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Calculus, MET2 2009 VCAA 2

VCAA 2009 2a

A train is travelling at a constant speed of `w` km/h along a straight level track from `M` towards `Q.`

The train will travel along a section of track `MNPQ.`

Section `MN` passes along a bridge over a valley.

Section `NP` passes through a tunnel in a mountain.

Section `PQ` is 6.2 km long.

From `M` to `P`, the curve of the valley and the mountain, directly below and above the train track, is modelled by the graph of
 

`y = 1/200 (ax^3 + bx^2 + c)` where `a, b` and `c` are real numbers.
 

All measurements are in kilometres.

  1. The curve defined from `M` to `P` passes through `N (2, 0)`. The gradient of the curve at `N` is – 0.06 and the curve has a turning point at  `x = 4`.
  2.  i. From this information write down three simultaneous equations in `a`, `b` and `c`.   (3 marks)

  3. ii. Hence show that  `a = 1`, `b = – 6` and `c = 16`.   (2 marks)

  4. Find, giving exact values
  5.   i. the coordinates of `M and P`.   (2 marks)

  6.  ii. the length of the tunnel.   (1 mark)

  7. iii. the maximum depth of the valley below the train track.   (1 mark)

The driver sees a large rock on the track at a point `Q`, 6.2 km from `P`. The driver puts on the brakes at the instant that the front of the train comes out of the tunnel at `P`.

From its initial speed of `w` km/h, the train slows down from point `P` so that its speed `v` km/h is given by

`v = k log_e ({(d + 1)}/7)`,

where `d` km is the distance of the front of the train from `P` and `k` is a real constant.

  1. Find the value of `k` in terms of `w`.   (1 mark)

  2. Find the exact distance from the front of the train to the large rock when the train finally stops.   (2 marks)

Show Answers Only

a.i.    `1/200 (8a + 4b + c) = 0`

a.ii.   `1/200 (12a + 4b) = (– 3)/50`

a.iii.  `1/200 (48a + 8b) = 0`

`text(Proof)\ \ text{(See Worked Solutions)}`

    1. `M (2 + 2 sqrt 3, 0),\ \ P (2-2 sqrt 3, 0)`
    2. `2 sqrt 3\ text(km)`
    3. `2/25\ text(km)`
  2. `(– w)/(log_e (7))`
  3. `120`
  4. `0.2\ text(km)`
Show Worked Solution

 a.i.   `N (2, 0),`

`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`
 

`(dy)/(dx) (x = 2) = (– 3)/50,`

`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`
 

`(dy)/(dx)(x = 4) = 0,`

`1/200 (48a + 8b) = 0\ \ text{… (3)}`
 

  ii.   `text(Using the above equations,)`

♦ Mean mark part (a)(ii) 41%.
`12a + 4b` `=-12` `\ \ \ …\ (2^{′})`
`24 a + 4b` `=0` `\ \ \ …\ (3^{′})`

 
`text{Solve simultaneous equations (By CAS):}`

`a=1, \ b=-6, \ c=16`
 

`\text{Solving manually:}`

`(3^{′})-(2^{′}):`

`12a=12 \ \Rightarrow\ \ a=1`

`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3^{′}):`

`124(1)+4b=0 \ \Rightarrow\ \ b=-1`

`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1):`

`8(1) + 4 (– 6) + c=0\ \ \Rightarrow\ \ c=16`
 

b.i.   `text(For)\ \ x text(-intercepts),`

`text(Solve:)\ \ x^3 + -6x^2 + 16` `= 0\ \ text(for)\ x,`
 `x= 2-2 sqrt 3, 2, 2 + 2 sqrt 3`  

 
`:. M (2 + 2 sqrt 3, 0),\ \ P (2-2 sqrt 3, 0)`
 

  ii.   `N (2, 0)`

♦ Mean mark part (a) 41%.
`bar (NP)` `=\ text(Tunnel length)`
  `= 2-(2-2 sqrt 3)`
  `= 2 sqrt 3\ text(km)`

 

   iii.  `text(Solve)\ \ frac (dy) (dx) = 0\ \ text(for)\ \ x in (2, 2 + 2 sqrt 3)`

♦ Mean mark part (b) 50%.

`x = 4`

`text(When)\ \ x=4,\ \ y = – 2/25\ text(km) = 80\ text{m (below track)}`

`:.\ text(Max depth below is)\ \ 80\ text(m.)`
 

c.  `text(Solution 1)`

♦ Mean mark part (c) 35%.

`text(Let)\ \ v(d) = k log_e ({(d + 1)}/7)`

`text(S)text(ince)\ \ v=w\ \ text(when)\ \ d=0\ \ text{(given),}`

`k log_e ({(0 + 1)}/7)` `=w`
`k` `=w/log_e (1/7)`
  `=(-w)/log_e7`

 
`text(Solution 2)`

`text(Solve:)\ \ v (0)` `= w\ \ text(for)\ \ k`
`:. k` `= (– w)/(log_e (7))`

 

d.  `v (2.5) = k log_e(1/2)`

♦ Mean mark part (d) 40%.
`text(Solve)\ \ v(2.5)` `= (120 log_e (2))/(log_e (7))\ \ text(for)\ \ k,`
`:. k` `= (– 120)/(log_e (7))`
`(– w)/(log_e (7))` `= (– 120)/(log_e (7))`
`:. w` `= 120`

 

e.   `text(Define)\ \ v (d) = (– 120)/(log_e (7)) log_e ((d + 1)/7)`

♦♦ Mean mark part (e) 32%.
`text(Solve:)\ \ v(d)` `= 0\ \ text(for)\ d,`
`:. d` `= 6\ text(km from)\ \ P`

 
`:.\ text(Distance between train and)\ \ Q`

`= 6.2-6`

`= 0.2\ text(km)`