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Calculus, EXT1 C1 2024 HSC 11f

The volume of a sphere of radius \(r\) cm, is given by  \(V=\dfrac{4}{3} \pi r^3\),  and the volume of the sphere is increasing at a rate of \(10 \text{ cm}^3 \text{ s}^{-1}\).

Show that the rate of increase of the radius is given by  \(\dfrac{d r}{d t}=\dfrac{5}{2 \pi r^2} \text{ cm s}^{-1}\).   (2 marks)

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  \(\dfrac{d r}{d t}\) \(=\dfrac{d V}{d t} \times \dfrac{d r}{d V}\)
    \(=10 \times \dfrac{1}{4 \pi r^2}\)
    \(=\dfrac{5}{2 \pi r^2} \text{ cm/s}\)

Show Worked Solution

\(V=\dfrac{4}{3} \pi r^3 \ \Rightarrow \ \dfrac{d V}{d r}=4 \pi r^2\)

\(\dfrac{d V}{d t}=10\)

  \(\dfrac{d r}{d t}\) \(=\dfrac{d V}{d t} \times \dfrac{d r}{d V}\)
    \(=10 \times \dfrac{1}{4 \pi r^2}\)
    \(=\dfrac{5}{2 \pi r^2} \text{ cm/s}\)

Calculus, EXT1 C1 EQ-Bank 2

A wheat silo in the shape of a right cone is pictured below.

The silo has a height of 12 metres and a base radius of 5 metres. Wheat is poured into the top of the silo at a rate of 3 cubic metres per minute.

At any given time, the wheat filling the silo will have a vertical height of \(h\) metres from the bottom tip and a radius \(r\).
 

  1. Explain why  \(r=\dfrac{5h}{12} \)   (1 mark)
  2. Find an expression for \(\dfrac{dh}{dt}\).   (2 marks)
  3. At what rate is the height of the wheat in the cone increasing when the height of the wheat in the cone is 7 metres from the bottom?   (1 mark)

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\(\dfrac{432}{1225\pi}\ \text{m/minute} \)

Show Worked Solution

a.   \(\text{Using similar triangles:}\)

\(\dfrac{r}{h}=\dfrac{5}{12}\ \ \Rightarrow \ r=\dfrac{5h}{12} \)
 

b.   \(V =\dfrac{1}{3} \pi r^2 h=\dfrac{1}{3} \pi\left(\dfrac{5h}{12}\right)^2 \times h = \dfrac{25\pi}{432} \times h^{3} \)

\(\dfrac{dV}{dh} = \dfrac{3 \times 25\pi}{432} \times h^{2} = \dfrac{25\pi}{144} \times h^{2}\)

\(\Rightarrow \dfrac{dh}{dV}= \dfrac{144}{25\pi \times h^{2}}\)

\(\Rightarrow \dfrac{dV}{dt}=3\ \ \text{(given)}\)
 

\(\dfrac{d h}{d t} =\dfrac{d h}{dV} \times \dfrac{dV}{dt} = 3 \times \dfrac{144}{25\pi \times h^{2}} \)

 
c.   \(\text {When}\ \  h=7:\)

\(\dfrac{d h}{d t} = 3 \times \dfrac{144}{25\pi \times 7^{2}}=\dfrac{432}{1225\pi}\ \text{m/minute} \)

Calculus, EXT1 C3 2023 HSC 13a

A hemispherical water tank has radius \(R\) cm. The tank has a hole at the bottom which allows water to drain out.

Initially the tank is empty. Water is poured into the tank at a constant rate of  \(2 k R\) cm³ s\(^{-1}\), where \(k\) is a positive constant.

After \(t\) seconds, the height of the water in the tank is \(h\) cm, as shown in the diagram, and the volume of water in the tank is \(V\) cm³.
  

It is known that  \(V= \pi \Big{(} R h^2-\dfrac{h^3}{3}\Big{)}. \)    (Do NOT prove this.)

