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Trigonometry, 2ADV T3 SM-Bank 16

Sammy visits a giant Ferris wheel. Sammy enters a capsule on the Ferris wheel from a platform above the ground. The Ferris wheel is rotating anticlockwise. The capsule is attached to the Ferris wheel at point `P`. The height of `P` above the ground, `h`, is modelled by  `h(t) = 65 - 55cos((pit)/15)`, where `t` is the time in minutes after Sammy enters the capsule and `h` is measured in metres.

Sammy exits the capsule after one complete rotation of the Ferris wheel.
 


 

  1. State the minimum and maximum heights of `P` above the ground.  (1 mark)

  2. For how much time is Sammy in the capsule?  (1 mark)

  3. Find the rate of change of `h` with respect to `t` and, hence, state the value of `t` at which the rate of change of `h` is at its maximum.  (2 marks)

Show Answers Only
  1. `h_text(min) = 10\ text(m), h_text(max) = 120\ text(m)`
  2. `30\ text(min)`
  3. `t = 7.5`
Show Worked Solution
i.    `h_text(min)` `= 65 – 55` `h_text(max)` `= 65 + 55`
    `= 10\ text(m)`   `= 120\ text(m)`

 

ii.   `text(Period) = (2pi)/(pi/15) = 30\ text(min)`

 

iii.    `h′(t)` `=65 – 55cos((pit)/15)`
  `h′(t)` `=pi/15 xx 55sin(pi/15 t)`
    `= (11pi)/3\ sin(pi/15 t)`
     

`text(S)text(ince)\ \ sin(pi/15 t)_text(max) = sin (pi/2),`

 
`:. h′(t)_text(max)\ \ text(occurs when)`

`(pi t)/15` `=pi/2`  
`:. t` `=pi/2 xx 15/pi`  
  `=15/2\ text(minutes)\ \ (0<=t<=30)`  

Calculus, 2ADV C3 SM-Bank 13

The figure shown represents a wire frame where `ABCE` is a convex quadrilateral. The point `D` is on line segment `EC` with  `AB = ED = 2\ text(cm)` and  `BC = a\ text(cm)`, where `a` is a positive constant.

`/_ BAE = /_ CEA = pi/2`

Let  `/_ CBD = theta`  where  `0 < theta < pi/2.`
 

 vcaa-2011-meth-10a
 

  1. Find `BD` and `CD` in terms of `a` and `theta`.  (2 marks)

  2. Find the length, `L` cm, of the wire in the frame, including length `BD`, in terms of `a` and `theta`.  (1 mark)

  3. Find  `(dL)/(d theta)`,  and hence show that  `(dL)/(d theta) = 0` when  `BD = 2CD`.  (2 marks)

  4. Find the maximum value of `L` if  `a = 3 sqrt 5`.  (1 mark)

Show Answers Only
  1. `BD = a cos theta,\ \ \ CD = a sin theta`
  2. `L = 4 + a + 2 a cos theta + a sin theta`
  3. `text(Proof)\ text{(See Worked Solutions)}`
  4. `L_max = 19 + 3 sqrt 5`
Show Worked Solution

i.   `text(In)\ \ Delta BCD,`

`cos theta` `= (BD)/a`
`:. BD` `= a cos theta`
`sin theta` `= (CD)/a`
`:. CD` `= a sin theta`

 

ii.   `L` `= 4 + 2 BD + CD + a`
    `= 4 + 2a cos theta + a sin theta + a`
    `= 4 + a + 2a cos theta + a sin theta`

 

iii.   `text(Noting that)\ a\ text(is a constant:)`

♦ Mean mark (Vic) 35%.

`(dL)/(d theta)= – 2 a sin theta + a cos theta`

`text(When)\ \ (dL)/(d theta) = 0`,

`- 2 a sin theta+ a cos theta` `= 0`
`a cos theta` `= 2 a sin theta`
`:.  BD` `= 2CD\ \ text{(using part (a))}`

 

iv.  `text(SP’s when)\ \ (dL)/(d theta)=0,`

♦♦♦ Mean mark (Vic) 5%.
`- 2 a sin theta+ a cos theta` `= 0`
`sin theta` `=1/2 cos theta`
`tan theta` `=1/2`

