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Calculus, SPEC2 2024 VCAA 3

A pollutant, at time \(t=0\) days, begins to enter a pond of still, unpolluted water at a rate of  \(\dfrac{d V}{d t}=\dfrac{8 t}{240+5 t^4}\), where \(V\) is the volume of pollutant, in cubic metres, in the pond after \(t\) days.

The pollutant does not dissolve or mix, and spreads across the pond, maintaining the shape of a thin circular disc of radius \(r(t)\) metres and constant depth of 1 millimetre.

  1. What is the maximum rate, in cubic metres per day, at which the pollutant will enter the pond, and for what value of \(t\) will this rate occur?   (1 mark)

  2. At what rate is the radius of the disc increasing after \(t=4\) days, where it may be assumed that the radius of the disc is 6.54 m ?
  3. Give your answer in metres per day correct to two decimal places.   (3 marks)

    1. Use the substitution  \(u=\sqrt{5} \, t^2\)  to express  \(\displaystyle \int \frac{8 t}{240+5 t^4} d t\)  as an integral involving only the variable \(u\).   (1 mark)

    2. Hence, or otherwise, find, in terms of \(t\), the total volume \(V \) m\(^3\) of pollutant that has entered the pond after \(t\) days.
    3. Give your answer in the form \(\dfrac{1}{a \sqrt{b}} \arctan \left(\dfrac{t^c}{d \sqrt{b}}\right)\),  where  \(a, b, c, d \in Z^{+}\).    (1 mark)

  5. What surface area of the pond would the coverage of the pollutant approach?
  6. Give your answer in square metres correct to two decimal places.  (2 marks)

  7. The clean-up of the pond begins after five days, where the pollutant is removed at a constant rate of 0.05 cubic metres per day until the pond is free of pollutant. However, efforts to stem the flow are unsuccessful and the pollutant continues to enter the pond at a rate of  \(\dfrac{8 t}{240+5 t^4}\)  cubic metres per day.
  8. After how many days, from the start of the clean-up, will the pond be free of pollutant? Give your answer in days correct to one decimal place.  (2 marks)

Show Answers Only

a.    \\(\dfrac{dV}{dt} \text{(max)}=\dfrac{1}{20} \ \text{m\(^3\)/day.}\)

b.  \(\dfrac{dr}{dt}=\dfrac{2}{95} \times \dfrac{500}{\pi \times 6.54}=0.51 \ \text{m/day}\)

c.i.  \(\displaystyle \int \frac{8t}{240+5 t^4} \, dt=\int \frac{8}{240+u^2} \cdot \frac{du}{2 \sqrt{5}}=\frac{4}{\sqrt{5}} \int \frac{du}{240+u^2}\)

c.ii.  \(V=\displaystyle \frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{t^2}{4 \sqrt{3}}\right)\)

d.   \(\dfrac{100\pi}{\sqrt{3}}\)

e.   \(t=3.4 \ \text{days}\)

Show Worked Solution

a.    \(\dfrac{dV}{dt}=\dfrac{8t}{240+5t^4}\)

\(\text {Max rate when} \ \ \dfrac{d^2V}{dt^2}=0\)

\(\text {Solve:  \(\ \dfrac{d^{2}V}{dt^2}=\dfrac{1920-120 t^4}{\left(240+5t^4\right)^2}=0 \ \)  (by CAS)}\)

\(\Rightarrow t=2 \quad (t>0)\)

\(\dfrac{dV}{dt} \text{(max)}=\dfrac{16}{240+5 \times 2^4}=\dfrac{1}{20} \ \text{m\(^3\)/day.}\)
 

b.    \(r(t) \ \text{is measured in metres,} \ \  h=1 \ \text{mm}=\dfrac{1}{1000} \ \text{m}\)

\(V=\pi r^2 h=\dfrac{\pi r^2}{1000} \Rightarrow \dfrac{dV}{dr}=\dfrac{\pi r}{500}\)

\(\text{At} \ \ t=4, \ \ \dfrac{dV}{dt}=\dfrac{8 \times 4}{240+5 \times 4^4} =\dfrac{2}{95}\)