While water flows into the tank and also drains out of the bottom, the rate of change of the volume of water in the tank is given by  \(\dfrac{d V}{d t}=k(2 R-h)\).

  1. Show that  \(\dfrac{d h}{d t}=\dfrac{k}{\pi h}\).  (2 marks)

  2. Show that the tank is full of water after  \(T=\dfrac{\pi R^2}{2 k}\) seconds.  (2 marks)

  3. The instant the tank is full, water stops flowing into the tank, but it continues to drain out of the hole at the bottom as before.
  4. Show that the tank takes 3 times as long to empty as it did to fill.  (3 marks)

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  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(V=\pi \Big{(}Rh^2-\dfrac{h^3}{3} \Big{)} \)

\(\dfrac{dV}{dh} = \pi(2Rh-h^2) \)

\(\dfrac{dV}{dt} = k(2R-h)\ \ \ \text{(given)} \)

\(\dfrac{dh}{dt}\) \(= \dfrac{dV}{dt} \cdot \dfrac{dh}{dV} \)  
  \(=k(2R-h) \cdot \dfrac{1}{\pi} \cdot \dfrac{1}{h(2R-h)} \)  
  \(= \dfrac{k}{\pi h} \)  

 
ii.    \(\dfrac{dt}{dh} = \dfrac{\pi h}{k} \)

\(t\) \(= \displaystyle \int \dfrac{dt}{dh}\ dh \)  
  \(= \dfrac{\pi}{k} \displaystyle \int h\ dh \)  
  \(= \dfrac{\pi}{k} \Big{[} \dfrac{h^2}{2} \Big{]} +c \)  

 
\(\text{When}\ \ t=0, h=0 \)

\(\Rightarrow c=0 \)

\( t= \dfrac{\pi h^2}{2k} \)

 
\(\text{Tank is full at time}\ T\ \text{when}\ \ h=R: \)

\( T= \dfrac{\pi R^2}{2k}\ \text{seconds} \)

♦ Mean mark (ii) 41%.

iii.   \(\text{Net water flow}\ = k(2R-h)\ \ \text{(given)} \)

\(\text{Flow in}\ =2kR\ \ \text{(given)} \)

\(\text{Flow out}\ = k(2R-h)-2kR=-kh \)
 

\( \dfrac{dh}{dt}= \dfrac{-kh}{\pi h(2R-h)} = \dfrac{-k}{\pi (2R-h)} \)

♦♦♦ Mean mark (iii) 20%.
 

\(\dfrac{dt}{dh}\) \(=\dfrac{- \pi (2R-h)}{k} \)  
\( \displaystyle \int k\ dt\) \(=- \pi \displaystyle \int (2R-h)\ dh \)  
\(kt\) \(=- \pi \Big{(} 2Rh-\dfrac{h^2}{2} \Big{)}+c \)  

 
\(\text{When}\ \ t=0, \ h=R: \)

\(0\) \(=- \pi \Big{(}2R^2-\dfrac{R^2}{2} \Big{)} + c\)  
\(c\) \(= \pi \Big{(} \dfrac{3R^2}{2} \Big{)} \)  

 
\(\text{Find}\ t\ \text{when}\ h=0: \)

\(kt\) \(=- \pi(0) + \pi \dfrac{3R^2}{2} \)  
\(t\) \(= \dfrac{3 \pi R^2}{2k} \)  
  \(= 3 \times \dfrac{\pi R^2}{2k} \)  

 
\(\therefore\ \text{Tank takes 3 times longer to empty than fill.} \)

Calculus, EXT1 C1 2021 HSC 11e

A spherical bubble is moving up through a liquid. As it rises, the bubble gets bigger and its radius increases at the rate of 0.2 mm/s.