 

 vcaa-2011-meth-10ai

`text(If)\ \ tan theta=1/2,\ \ cos theta = 2/sqrt5,\ \ sin theta = 1/sqrt5`

`L_(max)` `= 4 + a + 2a cos theta + a sin theta`
  `= 4 + (3 sqrt 5) + 2 (3 sqrt 5) (2/sqrt 5) + (3 sqrt 5) (1/sqrt 5)`
  `= 4 + 3 sqrt 5 + 12 + 3`
  `= 19 + 3 sqrt 5\ text(cm)`

Calculus, 2ADV C3 2007 HSC 10b

The noise level, `N`, at a distance `d` metres from a single sound source of loudness `L` is given by the formula

`N = L/d^2.`

Two sound sources, of loudness `L_1` and `L_2` are placed `m` metres apart.
 

The point `P` lies on the line between the sound sources and is `x` metres from the sound source with loudness `L_1.`

  1. Write down a formula for the sum of the noise levels at `P` in terms of `x`.  (1 mark)

  2. There is a point on the line between the sound sources where the sum of the noise levels is a minimum.

     

    Find an expression for `x` in terms of `m`, `L_1` and `L_2` if `P` is chosen to be this point.  (4 marks)

Show Answers Only
  1. `N = L_1/x^2 + L_2/(m-x)^2`
  2. `x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`
Show Worked Solution
i.  

`N = L/d^2`

`text(Noise from)\ L_1` `= L_1/x^2`
`text(Noise from)\ L_2` `= L_2/(m-x)^2`
`:. N` `= L_1/x^2 + L_2/(m-x)^2`

 

ii.  `N = L_1\ x^-2 + L_2 (m – x)^-2`

`(dN)/(dx)` `= -2 L_1 x^-3 + -2 L_2 (m – x)^-3 xx d/(dx) (m – x)`
  `= (-2 L_1)/x^3 + (2 L_2)/(m – x)^3`

 

`text(Max or min when)\ (dN)/(dx) = 0`

`(2 L_1)/x^3` `= (2 L_2)/(m – x)^3`
`2 L_1 (m – x)^3` `= 2 L_2\ x^3`
`L_1 (m – x)^3` `= L_2\ x^3`
`root 3 L_1 (m – x)` `= root 3 L_2\ x`
`root 3 L_1\ m – root 3 L_1\ x` `= root 3 L_2\ x`
`root 3 L_2\ x + root 3 L_1\ x` `= root 3 L_1\ m`
`x (root 3 L_2 + root 3 L_1)` `= root 3 L_1\ m`
`x` `= (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

 

`(dN)/(dx)` `= -2 L_1\ x^-3 + 2 L_2 (m – x)^-3`
`(d^2N)/(dx^2)` `= 6 L_1\ x^-4 – 6 L_2 (m – x)^-4 xx -1`
  `= (6 L_1)/x^4 + (6 L_2)/(m – x)^4 > 0`

 

`:.\ text(A minimum occurs when)`

`x = (root 3 L_1\ m)/{(root 3 L_2 + root 3 L_1)}`

Calculus, 2ADV C3 2009 HSC 9b

An oil rig,  `S`,  is 3 km offshore. A power station,  `P`,  is on the shore. A cable is to be laid from `P`  to  `S`.  It costs $1000 per kilometre to lay the cable along the shore and $2600 per kilometre to lay the cable underwater from the shore to  `S`.

The point  `R`  is the point on the shore closest to  `S`,  and the distance  `PR`  is 5 km.

The point  `Q`  is on the shore, at a distance of  `x`  km from  `R`,  as shown in the diagram.


  

  1. Find the total cost of laying the cable in a straight line from  `P`  to  `R`  and then in a straight line from  `R`  to  `S`.   (1 mark)

  2. Find the cost of laying the cable in a straight line from  `P`  to  `S`.  (1 mark)

  3. Let  `$C`  be the total cost of laying the cable in a straight line from  `P`  to  `Q`,  and then in a straight line from `Q`  to  `S`.
     