\(\dfrac{dV}{dt}=\dfrac{dV}{dr} \cdot \dfrac{dr}{dt}\)

\(\text{Find } \dfrac{dr}{dt} \ \text {when}\ \  r=6.54:\)

\(\dfrac{2}{95}=\dfrac{\pi r}{500} \times \dfrac{dr}{dt}\)

\(\dfrac{dr}{dt}=\dfrac{2}{95} \times \dfrac{500}{\pi \times 6.54}=0.51 \ \text{m/day}\)
 

c.i.  \(\displaystyle \int \frac{8t}{240+5t^4} \, dt\)

\(u=\sqrt{5} t^2, \ \dfrac{du}{dt}=2 \sqrt{5} t \ \Rightarrow \ dt=\dfrac{du}{2 \sqrt{5} t}\)

\(\displaystyle \int \frac{8t}{240+5 t^4} \, dt=\int \frac{8}{240+u^2} \cdot \frac{du}{2 \sqrt{5}}=\frac{4}{\sqrt{5}} \int \frac{du}{240+u^2}\)
 

c.ii.  \(V\) \(=\displaystyle\frac{4}{\sqrt{5}} \int_0^{\sqrt{5} t^2} \frac{1}{240+u^2} \, du\)
    \(=\displaystyle\frac{4}{\sqrt{5 \times 240}} \int_0^{\sqrt{5} t^2} \frac{\sqrt{240}}{240+u^2} \, du\)
    \(=\displaystyle \frac{1}{\sqrt{75}}\left[\tan ^{-1}\left(\frac{u}{\sqrt{240}}\right)\right]_0^{\sqrt{5} t^{2}}\)
    \(=\displaystyle \frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{t^2}{4 \sqrt{3}}\right)\)

 

d.    \(V=\text{S.A.} \times \text{height}=\text{S.A.} \times \dfrac{1}{1000}\)

\(\text{S.A.}=1000 \times V=\dfrac{200}{\sqrt{3}} \tan^{-1} \left(\dfrac{t^2}{4 \sqrt{3}}\right)\)

\(\text{As} \ t \rightarrow \infty, \tan^{-1}\left(\dfrac{t^2}{4 \sqrt{3}}\right) \rightarrow \dfrac{\pi}{2}\)

\(\text{As} \ t \rightarrow \infty, \ \text{S.A.} \ \rightarrow \dfrac{200}{\sqrt{3}} \cdot \dfrac{\pi}{2} = \dfrac{100\pi}{\sqrt{3}}\)

♦♦♦ Mean mark (d) 22%.

e.    \(\text{After 5 days}\)

\(V=\dfrac{1}{5 \sqrt{3}} \tan^{-1}\left(\dfrac{25}{4 \sqrt{3}}\right)\)

\(\text{Pollutant removed (after day 5)}=\displaystyle \int_{5}^{t+5}\left(\dfrac{8 x}{240+5x^4}-0.05\right)\)

\(\text{Find \(t\) such that:}\)

\(\displaystyle \int_5^{t+5}\left(\frac{8 x}{240+5 x^4}-0.05\right)+V(5)=0\)

\(t=3.4 \ \text{days}\)

♦♦♦ Mean mark (e) 13%.

Calculus, SPEC2 2021 VCAA 3

A thin-walled vessel is produced by rotating the graph of  `y = x^3-8`  about the `y`-axis for  `0 <= y <= H`.

All lengths are measured in centimetres.

    1. Write down a definite integral in terms of `y` and `H` for the volume of the vessel in cubic centimetres.   (1 mark)

    2. Hence, find an expression for the volume of the vessel in terms of `H`.   (1 mark)

Water is poured into the vessel. However, due to a crack in the base, water leaks out at a rate proportional to the square root of the depth `h` of water in the vessel, that is  `(dV)/(dt) = -4sqrth`, where `V` is the volume of water remaining in the vessel, in cubic centimetres, after `t` minutes.

