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Calculus, EXT1 C1 EQ-Bank 2

75 Tasmanian Devils are placed in a Devil's Ark sanctuary that can support a maximum population of 500 devils. The increase in the devil population is proportional to the difference between the devil population and the number of devils that the sanctuary can support.

  1. Show that  \(D=500-A e^{k t}\)  is a solution to the differential equation \(\dfrac{d D}{d t}=k(D-500)\), where \(D\) is the current devil population, \(t\) is time in years and \(k\) is the constant of proportionality.   (1 mark)

  2. After two years, the population has grown to 245. Show that  \(k=-0.255\), correct to three significant figures.   (2 marks)

  3. Hence, determine how many devils will be on the island after 5 years.  (1 mark)

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a.    \(D\) \(=500-A e^{k t}\)
  \(\dfrac{dD}{dt}\) \(=-kAe^{kt}\)
    \(=k(500-Ae^{kt}-500)\)
    \(=k(D-500)\)

 
b.    \(\text{Find}\ A,\ \text{given}\ \ t=0\ \ \text{when}\ \ D=75:\)

\(\begin{aligned} 75 & =500-A e^0 \\
A & =500-75 \\
& =425
\end{aligned}\)

\(\text{Find}\ k,\ \text{given}\ \ t=2\ \ \text{when}\ \ D=245:\)

\(\begin{aligned}
245 & =500-425 e^{2k} \\
e^{2k} & =\dfrac{500-245}{425} \\
2k & = \log_{e}(0.6) \\
k & =\dfrac{1}{2}\log_{e}(0.6)\\
 & =-0.2554… \\
& =-0.255 \ \text{(3 sig. figures as required)}
\end{aligned}\)

 
c.   \(\text{381 devils}\)

Show Worked Solution

a.    \(D\) \(=500-A e^{k t}\)
  \(\dfrac{dD}{dt}\) \(=-kAe^{kt}\)
    \(=k(500-Ae^{kt}-500)\)
    \(=k(D-500)\)

 
b.    \(\text{Find}\ A,\ \text{given}\ \ t=0\ \ \text{when}\ \ D=75:\)

\(\begin{aligned} 75 & =500-A e^0 \\
A & =500-75 \\
& =425
\end{aligned}\)

\(\text{Find}\ k,\ \text{given}\ \ t=2\ \ \text{when}\ \ D=245:\)

\(\begin{aligned}
245 & =500-425 e^{2k} \\
e^{2k} & =\dfrac{500-245}{425} \\
2k & = \log_{e}(0.6) \\
k & =\dfrac{1}{2}\log_{e}(0.6)\\
 & =-0.2554… \\
& =-0.255 \ \text{(3 sig. figures as required)}
\end{aligned}\)

 
c.   \(\text{Find}\ D\ \text{when}\ \ t=5:\)

\(D\) \(=500-425e^{5 \times -0.255} \)  
  \(=381.24…\)  
  \(=381\ \text{devils (nearest whole)}\)  

Calculus, EXT1 C1 2017 HSC 14c

The concentration of a drug in a body is  `F(t)`, where `t` is the time in hours after the drug is taken.

Initially the concentration of the drug is zero. The rate of change of concentration of the drug is given by   

`F′(t) = 50e^(−0.5t) - 0.4F(t)`.

  1. By differentiating the product  `F(t)e^(0.4t)` show that

     

    `qquadd/(dt)(F(t)e^(0.4t)) = 50e^(−0.1t)`.  (2 marks)

  2. Hence, or otherwise, show that  `F(t) = 500(e^(−0.4t) - e^(−0.5t))`.  (2 marks)

  3. The concentration of the drug increases to a maximum.

     

    For what value of `t` does this maximum occur?  (2 marks)

Show Answers Only
  1. `text(Show Worked Solutions)`
  2. `text(Show Worked Solutions)`
  3. `10 ln\ 5/4`
Show Worked Solution

i.  `text(Show)\ \ d/(dt)(F(t)e^(0.4t)) = 50e^(−0.1t)`

`F′(t) = 50e^(−0.5t) – 0.4F(t)\ …\ text{(1)  (given)}`

 
`text(Using product rule:)`

`d/(dt)(F(t)e^(0.4t))` `= F′(t)e^(0.4t) + 0.4e^(0.4t)F(t)`
  `= e^(0.4t)(F′(t) + 0.4F(t))`
  `= e^(0.4t) · 50e^(−0.5t)\ \ text{(using (1) above)}`
  `= 50e^(−0.1t)\ \ …\ text(as required)`