At what rate is the volume of the bubble increasing when its radius reaches 0.6 mm? Express your answer in mm³/s rounded to one decimal place.  (2 marks)

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`0.9\ text{mm³/s}`

Show Worked Solution

`V= 4/3 pi r^3\ \ =>\ \ (dV)/(dr) = 4pir^2`

`(dr)/dt = 0.2\ text(mm/s)`

`text(Find)\ \ (dV)/(dt) \ \ text(when)\ \ r = 0.6:`

`(dV)/(dt)` `= (dV)/(dr) · (dr)/(dt)`
  `= 4pi(0.6)^2 · 0.2`
  `= 0.9047…`
  `= 0.9\ text{mm³ /s  (1 d.p.)}`

Calculus, EXT1 C1 2016 HSC 12a

The diagram shows a conical soap dispenser of radius 5 cm and height 20 cm.
 

     ext1-2016-hsc-q12
 

At any time `t` seconds, the top surface of the soap in the container is a circle of radius `r` cm and its height is `h` cm.

The volume of the soap is given by  `v = 1/3 pir^2h`.

  1.  Explain why  `r = h/4`.  (1 mark)

  2.  Show that  `(dv)/(dh) = pi/16 h^2`.  (1 mark)

The dispenser has a leak which causes soap to drip from the container. The area of the circle formed by the top surface of the soap is decreasing at a constant rate of  `0.04\ text(cm² s)^-1`.
 

  1.  Show that  `(dh)/(dt) = (−0.32)/(pih)`.  (2 marks)

  2.  What is the rate of change of the volume of the soap, with respect to time, when `h = 10`?  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `-0.2\ text(cm³ s)^-1`
Show Worked Solution
i.   ext1-hsc-2016-12a

`text(Using similar triangles,)`

`r/h` `= 5/20`
 `:. r` `= h/4\ text(… as required)`

 

ii.   `v` `= 1/3 pi r^2 h`
    `= 1/3 pi · (h/4)^2 h`
    `= (pi h^3)/48`
  `:. (dv)/(dh)` `= 3 xx (pi h^2)/48`
    `= (pi h^2)/16\ text(… as required.)`

 

iii.   `(dA)/(dt)` `= -0.04\ text(cm² s)^-1`
  `A` `= pi r^2`
    `= (pi h^2)/16`
  `:. (dA)/(dh)` `= (pi h)/8`

 

`(dA)/(dt)` `= (dA)/(dh) xx (dh)/(dt)`
`-0.04` `= (pi h)/8 xx (dh)/(dt)`
`:. (dh)/(dt)` `= (-0.32)/(pi h)\ text(… as required.)`

 

iv.   `(dv)/(dt)` `= (dv)/(dh) · (dh)/(dt)`
    `= (pi h^2)/16 · (-0.32)/(pi h)`
    `= (-0.32 h)/16`

 

`text(When)\ \ h =10,`

`(dv)/(dt)` `= (-0.32 xx 10)/16`
  `= -0.2\ text(cm³ s)^-1`

Calculus, EXT1 C1 2006 HSC 5c

2006 5c
 

A hemispherical bowl of radius  `r\ text(cm)`  is initially empty. Water is poured into it at a constant rate of  `k\ text(cm³)`  per minute. When the depth of water in the bowl is  `x\ text(cm)`, the volume, `V\ text(cm³)`, of water in the bowl is given by

`V = pi/3 x^2 (3r - x).`    (Do NOT prove this)

  1. Show that  `(dx)/(dt) = k/(pi x (2r - x).`  (2 marks)

  2. Hence, or otherwise, show that it takes 3.5 times as long to fill the bowl to the point where  `x = 2/3r`  as it does to fill the bowl to the point where  `x = 1/3r.`  (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
Show Worked Solution

i.  `text(Show)\ \ (dx)/(dt) = k/(pi x (2r – x))`

`(dV)/(dt)` `= k`
`V` `= pi/3 x^2 (3r – x)`
  `= r pi x^2 – pi/3 x^3`
`(dV)/(dx)` `= 2 pi r x – pi x^2`
  `= pi x (2r – x)`
   