    1. Show that  `(dh)/(dt) = (-4sqrth)/(pi(h + 8)^(2/3))`.   (2 marks)

    2. Find the maximum rate, in centimetres per minute, at which the depth of water in the vessel decreases, correct to two decimal places, and find the corresponding depth in centimetres.   (2 marks)

    3. Let  `H = 50`  for a particular vessel. The vessel is initially full and water continues to leak out at a rate of  `4 sqrth`  cm³ min`\ ^(-1)`.
    4. Find the maximum rate at which water can be added, in cubic centimetres per minute, without the vessel overflowing.   (1 mark)

  2. The vessel is initially full where  `H = 50`  and water leaks out at a rate of  `4sqrth`  cm³ min`\ ^(-1)`. When the depth of the water drops to 25 cm, extra water is poured in at a rate of  `40sqrt2`  cm³ min`\ ^(-1)`.
  3. Find how long it takes for the vessel to refill completely from a depht of 25 cm. Give your answer in minutes, correct to one decimal place.   (3 marks)

Show Answers Only
    1. `V = pi int_0^H (y + 8)^(2/3)\ dy`
    2. `V = (3pi)/5 [(H + 8)^(5/3)-32]`
    1. `text(See Worked Solutions)`
    2. `0.62\ text(cm/min when)\ \ h = 24.`
    3. `4sqrt50\ \ text(cm³/min)`
  3. `31.4\ text(mins)`
Show Worked Solution

a.i.   `y = x^3-8 \ => \ x = root3(y + 8) \ => \ x^2 = (y + 8)^(2/3)`

`:. V = pi int_0^H (y + 8)^(2/3)\ dy`

 

a.ii.   `V = (3pi)/5 [(H + 8)^(5/3)-32]`

 

b.i.   `(dV)/(dt) = -4sqrth`

`(dV)/(dh) = pi(h + 8)^(2/3) \ => \ (dh)/(dV) =1/(pi(h + 8)^(2/3))`

`(dh)/(dt)` `= (dV)/(dt) * (dh)/(dV)`
  `= (-4sqrth)/(pi(h + 8)^(2/3))`

 

b.ii.   `text(Solve)\ (d^2h)/(dt^2) = 0\ \ text(for)\ \ h\ \ text{(by CAS):}`

♦ Mean mark part (b)(ii) 42%.

`(d^2h)/(dt^2) = (2(h-24))/(3sqrth pi (h + 8)^(5/3))=0`

`=>h = 24`

`text(At)\ \ h=24, \ (dh)/(dt) = -0.62\ text(cm/min)`

`:.\ text(Max rate at which depth decreases is)`

`0.62\ text(cm/min when)\ \ h = 24.`

♦♦ Mean mark part (b)(iii) 24%.

 

b.iii.   `text(At)\ \ H = 50, text(vessel is full and losing water at)\ \ 4sqrt50\ \ text(cm³/min)`

`:. text(Water can be added at a max-rate of)\ \ 4sqrt50\ \ text(cm³/min and)`

`text(vessel will not overflow.)`

 

c.   `(dV)/(dt) = 40sqrt2-4sqrth`

♦♦♦ Mean mark part (c) 16%.

`(dV)/(dh) · (dh)/(dt) = 40sqrt2-4sqrth`

`pi(h + 8)^(2/3) · (dh)/(dt)` `= 40sqrt2-4sqrth`
`(dh)/(dt)` `= (40sqrt2-4sqrth)/(pi(h + 8)^(2/3)`
`(dt)/(dh)` `= (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)`
`t` `= int (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`

 
`text(Time of vessel to refill from)\ \ h = 25\ \ text(to)\ \ h = 50:`

`t` `= int_25^50 (pi(h + 8)^(2/3))/(40sqrt2-4sqrth)\ dh`
  `~~ 31.4\ text(mins)`

Calculus, SPEC1 2019 VCAA 10 MC


 

Sand falls from a chute to form a pile in the shape of a right circular cone with semi-vertex angle 60°. Sand is added to the pile at a rate of 1.5 m³ per minute.