 

ii.   `text(Show)\ \ F(t) = 500(e^(−0.4t) – e^(−0.5t))`

♦♦ Mean mark 37%.
`F(t)e^(0.4t)` `= int 50e^(−0.1t)\ dt`
  `= 50/(−0.1) · e^(−0.1t) + c`
  `= −500 e^(−0.1t) + c`

 
`text(When)\ \ t = 0, \ F(t) = 0`

`0` `= −500 e^0 + c`
`c` `= 500`

 

`F(t)e^(0.4t)` `= 500 – 500e^(−0.1t)`
`:. F(t)` `= 500e^(−0.4t) – 500e^(−0.5t)`
  `= 500 (e^(−0.4t) – e^(−0.5t))`

 

iii.    `F(t)` `= 500(e^(−0.4t) – e^(−0.5t))`
  `F′(t)` `= 500(−0.4e^(-0.4t) + 0.5e^(−0.5t))`

 

`text(Find)\ t\ text(when)\ F′(t) = 0 :`

♦♦ Mean mark 39%.
`0.4e^(−0.4t)` `= 0.5e^(−0.5t)`
`(e^(−0.4t))/(e^(−0.5t))` `= 0.5/0.4`
`e^(0.1t)` `= 5/4`
`0.1t` `= ln (5/4)`
`:. t_text(max)` `= 10 ln (5/4)`

Calculus, EXT1 C1 2016 HSC 12b

In a chemical reaction, a compound `X` is formed from a compound `Y`. The mass in grams of `X` and `Y` are `x(t)` and `y(t)` respectively, where `t` is the time in seconds after the start of the chemical reaction.

Throughout the reaction the sum of the two masses is 500 g. At any time `t`, the rate at which the mass of compound `X` is increasing is proportional to the mass of compound `Y`.

At the start of the chemical reaction, `x = 0`  and  `(dx)/(dt) = 2`.

  1.  Show that  `(dx)/(dt) = 0.004(500 - x)`.  (3 marks)

  2.  Show that  ` x = 500 - Ae^(−0.004t)`  satisfies the equation in part (i), and find the value of `A`.  (2 marks)

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

i.   `x + y = 500\ \ text{(given)}`

♦ Mean mark (part i) 43%.

`(dx)/(dt)` `= ky`
  `= k(500 – x)`

 
`text(When)\ \ t = 0,\ \ x = 0,\ \ (dx)/(dt) = 2`

`2` `= k (500 – 0)`
`:. k` `= 0.004`

 
`:. (dx)/(dt) = 0.004 (500 – x)\ text(… as required)`

 

ii.    `x` `= 500 – Ae^(-0.004t)`
  `Ae^(-0.004t)` `= 500 – x`

 

`(dx)/(dt)` `= 0.004 Ae^(-0.004t)`
  `= 0.004 (500 – x)`

 
`text(When)\ \ t = 0,\ \ x = 0,`

`0` `= 500 – Ae^0`
`:. A` `= 500`

Calculus, EXT1 C1 2010 HSC 2b

The mass `M` of a whale is modelled by

`M=36-35.5e^(-kt)` 

where  `M`  is measured in tonnes,  `t`  is the age of the whale in years and  `k`  is a positive constant.

  1. Show that the rate of growth of the mass of the whale is given by the differential equation
     
    `qquad qquad (dM)/(dt)=k(36-M)`    (1 mark)

  2. When the whale is 10 years old its mass is 20 tonnes.

     

    Find the value of  `k`,  correct to three decimal places.    (2 marks)

  3. According to this model, what is the limiting mass of the whale?    (1 mark)

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  1. `text{Proof (See Worked Solutions)}`
  2. `0.080`
  3. `36\ text(tonnes)`
Show Worked Solution

i.  `M=36-35.5e^(-kt)`

IMPORTANT: Know this standard proof well and be able to produce it quickly.

`35.5e^(-kt)=36-M`

`:. (dM)/(dt)` `=-kxx-35.5e^(-kt)`
  `=kxx35.5e^(-kt)`
  `=k(36-M)\ \ \ text(… as required)`

 

ii.  `text(Find)\ \ k`

`text(When)\ \ t=10,\ \ M=20`

`M` `=36-35.5e^(-kt)`
`20` `=36-35.5e^(-10k)`
`35.5e^(-10k)` `=16`
`lne^(-10k)` `=ln(16/35.5)`
`-10k` `=ln(16/35.5)`
`:. k` `=-ln(16/35.5)/10`
  `=0.07969…`
  `=0.080\ \ text{(to 3 d.p.)}`

 

iii.  `text(As)\ t->oo,  e^(-kt)=1/e^(kt)\ ->0,\ \ k>0`

`M->36`

`:.\ text(The whale’s limiting mass is 36 tonnes.)`