SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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)

Show Answers Only
  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 C3 EQ-Bank 9

The rate of fuel running out of a leaking tank can be modelled by the following equation

`(dh)/(dt)=-Ae^(-0.2t)`  where  `h`  is the height of the fuel in the tank after `t` hours?

Initially, the height of the fuel in the tank is  4 metres and after 1.5 hours, it has fallen to 3 metres.

At what height of fuel in the tank will it eventually stabilise, giving your answer to the nearest centimetre?  (4 marks)

Show Answers Only

`text{14 centimetres.}`

Show Worked Solution
`(dh)/(dt)` `=-Ae^(-0.2t)`  
`h` `=int-Ae^(-0.2t)\ dt`  
  `=5Ae^(-0.2t)+C`  

 

`text{As}\ \ t->oo, 5Ae^(-0.2t)->0`

`text{i.e.}\ h\ text{eventually stabilises at}\ C.`

`text{Find}\ A:`

`text{When}\ \ t=0,\ \ h=4`

`4` `=5Ae^(-0.2t)+C`  
`4` `=5Ae^0+C`  
`A` `=(4-C)/5`  

 

`text{When}\ \ t=1.5,\ \ h=3`

`3` `=5xx(4-C)/5 e^(-0.2xx1.5)+C`  
`3` `=4/e^0.3-C/e^0.3+C`  
`(3e^0.3-4)/e^0.3` `=C(1-1/e^0.3)`  
`(3e^0.3-4)/e^0.3` `=C((e^0.3-1)/e^0.3)`  
`C` `=(3e^0.3-4)/e^0.3 xx e^0.3/(e^0.3-1)`  
  `=0.1417…\ text{m}`  
  `=14\ text{cm (nearest cm)`  

 

`:.\ text{Height of fuel in the tank will stabilise at 14 centimetres.}`