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Calculus, SPEC2 2022 VCAA 9 MC

Euler's method is used to find an approximate solution to the differential equation  `\frac{d y}{d x}=2 x^2`.

Given that  `x_0=1, y_0=2`  and  `y_2=2.976`, the value of the step size `h` is

  1. 0.1
  2. 0.2
  3. 0.3
  4. 0.4
  5. 0.5
Show Answers Only

`B`

Show Worked Solution

`y_{n+1}=y_{n}+y_{n}^{\prime}*h`

`y_1=y_0+y_0^{′}*h = 2+2 xx 1^2 * h=2+2h`

`y_2=y_1+y_1^{′}*h = (2+2h)+2 xx (1+h)^2*h = 2h(1+h)^2+2(1+h)`

`text{Solve:}\ 2h(1+h)^2+2(1+h)=2.976,\ \text{for}\ h:`

`h=0.2`

`=>B`

Calculus, SPEC2 2021 VCAA 8 MC

Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation  `(dy)/(dx) = ysin(x)`.

Given that  `y = 2`  when  `x = 1`, the value of `y`, correct to three decimal places, when  `x = 1.2` is

  1. 2.168
  2. 2.178
  3. 2.362
  4. 2.370
  5. 2.381
Show Answers Only

`C`

Show Worked Solution

`text(Using)\ \ y_0 = 2, \ x_0 = 1, \ h = 0.1:`

`y_1` `= y_0 + h(y_0sinx_0)`
  `= 2 + 0.1(2sin1)`
  `= 2.1683`

 

`y_2` `= 2.1683 + 0.1(2.1683 xx sin 1.1)`
  `= 2.362`

 
`=>\ C`

Calculus, SPEC2 2020 VCAA 12 MC

If  `(dy)/(dx) = e^(cos(x))`  and  `y_0 = e`  when  `x_0 = 0`, then, using Euler's formula with step size 0.1, `y_3` is equal to

  1. `e + 0.1(1 + e^(cos(0.1)))`
  2. `e + 0.1(1 + e^(cos(0.1)) + e^(cos(0.2)))`
  3. `e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)))`
  4. `e + 0.1(e^(cos(0.1)) + e^(cos(0.2)) + e^(cos(0.3)))`
  5. `e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)) + e^(cos(0.3)))`
Show Answers Only

`C`

Show Worked Solution

`text(When)\ \ y_0 = e, x_0 = 0`

`y_1 = e + 0.1 e^(cos(0)) = e + 0.1e`

`y_2` `= e + 0.1e + 0.1e^(cos(0.1))`
  `= e + 0.1(e + e^(cos(0.1)))`
`y_3` `= e + 0.1(e + e^(cos(0.1))) + 0.1e^(cos(0.2))`
  `= e + 0.1(e + e^(cos(0.1)) + e^(cos(0.2)))`

 
`=>C`

Calculus, SPEC2-NHT 2019 VCAA 10 MC

Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation  `1/y(dy)/(dx) = cos(x)`, with  `y = 2`  when  `x = 0`.

When  `x = 0.2`, the value obtained for `y`, correct to four decimal places, is

  1.  2.2000
  2.  2.3089
  3.  2.3098
  4.  2.4189
  5.  2.4199
Show Answers Only

`D`

Show Worked Solution
`1/y · (dy)/(dx)` `= cosx`
`(dy)/(dx)` `= ycosx`

 
`text(Using)\ \ y_0 = 2, x_0 = 0, h = 0.1:`

`y_1` `= y_0 + h(y_0cosx_0)`
  `= 2 + 0.1(2cos0)`
  `= 2.2`
`y_2` `= 2.2 + 0.1(2.2cos(0.1))`
  `= 2.4189`

 
`=>\ D`

Calculus, SPEC2 2012 VCAA 9 MC

Euler's formula is used to find  `y_2`, where  `(dy)/(dx) = cos(x),\ x_0 = 0, \ y_0 = 1`  and  `h = 0.1`.

The value of `y_2` correct to four decimal places is

A.   1.1000 and this is an underestimate of  `y(0.2)`

B.   1.1995 and this is an overestimate of  `y(0.2)`

C.   1.1995 and this is an underestimate of  `y(0.2)`

D.   1.2975 and this is an underestimate of  `y(0.2)`

E.   1.2975 and this is an overestimate of  `y(0.2)`

Show Answers Only

`B`

Show Worked Solution

`x_0 = 0,\ \ y_0 = 1,\ \ h=0.1`

`y_1` `= y_0 + h* (dy)/(dx)|_{(0,1)}`
  `= 1 + 0.1 xx cos(0)`
  `= 1.1`
   
`y_2` `= 1.1 + 0.1 xx cos(0.1)`
  `~~ 1.1995`

 
`text(S)text(ince)\ \ sin(x)\ \ text(is concave down in the region)\ \ x=0.2,`

`1.1995\ text(is an overestimate.)`

`=> B`

Calculus, SPEC2 2011 VCAA 18 MC

The amount of chemical `x` in a tank at time  `t`  is given by the differential equation  `dx/dt = -10/(10 - t)`  and when  `t = 0`, `\ x_0 = 5`. Euler's method is used with a step size of 0.5 in the values of  `t`.