The rate at which the height  `h`  metres of the pile is increasing, in metres per minute, when the height of the pile is 0.5 m, correct to two decimal places, is

  1. 0.21
  2. 0.31
  3. 0.64
  4. 3.82
  5. 3.53
Show Answers Only

`C`

Show Worked Solution

`V = 1/3 pir^2h`

`tan 60° = r/h\ =>\ r = sqrt3 h`

`V = 1/3 pi(sqrt3 h)^2h = pi h^3`

`(dV)/(dh)` `= 3pih^2`
`(dV)/(dt)` `= 3/2`
`(dh)/(dt)` `= (dh)/(dV) · (dV)/(dt)`
  `= 1/(3pih^2) · 3/2`
  `= 1/(2pih^2)`

 
`text(When)\ \ h = 0.5`

`(dh)/(dt) = 1/(2pi(0.5)^2) = 0.636…`

`=>C`

Calculus, SPEC2-NHT 2017 VCAA 1

  1. i.  Use an appropriate double angle formula with  `t = tan((5 pi)/12)`  to deduce a quadratic equation of the form  `t^2 + bt + c = 0`, where `b` and `c` are real values.   (2 marks)

  2. ii. Hence show that  `tan((5 pi)/12) = 2 + sqrt 3`.   (1 mark)

Consider  `f: [sqrt 3, 6 + 3 sqrt 3] -> R,\ \ f(x) = arctan (x/3)-pi/6`.

  1. Sketch the graph of `f` on the axes below, labelling the end points with their coordinates.   (3 marks)


 

  1. The region between the graph of  `f`  and the `y`-axis is rotated about the `y`-axis to form a solid of revolution.
  2. i.  Write down a definite integral in terms of  `y`  that gives the volume of the solid formed.   (2 marks)

  3. ii. Find the volume of the solid, correct to the nearest integer.   (1 mark)

  1. A fish pond that has a shape approximately like that of the solid of revolution in part c. is being filled with water. When the depth is `h` metres, the volume, `V\ text(m)^3`, of water in the pond is given by

     

    `qquad V = tan(h + pi/6)-h-sqrt 3/3`

     

    If water is flowing into the pond at a rate of 0.03 m³ per minute, find the rate at which the depth is increasing when the depth is 0.6 m. Give your answer in metres per minute, correct to three decimal places.   (3 marks)

Show Answers Only
  1. i.  `t^2-2t sqrt 3-1 = 0`
  2. ii. `text(Proof)\ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
  4. i.  `V = pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`
  5. ii. `67`
  6. `0.007\ text(m/min)`
Show Worked Solution

a.i.   `tan (theta/2) = tan ((5pi)/12) = t`

`tan (theta)` `=(2 tan (theta/2))/(1-tan^2(theta/2)`
`tan((5 pi)/6)` `= (2tan ((5pi)/12))/(1-tan^2((5pi)/2)`
`-1/sqrt 3` `= (2t)/(1-t^2), quad (t != +- 1)`
`-(1-t^2)` `= 2t sqrt 3`
`-1 + t^2` `= 2t sqrt 3`

 
`:. t^2-2 sqrt 3 t-1 = 0`

 

a.ii.    `t^2-2 sqrt 3 t + (-sqrt 3)^2 +3 -4 = 0`
  `(t-sqrt 3)^2` `= 4`
  `t-sqrt 3` `= +- 2`
  `t` `= sqrt 3 +-2`

 
`=> tan ((5 pi)/12)\ \ text(is in 1st quadrant,)`

`:. t` `= tan ((5 pi)/12) = 2 + sqrt 3`

 

b.    `f(sqrt 3)` `= 0`
`f(6 + 3 sqrt 3)` `=pi/4`

`text(Graphing)\ \ f(x) = arctan (x/3)-pi/6\ \ text(on CAS will)`

`text(show the shape of the graph.)`
 

 

c.i.    `y` `= tan^(-1)(x/3)-pi/6`
  `y + pi/6` `= tan^(-1)(x/3)`
  `x/3` `= tan(y + pi/6)`
  `x` `= 3 tan (y + pi/6)`
  `x^2` `= 9 tan^2 (y + pi/6)`

 

`:. V` `= pi int_0^(pi/4) x^2\ dy`  
  `=pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`  

