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Calculus, EXT1* C1 2012 HSC 14c

Professor Smith has a colony of bacteria. Initially there are 1000 bacteria. The number of bacteria,  `N(t)`,  after  `t`  minutes is given by

`N(t)=1000e^(kt)`. 

  1. After 20 minutes there are 2000 bacteria.

     

    Show that `k=0.0347`  correct to four decimal places.   (1 mark)

  2. How many bacteria are there when  `t=120`?    (1 mark)

  3. What is the rate of change of the number of bacteria per minute, when  `t=120`?     (1 mark)

  4. How long does it take for the number of bacteria to increase from 1000 to 100 000?    (2 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions).}`
  2. `64\ 328`
  3. `2232`
  4. `133\ text(minutes)` 
Show Worked Solution

i.   `N=1000e^(kt)\ text(and given)\ N=2000\ text(when)\  t=20`

`2000` `=1000e^(20xxk)`
`e^(20k)` `=2`
`lne^(20k)` `=ln2`
`20k` `=ln2`
`:.k` `=ln2/20`
  `=0.0347\ \ text{(to 4 d.p.)  … as required}`

 

 ii.  `text(Find)\ \ N\ \ text(when)\  t=120`

NOTE: Students could have used the exact value of  `k=ln2/20`  in parts (ii), (iii) and (iv), which would yield the answers (ii) 64,000, (iii) 2,218, and (iv) 133 minutes.
`N` `=1000e^(120xx0.0347)`
  `=64\ 328.321..`
  `=64\ 328\ \ text{(nearest whole number)}`

 
`:.\ text(There are)\  64\ 328\ text(bacteria when t = 120.)`
 

iii.  `text(Find)\ (dN)/(dt)\ text(when)\  t=120`

MARKER’S COMMENT: This part proved challenging for many students. Differentiation is required to find the rate of change in these type of questions.
`(dN)/(dt)` `=0.0347xx1000e^(0.0347t)`
  `=34.7e^(0.0347t)`

 
`text(When)\ \ t=120`

`(dN)/(dt)` `=34.7e^(0.0347xx120)`
  `=2232.1927…`
  `=2232\ \ text{(nearest whole)}`

 
`:.\ (dN)/(dt)=2232\ text(bacteria per minute at)\  t=120`

 

iv.  `text(Find)\ \ t\ \ text(such that)\  N=100,000`

`=>100\ 000` `=1000e^(0.0347t)`
`e^(0.0347t)` `=100`
`lne^(0.0347t)` `=ln100`
`0.0347t` `=ln100`
`t` `=ln100/0.0347`
  `=132.7138…`
  `=133\ \ text{(nearest minute)}`

 
`:.\ N=100\ 000\ text(when)\  t=133\ text(minutes.)`

Financial Maths, 2ADV M1 2011 HSC 3a

A skyscraper of 110 floors is to be built. The first floor to be built will cost $3 million. The cost of building each subsequent floor will be $0.5 million more than the floor immediately below.

  1. What will be the cost of building the 25th floor?     (2 marks)

  2. What will be the cost of building all 110 floors of the skyscraper?     (2 marks)

Show Answers Only
  1. `$15\ text(million)`
  2. `$3,327.5\  text(million)`
Show Worked Solutions
i.     `T_1` `=a=3`
`T_2` `=a+d=3.5`
`T_3` `=a+2d=4`

 
`=>\ text(AP where)\ \ a=3\ \ d=0.5` 

MARKER’S COMMENT: Better responses listed the sequence of terms in the series as illustrated.
`T_25` `=a+24d`
  `=3+24(0.5)`
  `=15`

 
 `:.\ text(The 25th floor costs  $15,000,000.)`

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This allows markers to allocate part marks to students who had errors in calculation.

 

ii.     `S_110` `=\ text(Total cost of 110 floors)`
  `=n/2(2a+(n-1)d)` 
  `=110/2(2xx3\ 000\ 000+(110-1)500\ 000)`
  `=55(6\ 000\ 000+49\ 500\ 000)`
  `=$3327.5\  text(million)`

 

`:.\ text{The total cost of 110 floors is $3327.5 million}`

Financial Maths, 2ADV M1 2012 HSC 12c

Jay is making a pattern using triangular tiles. The pattern has 3 tiles in the first row, 5 tiles in the second row, and each successive row has 2 more tiles than the previous row.

2012 12c

  1. How many tiles would Jay use in row 20?     (2 marks)

  2. How many tiles would Jay use altogether to make the first 20 rows?     (1 mark)

  3. Jay has only 200 tiles. How many complete rows of the pattern can Jay make?     (2 marks)

Show Answers Only
  1. `\ 41`
  2. `440`
  3. `13\ text(rows)`
Show Worked Solutions
i.    `T_1` `=a=3`
  `T_2` `=a+d=5`
  `T_3` `=a+2d=7`

 
`=>\ text(AP where)\ \ a=3,\ \ d=2`

`\ \ \ \ \ vdots`

`T_20` `=a+19d`
  `=3+19(2)`
  `=41`

 

`:.\ text(Row 20 has 41 tiles.)`

 

MARKER’S COMMENT: Better responses stated the formula BEFORE any calculations were performed. This enabled students to get some marks if they made an error in their working.
ii.    `S_20` `=\ text(the total number of tiles in first 20 rows)`
`S_20` `=n/2(a+l)`
  `=20/2(3+41)`
  `=440`

 

`:.\ text(There are 440 tiles in the first 20 rows.)`

 

 iii.   `text(If Jay only has 200 tiles, then)\ \ S_n<=200`

NOTE: Examiners often ask questions requiring `n` to be found using the formula `S_n=n/2[2a+(n-1)d]` as this requires the solving of a quadratic, and interpretation of the answer.
`n/2(2a+(n-1)d)` `<=200`
`n/2(6+2n-2)` `<=200`
`n(n+2)` `<=200`
`n^2+2n-200` `<=0`
`n` `=(-2+-sqrt(4+4*1*200))/(2*1)`
  `=(-2+-sqrt804)/2`
  `=-1+-sqrt201`
  `=13.16\ \ text{(answer must be positive)}`

 

`:.\ text(Jay can complete 13 rows.)`