`(dV)/(dt)` `= (dV)/(dx) * (dx)/(dt)`
`k` `= pi x (2r – x) * (dx)/(dt)`
`:. (dx)/(dt)` `= k/(pi x (2r – x))\ \ text(…  as required)`

 

ii.  `(dx)/(dt)` `= k/(pi x (2r – x))`
`(dt)/(dx)` `= 1/k pi x (2r – x)`
`t` `= 1/k int 2 pi r x – pi x^2\ dx`
  `= 1/k [pi r x^2 – 1/3 pix^3] + c`

 
`text(When)\ \ t = 0,\ \ \ x = 0`

`:.\ c = 0`

 `:.t = 1/k [pi r x^2 – 1/3 pi x^3]`

 
`text(Find)\ \ t_1,\ \ text(when)\ \ x = 1/3r`

`t_1` `= 1/k [pi r (r/3)^2 – 1/3 pi (r/3)^3]`
  `= 1/k [(pi r^3)/9 – (pi r^3)/81]`
  `= 1/k ((9 pi r^3)/81 – (pi r^3)/81)`
  `= (8 pi r^3)/(81k)`

 
`text(Find)\ \ t_2\ \ text(when)\ \ x = 2/3r`

`t_2` `= 1/k [pi r((2r)/3)^2 – 1/3 pi ((2r)/3)^3]`
  `= 1/k [(4 pi r^3)/9 – (8 pi r^3)/81]`
  `= 1/k ((36 pi r^3)/81 – (8 pi r^3)/81)`
  `= (28 pi r^3)/(81k)`
  `= 3.5 xx (8 pi r^3)/(81k)`
  `= 3.5 xx t_1`

 
`:.\ text(It takes 3.5 times longer to fill the bowl.)`

Calculus, EXT1 C1 2009 HSC 5b

The cross-section of a 10 metre long tank is an isosceles triangle, as shown in the diagram. The top of the tank is horizontal.
 

 
 

When the tank is full, the depth of water is 3 m. The depth of water at time `t` days is `h` metres.   

  1. Find the volume, `V`, of water in the tank when the depth of water is `h` metres.     (1 mark)

  2. Show that the area, `A`, of the top surface of the water is given by  `A = 20 sqrt3 h`.   (1 mark)

  3. The rate of evaporation of the water is given by  `(dV)/(dt) = - kA`, where `k` is a positive constant. 

     

    Find the rate at which the depth of water is changing at time `t`.   (2 marks)

  4. It takes 100 days for the depth to fall from 3 m to 2 m. Find the time taken for the depth to fall from 2 m to 1 m.   (1 mark)

Show Answers Only
  1. `10 sqrt 3 h^2\ \ text(m³)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `-k\ \ \ text(metres per day)`
  4. `100\ text(days)`
Show Worked Solution
MARKER’S COMMENT: Students who drew a diagram and included their working calculations on it were the most successful.
(i) 
  `text(Let)\ A = text(area of front)`
`tan 30^@` `= h/x`
`x` `= h/(tan 30^@)`
  `= sqrt3 h`
`:.\ A` `= 2 xx 1/2 xx sqrt 3 h xx h`
  `= sqrt 3 h^2\ \ text(m²)`

 

`V` `= Ah`
  `= sqrt 3 h^2 xx 10`
  `= 10 sqrt3 h^2\ \ text(m³)`

 

(ii)    `text(Area of surface)`
  `= 10 xx 2 sqrt 3 h`
  `= 20 sqrt 3 h\ \ text(m²)`

 

(iii)    `(dV)/(dt)` `= -kA`
    `= -k\ 20 sqrt3 h`
`V` `= 10 sqrt3 h^2`
`(dV)/(dh)` `= 20 sqrt 3 h`

 

`text(Find)\ (dh)/(dt)`

MARKER’S COMMENT: Half marks awarded for stating an appropriate chain rule, even if the following calculations were incorrect. Show your working!
`(dV)/(dt)` `= (dV)/(dh) * (dh)/(dt)`
`(dh)/(dt)` `= ((dV)/(dt))/((dV)/(dh))`
  `= (-k * 20 sqrt 3 h)/(20 sqrt 3 h)`
  `= -k`