The value of `x` correct to two decimal places when  `t = 1`  is found to be

A.   3.95

B.   3.97

C.   4.50

D.   5.50

E.   6.03

Show Answers Only

`B`

Show Worked Solution
`x_1 ~~ x(0.5)` `= 5 + 0.5 xx (−10/(10 – 0))`
  `= 5 – 0.5`
  `= 4.5`

 

Mean mark 51%.

`x_2 ~~ x(1)` `= 4.5 + 0.5 xx (−10/(10 – 0.5))`
  `= 4.5 – 10/19`
  `~~ 3.97`

`=> B`

Calculus, SPEC2 2013 VCAA 11 MC

Consider the differential equation  `(dy)/(dx) = 1/(3 + 3x + x^2)`, with  `y_0 = 1`  when  `x_0 = 0`.

Using Euler's method with a step size of 0.1, the value of  `y_2` correct to three decimal places, is

A.   1.033

B.   1.063

C.   1.064

D.   1.065

E.   1.066

Show Answers Only

`C`

Show Worked Solution
`y_1` `= y_0 + h xx (dy)/(dx)|_{(x_0,y_0)}`
  `= 1 + 0.1 xx 1/(3 + 3 xx 0 + 0)`
  `= 1.03`

 

`y_2` `= y_1 + h xx (dy)/(dx)|_{(x_1,y_1)}`
  `= 1.03 + 0.1(1/(3 + 0.3 + 0.01))`
  `~~ 1.064`

 
`=> C`

Calculus, SPEC2 2016 VCAA 9 MC

If  `f(x) = (dy)/(dx) = 2x^2 - x`, where  `y_0 = 0 = y(2)`, then  `y_3`  using Euler’s formula with step size 0.1 is

A.  `0.1\ f(2)`

B.  `0.6 + 0.1\ f(2.1)`

C.  `1.272 + 0.1\ f(2.2)`

D.  `2.02 + 0.1\ f(2.3)`

E.  `2.02 + 0.1\ f(2.2)` 

Show Answers Only

`C`

Show Worked Solution
`y_1` `= y_0 + 0.1(2(2)^2 – 2)`
  `= 0.6`
`y_2` `= 0.6 + 0.1 (2(2.1)^2 – 2.1)`
  `= 1.272`
`:.  y_3` `= 1.272 + 0.1\ f(2.2)`

 
`=>  C`

Calculus, SPEC2-NHT 2017 VCAA 3

Bacteria are spreading over a Petri dish at a rate modelled by the differential equation

`(dP)/(dt) = P/2 (1 - P),\ 0 < P < 1`

where  `P`  is the proportion of the dish covered after  `t`  hours.

    1. Express  `2/(P(1 - P))`  in partial fraction form.  (1 mark)
    2. Hence show by integration that  `(t - c)/2= log_e(P/(1 - P))`, where  `c`  is a constant of integration.  (2 marks)
    3. If half of the Petri dish is covered by the bacteria at  `t = 0`, express  `P`  in terms of  `t`.  (2 marks)

After one hour, a toxin is added to the Petri dish, which harms the bacteria and reduces their rate of growth. The differential equation that models the rate of growth is now

`(dP)/(dt) = P/2 (1 - P) - sqrt P/20`  for  `t >= 1`

  1. Find the limiting value of  `P`, which is the maximum possible proportion of the Petri dish that can now be covered by the bacteria. Give your answer correct to three decimal places.  (2 marks)
  2. The total time, `T`  hours, measured from time  `t = 0`, needed for the bacteria to cover 80% of the Petri dish is given by
     

     

    `qquad qquad T = int_q^r (1/(P/2(1 - P) - sqrt P/20)) dP + s`
     

     

    where  `q, r and s in R`.