    Show that  `C=1000(5-x+2.6sqrt(x^2+9))`.    (2 marks)

  4. Find the minimum cost of laying the cable.    (4 marks)

  5. New technology means that the cost of laying the cable underwater can be reduced to $1100 per kilometre.

     

    Determine the path for laying the cable in order to minimise the cost in this case.    (2 marks)

Show Answers Only
  1. `$12\ 800`
  2. `$15\ 160`
  3. `text{Proof (See Worked Solutions)}`
  4. `$12\ 200`
  5. `P\ text(to)\  S\ text(in a straight line.)`
Show Worked Solution
♦♦ Although specific data is unavailable for question parts, mean marks were 35% for Q9 in total.
i.   `text(C)text(ost)` `=(PRxx1000)+(SRxx2600)`
  `=(5xx1000)+(3xx2600)`
  `=12\ 800`

 
`:. text(C)text(ost is)\   $12\ 800`

 

ii.   `text(C)text(ost)=PSxx2600`

`text(Using Pythagoras:)`

`PS^2` `=PR^2+SR^2`
  `=5^2+3^2`
  `=34`
`PS` `=sqrt34`

 

`:.\ text(C)text(ost)` `=sqrt34xx2600`
  `=15\ 160.474…`
  `=$15\ 160\ \ text{(nearest dollar)}`

 

iii.  `text(Show)\ \ C=1000(5-x+2.6sqrt(x^2+9))`

`text(C)text(ost)=(PQxx1000)+(QSxx2600)`

`PQ` `=5-x`
`QS^2` `=QR^2+SR^2`
  `=x^2+3^2`
`QS` `=sqrt(x^2+9)`
`:.C` `=(5-x)1000+sqrt(x^2+9)\ (2600)`
  `=1000(5-x+2.6sqrt(x^2+9))\ \ text(…  as required)`

 

iv.   `text(Find the MIN cost of laying the cable)`

`C` `=1000(5-x+2.6sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+2.6xx1/2xx2x(x^2+9)^(–1/2))`
  `=1000(–1+(2.6x)/(sqrt(x^2+9)))`

`text(MAX/MIN when)\ (dC)/(dx)=0`

IMPORTANT: Tougher derivative questions often require students to deal with multiple algebraic constants. See Worked Solutions in part (iv).

`1000(–1+(2.6x)/(sqrt(x^2+9)))=0`

`(2.6x)/sqrt(x^2+9)` `=1`
`2.6x` `=sqrt(x^2+9)`
`(2.6)^2x^2` `=x^2+9`
`x^2(2.6^2-1)` `=9`
`x^2` `=9/5.76`
  `=1.5625`
`x` `=1.25\ \ \ \ (x>0)`
MARKER’S COMMENT: Check the nature of the critical points in these type of questions. If using the first derivative test, make sure some actual values are substituted in.

`text(If)\ \ x=1,\ \ (dC)/(dx)<0`

`text(If)\ \ x=2,\ \ (dC)/(dx)>0`

`:.\ text(MIN when)\ \ x=1.25`

`C` `=1000(5-1.25+2.6sqrt(1.25^2+9))`
  `=1000(122)`
  `=12\ 200` 

 

`:.\ text(MIN cost is)\  $12\ 200\ text(when)\   x=1.25`

 

v.   `text(Underwater cable now costs $1100 per km)`

`=>\ C` `=1000(5-x)+1100sqrt(x^2+9)`
  `=1000(5-x+1.1sqrt(x^2+9))`
`(dC)/(dx)` `=1000(–1+1.1xx1/2xx2x(x^2+9)^(-1/2))`
  `=1000(–1+(1.1x)/sqrt(x^2+9))`

 
`text(MAX/MIN when)\ (dC)/(dx)=0`

`1000(–1+(1.1x)/sqrt(x^2+9))` `=0`
`(1.1x)/sqrt(x^2+9)` `=1`
`1.1x` `=sqrt(x^2+9)`
`1.1^2x^2` `=x^2+9`
`x^2(1.1^2-1)` `=9`
`x^2` `=9/0.21`
`x` `~~6.5\ text{km (to 1 d.p.)}`
  `=>\ text(no solution since)\ x<=5`

 
`text(If we lay cable)\ PR\ text(then)\ RS`

MARKER’S COMMENT: Many students failed to interpret a correct calculation of  `x>5`  as providing no solution.

`=>\ text(C)text(ost)=5xx1100+3xx1000=8500`

`text(If we lay cable directly underwater via)\ PS`

`=>\ text(C)text(ost)=sqrt34xx1100=6414.047…`

`:.\ text{MIN cost is  $6414 by cabling directly from}\ P\ text(to)\ S`.