 

c.ii.  `pi int_0^(pi/4) 9 tan^2 (y + pi/6)\ dy`

`=66.99…`

`=67\ text(u³)`

 

d.    `(dV)/(dt)` `= 0.03`
  `V` `= tan(h + pi/6)-h-sqrt 3/3`
  `(dV)/(dh)` `= tan^2 (h+pi/6)\ \ \ text{(by CAS)}`
  `(dh)/(dt)` `= (dh)/(dV) * (dV)/(dt)`
    `= 1/(tan^2 (h+pi/6)) xx 3/100`

  
`(dh)/(dt)|_(h = 0.6)~~0.007\ text(m/min)`

Calculus, SPEC2-NHT 2018 VCAA 1

Consider the function  `f`  with rule  `f(x) = 10 arccos (2-2x)`.

  1.  Sketch the graph of  `f`  over its maximal domain on the set of axes below. Label the endpoints with their coordinates.   (3 marks)

     


 

  1. A vase is to be modelled by rotating the graph of  `f`  about the `y`-axis to form a solid of revolution, where units of measurement are in centimetres.
    1.  Write down a definite integral in terms of `y` that gives the volume of the vase.   (2 marks)

    2.  Find the volume of the vase in cubic centimetres.  (1 mark)

  3. Water is poured into the vase at a rate of 20 cm³ s¯¹.
  4. Find the rate, in centimetres per second, at which the depth of the water is changing when the depth is  `5 pi`  cm.  (3 marks)

  5. The vase is placed on a table. A bee climbs from the bottom of the outside of the vase to the top of the vase.
  6. What is the minimum distance the bee will need to travel? Give your answer in centimetres, correct to one decimal place.  (1 mark)

Show Answers Only

  1. `text(See Worked Solutions)`
  2.  i. `V = pi int_0^(10 pi) (1-1/2 cos (y/10))^2 dy`
  3. ii. `V = (45 pi^2)/4 text(cm)^3`
  4. `20/pi text(cm s)^(-1)`
  5. `31.4\ text(cm)`

Show Worked Solution

a.  

`2-2x in [-1, 1]`

`-2x in [-3, -1]`

`:. x in [1/2, 3/2]`

`f(1/2)` `= 10 cos^(-1) (1)=0`
`f(3/2)` `= 10 cos^(-1) (1)=10pi`

 

b.i.   `y` `= 10 cos^(-1) (2-2x)`
  `y/10` `= cos^(-1) (2-2x)`
  `cos (y/10)` `= 2-2x`
  `2x` `= 2-cos (y/10)`
  `x` `= 1-1/2 cos (y/10)`

 

  `:. V` `= pi int_0^(10 pi) x^2\ dy`
    `= pi int_0^(10 pi) (1-1/2 cos (y/10))^2\ dy`

 

b.ii.   `V = (45 pi^2)/4 text(cm)^3`

 

c.  `(dV)/(dt) = 20\ text(cm³/s)\ \ \ text{(given)}`

`(dV)/(dh) = pi (1-1/2 cos (y/10))^2\ \ text(when)\ \ y=5pi` 

`=> (dV)/(dh) = pi`
 

`:. (dh)/(dt)` `= (dh)/(dV)*(dV)/(dt)`
  `= 1/pi * 20`
  `= 20/pi\ text(cm s)^(-1)`

 

d.   `f(x) = 10cos^(-1)(2-2x)`

`l=int_(1/2)^(3/2) sqrt(1 + (f′(x))^2)\ dx`

  `~~31.4\ text(cm)\ \ \ text{(by CAS)}`

Calculus, SPEC2 2018 VCAA 3

Part of the graph of  `y = 1/2 sqrt(4x^2-1)`  is shown below.
 


 

The curve shown is rotated about the `y`-axis to form a volume of revolution that is to model a fountain, where length units are in metres.

  1. Show that the volume, `V` cubic metres, of water in the fountain when it is filled to a depth of `h` metres is given by  `V = pi/4(4/3h^3 + h)`.   (2 marks)

  2. Find the depth `h` when the fountain is filled to half's its volume. Give your answer in metres, correct to two decimal places.   (2 marks)

The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of  `0.05 sqrt h`  cubic metres per second when the depth is `h` metres.