 

`:.\ text(The water depth is changing at a rate)`

`text(of)\ -k\ text(metres per day.)`

 

♦♦♦ Exact data for part (iv) not available.
COMMENT: Interpreting a constant rate of change was very poorly understood!
(iv)    `text(S)text(ince)\ \ (dh)/(dt)\ \ text(is a constant, each metre)`
  `text(takes the same time.)`

 
`:.\ text(It takes 100 days to fall from 2 m to 1 m.)`

Calculus, EXT1 C1 2011 HSC 7a

The diagram shows two identical circular cones with a common vertical axis.  Each cone has height `h` cm and semi-vertical angle 45°.
 

2011 7a

The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered  is given by  

`(dl)/(dt) = 10`,

where `l` cm is the distance the upper cone has descended into the water after `t` seconds.

As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time `t` is  `V` cm³.

  1. Show that  `V = pi/3(h^3\ - l^3)`.   (1 mark)

  2. Find the rate at which `V` is changing with respect to time when  `l = 2`.     (2 marks)

  3. Find the rate at which `V` is changing with respect to time when the lower cone has lost  `1/8`  of its water. Give your answer in terms of `h`.   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `-40 pi\ \ text(cm³)// text(sec)`
  3. `(-5pih^2)/2\ text(cm³)// text(sec)`
Show Worked Solution

i.   `text(Show that)\ V = pi/3 (h^3\ – l^3)`

 ♦ Mean mark 42% 

`text(S)text(ince)\ \ tan45° = r/h=1`

`=>r=h`

`=>\ text(Radius of lower cone) = h`

`:.\ V text{(lower cone)}` `= 1/3 pi r^2 h`
  `= 1/3 pi h^3`

 
 `text(Similarly,)`

`V text{(submerged upper cone)} = 1/3 pi l^3`

`V text{(water left)}` `= 1/3 pi h^3\ – 1/3 pi l^3`
  `= pi/3 (h^3\ – l^3)\ \ \ text(… as required)`

 

ii.  `text(Find)\ (dV)/(dt)\ text(at)\ l = 2`

`(dV)/(dt)= (dV)/(dl) xx (dl)/(dt)\ …\ text{(1)}`

`=>(dl)/(dt)` `= 10\ text{(given)}`
`text(Using)\ \ V` `= pi/3 (h^3\ – l^3)\ \ \ \ text(from part)\ text{(i)}`
`=>(dV)/(dl)` `= -3 xx pi/3 l^2`
  `= -pi l^2`

  
`text(At)\ \ l = 2,`

`text{Substitute into (1) above}`

`(dV)/(dt)` `= -pi xx 2^2 xx 10`
  `= -40 pi\ \ \ text(cm³)//text (sec)`

 

iii.  `text(Find)\ \ (dV)/(dt)\ \ text(when lower cone has lost)\ 1/8 :`

♦♦♦ Mean mark 12%
MARKER’S COMMENT: Many unsuccessful answers attempted to find an alternate version of `(dV)/(dt)`. Part (ii) directed students directly toward the correct strategy.

`text(Find)\ \ l\ \ text(when)\ \ V = 7/8 xx 1/3 pi h^3`

`pi/3 (h^3\ – l^3)` `= 7/8 xx 1/3 pi h^3`
`h^3 -l^3` `= 7/8 h^3`
`l^3` `= 1/8 h^3`
`l` `= h/2`

   
`text(When)\ \ l = h/2 ,`

`(dV)/(dt)` `= -pi (h/2)^2 xx 10\ \ …\ text{(*)}`
  `= (-5pih^2)/2\ text(cm³)// text(sec)`

 
`:. V\ text(is decreasing at the rate of)\ \  (5 pi h^2)/2\ text(cm³)//text(sec).`