 5pi

   

• eagle

  21 Apr 08, 23:01

  Originally posted by wishboy:

  another qn

  solve the equation

  2 sin² x = 3 cos x , for 0 <= x <= 5pi

   

  2 sin² x = 2(1-cos² x)

  so equation becomes

  2 - 2cos² x = 3 cos x
  2cos² x + 3cos x - 2 = 0
  (2cos x - 1) (cos x + 2) = 0
  cos x = 1/2 or cos x = -2 (reject)
  so x = pi/3, 5pi/3, 7pi/3, 11pi/3, 13pi/3

• wishboy

  21 Apr 08, 23:24

  thx alot

• wishboy

  27 Apr 08, 19:34

  how to expand (1 + ax + x²)^10 ?

  EDIT: ok i got it =D

  ok the qn was

  Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),

  i) find the values of a and b.

  ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.

  i found tat a = 2, b = 190 for part (i)

  how do i do part (ii)?

• macshaobao

  28 Apr 08, 03:30

  how to expand (1 + ax + x²)^10 ?

  EDIT: ok i got it =D

  ok the qn was

  Given that the expansion of (1 + ax + x²)^10 up to the x² term is given by (1 + 20x + bx² + ...),

  i) find the values of a and b.

  ii) hence, without using the calculator, find the value of (1.0404)^10, giving your answer to 3 decimal places.

  i found tat a = 2, b = 190 for part (i)

  how do i do part (ii)?

   

  Given that your answer to part one is right, i.e., a=2,b=190,

  then (1 + ax + x²)^10  is also (1 + 2x + x²)^10 ,

  so u sub x =0.02  into (1 + 20x + bx² + ...) to get the value of (1.0404)^10... goddit?

• wishboy

  28 Apr 08, 21:40

  need to expand out the whole thing?

• weewee

  28 Apr 08, 22:22

  (1 + ax + x²)^10

  = ((1 + ax) + x²)^10

  = (1 + ax)^10 + 10(1+ax)^9 (x^2) + ....all the rest will be x with powers above 2

  = 1 + 10ax + 45a^2x^2 + 10(1 + ...)x^2 + ....

  = 1 + 10ax + (45a^2 + 10)x^2 + ...

  therefore by comparing of coefficients, 10a=20=>a=2

  45a^2+10=190

  (ii) By inspection sub in x = 0.02. The answer is an approximated solution so use curvy equality.

• wishboy

  28 Apr 08, 22:55

  Originally posted by weewee:

  (ii) By inspection sub in x = 0.02. The answer is an approximated solution so use curvy equality.

  
  wad is curvy equality?

• weewee

  29 Apr 08, 00:27

  Originally posted by wishboy:

  
  wad is curvy equality?

  x ≈ y means x is approximately equal to y.
• wishboy

  29 Apr 08, 00:35

  oic....
  thanks

• escadaraindrops

  1 May 08, 00:33

  this is linar law right? refer to the textbook, there is a similar example