  1.  i. Show that  `(dh)/(dt) = (4-5sqrt h)/(25 pi (4h^2 + 1))`.   (2 marks)

  2. ii. Find the rate, in metres per second, correct to four decimal places, at which the depth is increasing when the depth is 0.25 m.   (1 mark)

  3. Express the time taken for the depth to reach 0.25 m as a definite integral and evaluate this integral correct to the nearest tenth of a second.   (2 marks)

  4. After 25 seconds the depth has risen to 0.4 m.
    Using Euler's method with a step size of five seconds, find an estimate of the depth 30 seconds after the fountain began to fill. Give your answer in metres, correct to two decimal places.   (2 marks)

  5. How far from the top of the fountain does the water level ultimately stabilise? Give your answer in metres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `h~~ 0.59\ text(m)`
    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `0.0153\ text(ms)^(-1)`
  1. `9.8\ text(seconds)`
  2. `0.43\ text(m)`
  3. `0.23\ text(m)`
Show Worked Solution

a.   `V= pi int_0^h x^2\ dy`

`y` `=1/2 sqrt(4x^2-1)`
`2y` `=sqrt(4x^2-1)`
`4y^2` `= 4x^2-1`
`4x^2` `= 4y^2 + 1`
`x^2` `= y^2 + 1/4`

 

`:. V` `= pi int_0^h y^2 + 1/4\ dy`
  `= pi[y^3/3 + y/4]_0^h`
  `= pi(h^3/3 + h/4-0)`
  `= pi(1/4((4h^3)/3 + h))`
  `= pi/4((4h^3)/3 + h)\ \ …\ text(as required)`

 

b.    `V_text(max)` `= pi/4 (4/3 xx (sqrt 3/2)^3 + sqrt 3/2)`
    `= pi/4 (sqrt 3/2 + sqrt 3/2)`
    `= (pi sqrt 3)/4`

 

`1/2 V_text(max)` `= (pi sqrt 3)/8`
`(pi sqrt 3)/8` `= pi/4 (4/3 h^3 + h)`
`sqrt 3/2` `= 4/3 h^3 + h`
`:. h` `~~0.59\ text(m)`

  

c.i.   `((dV)/(dt))_text(in)` `= 0.04`
  `((dV)/(dt))_text(out)` `= 0.05 sqrt h`
  `(dV)/(dt)` `= 0.04-0.05 sqrt h`
    `= (4-5 sqrt h)/100`
     
  `(dV)/(dh)` `= pi/4(4h^2 + 1)`
  `:. (dh)/(dt)` `= (dh)/(dV) ⋅ (dV)/(dt)`
    `= 4/(pi(4h^2 + 1)) xx (4-5 sqrt h)/100`
    `= (4-5 sqrt h)/(25 pi (4h^2 + 1))`

 

c.ii.   `(dh)/(dt)|_(h = 0.25)` `= (4-5 sqrt(0.25))/(25 pi (4(0.25)^2 + 1))`
    `~~ 0.0153\ text(ms)^(-1)`

 

d.   `(dt)/(dh) = (25 pi (4h^2 + 1))/(4-5 sqrt h)`

`:. t(0.25)` `= int_0^0.25 (25 pi (4h^2 + 1))/(4-5 sqrt h)\ dh`
  `~~9.8\ text(seconds)`

 

e.   `text(When)\ \ t=25,\ \ h=0.4\ \ text{(given)}`

♦♦ Mean mark part (e) 30%.

`:. h(30)` `~~ h(25) + 5 xx (dh)/(dt)|_(h = 0.4)`
  `~~ 0.4 + 5 xx ((4-5 sqrt 0.4)/(25 pi (4(0.4)^2 + 1)))`
  `~~ 0.43\ text(m)`

 

f.    `(dV)/(dt) = 0`

♦♦ Mean mark part (f) 32%.

`0.04-0.05 sqrt h` `= 0`
`0.04` `= 0.05 sqrt h“
`sqrt h` `= 4/5`
`h` `= 16/25`

  

`d` `= h_max-16/25`
  `= sqrt 3/2-16/25`
  `~~ 0.23\ text(m)`