     

     

    Find the values of  `q, r` and `s`, giving the value of `q` correct to two decimal places.  (2 marks)

  1. Given that  `P = 0.75`  when  `t = 3`, use Euler’s method with a step size of 0.5 to estimate the value of `P` when  `t = 3.5`. Give your answer correct to three decimal places.  (3 marks)
Show Answers Only
    1. `2/(P(1 – P)) = 2/P + 2/(1 – P)`
    2. `text(Proof)\ text{(See Worked Solutions)}`
    3. `P = e^(t/2)/(1 + e^(t/2))`
  2. `P ~~ 0.894`
  3. `r = 0.8`
    `s = 1`
    `q ~~ 0.62`
  4. `~~ 0.775`
Show Worked Solution

a.i.   `2/(P(1 – P)) = A/P + B/(1 ⋅ P)`

`A(1 – P) + BP = 2`

`text(If)\ \ P = 0\ \ =>\ \ A = 2`

`text(If)\ \ P = 1\ \ =>\ \ B = 2`

`:. 2/(P(1 – P)) = 2/P + 2/(1 – P)\ \ \ text{(can also solve by CAS)}`

 

a.ii.  `(dt)/(dP) = 2/(P(1 – P)) = 2/P + 2/(1 – P)`

`t` `= int 2/P + 2/(1 – P)\ dP`
  `= 2 ln |P| – 2ln|1 – P| + c`
`(t-c)/2` `=ln|P| – ln|1-P|`
  `=ln |(P)/(1-P)|`
  `= ln (P/(1 – P))`

  
`text(S)text(ince)\ \ 0 < P < 1 :\ |P| = P\ and\ |1 – P| = 1 – P`


a.iii.  `text(When)\ \ t=0,\ P=0.5`

`(-c)/2` `= ln (0.5/0.5)`
`c` `= ln (1) = 0`

 

`t/2` `= ln (P/(1 – P))\ \ \ text{(solve manually or by CAS)}`
`e^(t/2)` `= P/(1 – P)`
`e^(t/2) (1 – P)` `= P`
`e^(t/2) – Pe^(t/2)` `= P`
`e^(t/2)` `= P(1 + e^(t/2))`
`:. P` `= e^(t/2)/(1 + e^(t/2))`

 

b.   `(dP)/(dt) = P/2 (1 – P) – sqrt P/20`

`text(Limiting value occurs when)\ \ (dP)/(dt) = 0,`

`P ~~ 0.894\ \ \ text{(by CAS)}`
 

`=>\ text(Lower solution values at levels already exceeded)`

 `text(are ignored.)`

 

c.   `(dt)/(dP) = 1/(P/2 (1 – P) – sqrt P/20)\ \ text(for)\ \ t>=1`

`text(When)\ \ t=1\ \ =>\ \ P=0.622`
 

`T = int_0.62^0.8 1/(P/2 (1 – P) – sqrt P/20) dP + 1`
 

`:. r = 0.8,\ s = 1 and q = 0.62`

 

d.   `P(3.5) ~~ P(3) + h* (dP)/(dt)|_(P= 0.75)`

   `= 0.75 + 0.5 (0.75/2 (1 – 0.75) – sqrt 0.75/20)`

   `~~ 0.775`
 

`:. P~~0.775\ \ text(when)\ \ t=3.5`

Calculus, SPEC2 2017 VCAA 9 MC

Consider  `(dy)/(dx) = 2x^2 + x + 1`, where  `y(1) = y_0 = 2`.

Using Euler's method with a step size of 0.1, an approximation to  `y(0.8) = y_2`  is given by

  1.  `0.94`
  2.  `1.248`
  3.  `1.6`
  4.  `2.4`
  5.  `2.852`
Show Answers Only

`B`

Show Worked Solution
`y(0.9)~~y_1` `= y_0 + h * (dy)/(dx)|_{(x_0,y_0)}`
  `= 2 – 0.1 xx (2 xx 1^2 + 1 + 1)`
  `= 2 – 0.1(4)`
  `= 2 – 0.4`
  `= 1.6`

♦ Mean mark 45%.

`y(0.8)~~y_2` `= y_1 + h*(dy)/(dx)|_{(x_1,y_1)}`
  `= 1.6 – 0.1(2(0.9)^2 + 0.9 + 1)`
  `= 1.248`

 
`=>B`

Calculus, SPEC2 2014 VCAA 11 MC

Let  `(dy)/(dx) = x^3 - xy`  and  `y = 2`  when  `x = 1`.

Using Euler’s method with a step size of 0.1, the approximation to  `y`  when  `x = 1.1`  is

A.   0.9

B.   1.0

C.   1.1

D.   1.9

E.   2.1

Show Answers Only

`D`

Show Worked Solution

`text(When)\ \ x=1,\ \ y=2.`

`text(Using Euler when)\ \ x=1.1`

`y(1.1)` `~~ 2 + 0.1 xx (1^3 – 1 xx 2)`
  `= 1.9`

 
`=> D`