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Calculus, SPEC2 2022 VCAA 8 MC

The direction field shown above best represents the differential equation

  1. `\frac{d y}{d x}=\frac{2 x}{y}`
  2. `\frac{d y}{d x}=-\frac{x}{2 y}`
  3. `\frac{d y}{d x}=-\frac{2 x}{y}`
  4. `\frac{d y}{d x}=\frac{y^2}{2}+x^2`
  5. `\frac{d y}{d x}=\frac{x^2}{2}+y^2`
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`C`

Show Worked Solution

`text{Slope lines are vertical when}\ y=0\ \text{(eliminate D and E)}`

`text{Slope lines are negative in 1st/3rd quadrants (eliminate A)}`

`text{Consider slope at point (1, 1) ⇒ gradient close to –2.}`

`=>C`

Calculus, SPEC2 2023 VCAA 7 MC

The direction field for a differential equation is shown above. On a certain solution curve of this differential equation, \(y=2\)  when  \(x=-1\).

The value of \(y\) on the same solution curve when  \(x=1.5\)  is closest to

  1. \(-0.5\)
  2. \(0\)
  3. \(0.5\)
  4. \(1.0\)
  5. \(1.5\)
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\(D\)

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\(\text{Trace a curve that does not cross slope lines from}\ (-1,2).\)

\(\text{Curve crosses}\ \ x=1\ \ \text{at}\ \ y\approx 1.\)

\(\Rightarrow D\)

Calculus, EXT1 C3 2022 HSC 12a

A direction field is to be drawn for the differential equation

`(dy)/(dx)=(x-2y)/(x^(2)+y^(2)). `

On the diagram, clearly draw the correct slopes of the direction field at the points `P, Q` and `R`.  (2 marks)
  

           

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Show Worked Solution

`text{At (–1, 1):}\ \ dy/dx=(-1-2)/(1+1)=-3/2`

`text{At (1, 1):}\ \ dy/dx=(1-2)/(1+1)=-1/2`

`text{At (2, 1):}\ \ dy/dx=(2-2)/(4+1)=0`
 

Calculus, SPEC2 2021 VCAA 10 MC

The differential equation that has the diagram above as its direction field is

  1. `(dy)/(dx) = y + 2x`
  2. `(dy)/(dx) = 2x - y`
  3. `(dy)/(dx) = 2y - x`
  4. `(dy)/(dx) = y - 2x`
  5. `(dy)/(dx) = x + 2y`
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`D`

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`text(By elimination:)`

`text(At)\ (1, 2), m = 0`

`->\ text(Eliminate)\ A, C, E`

`text(At)\ (0, 1),\ m\ text(is positive)`

`->\ text(Eliminate)\ B`

`=>\ D`

Calculus, EXT1 C3 2021 HSC 12a

The direction field for a differential equation is shown below.

The graph of a particular solution to the differential equation passes through the point `P`.

On the graph, sketch the graph of this particular solution.  (1 mark)

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Show Worked Solution

♦♦♦ Mean mark 18%!
MARKER COMMENT: A solution curve does not cross any tangent line.

Calculus, SPEC2 2020 VCAA 9 MC

`P(x, y)`  is a point on a curve. The `x`-intercept of a tangent to point  `P(x, y)`  is equal to the `y`-value at `P`.

Which one of the following slope fields best represents this curve?

A.   B.
C. D.
E.    
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`B`

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`text(The tangent to the curve passes through)`

♦♦ Mean mark 35%.

`(x, y)\ and\ (y, 0)`

`(dy)/(dx) = (0 – y)/(y – x) = y/(x – y)`
 

`text(When)\ \ x = 0:`

`(dy)/(dx) = y/(−y) = −1`

`=>B`

Calculus, SPEC2 2019 VCAA 9 MC

The differential equation that has the diagram above as its direction field is

  1. `(dy)/(dx) = sin(y - x)`
  2. `(dy)/(dx) = cos(y - x)`
  3. `(dy)/(dx) = sin(x - y)`
  4. `(dy)/(dx) = 1/(cos(y - x))`
  5. `(dy)/(dx) = 1/(sin(y - x))`
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`B`

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`text(Along line)\ y = x,\ text(gradient = 1)`

`:.\ text(Eliminate A, C, and E.)`
 

`text{At (1, 0),  0 < gradient < 1}`

`1/(cos(-1)) > 1`

`:.\ text(Eliminate D)`

`=>B`

Calculus, SPEC2 2012 VCAA 10 MC

The diagram that best represents the direction field of the differential equation  `(dy)/(dx) = xy`  is

A. B.
C. D.
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`A`

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`(dy)/(dx) = xy`

`text(When)\ \ x=0 \ or\  y=0\ \ =>\ text(gradient = 0)`

`text(In 1st and 3rd quartile)\ \ =>\ \ text(gradients positive)`

`text(In 2nd and 4th quartile)\ \ =>\ \ text(gradients negative)`

`=> A`

Calculus, SPEC2 2011 VCAA 17 MC

SPEC2 2011 VCAA 17 MC

The differential equation which best represents the above direction field is

A.   `(dy)/(dx) = (y - 2x)/(2y + x)`

B.   `(dy)/(dx) = (2x - y)/(y - 2x)`

C.   `(dy)/(dx) = (2y - x)/(y + 2x)`

D.   `(dy)/(dx) = (y - 2x)/(2y - x)`

E.   `(dy)/(dx) = (x - 2y)/(2y + x)`

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`A`

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`text(When)\ \ x=0\ \ => \ \ text(gradients are all positive)`

Almost half of all students answered incorrectly – mean mark 52%.

`text(Eliminate B and E.)`

`text(When)\ \ y=0\ \ => \ \ text(gradients are all negative)`

`text(Eliminate D.)`

`text(Option A will have zero gradient along)\ \ y=2x\ \ text{(correct)}`

`text(Option C will have zero gradient along)\ \ y=1/2 x\ \ text{(incorrect)}`

`=> A`

Calculus, SPEC2 2013 VCAA 12 MC

SPEC2 2013 VCAA 12 MC

The differential equation that best represents the above direction field is

A.   `(dy)/(dx) = x^2 - y^2`

B.   `(dy)/(dx) = y^2 - x^2`

C.   `(dy)/(dx) = y/x`

D.   `(dy)/(dx) = −x/y`

E.   `(dy)/(dx) = x/y`

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`E`

Show Worked Solution

`text(By inspection:)`

`text(When)\ \ x=0\ \ =>\ \ (dy)/(dx) = 0`

`text(When)\ \ y=0\ \ => (dy)/(dx) -> oo`

`:.\ text(Eliminate A, B and C)`
 

`text(Along)\ \ y = x\ \ =>\ \ (dy)/(dx) > 0`

`:.\ text(Eliminate D)`

`=> E`

Calculus, SPEC2 2014 VCAA 14 MC

The differential equation that is best represented by the above direction field is

A.   `(dy)/(dx) = 1/(x - y)`

B.   `(dy)/(dx) = y - x`

C.   `(dy)/(dx) = 1/(y - x)`

D.   `(dy)/(dx) = x - y`

E.   `(dy)/(dx) = 1/(y + x)`

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`C`

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`text(Method 1:)`

`text(Draw direction field of each option by CAS.)`
 

`text(Method 2:)`

`text(Consider quadrant 2,)`

`x < 0, \ \ y > 0, \ \  m > 0\ \ => text(Eliminate A, D and E)`
 

`text(Consider vertical gradients where)\ \ m=oo\ \ => text(Eliminate B)`

 
`=> C`

Calculus, SPEC2 2015 VCAA 13 MC

SPEC2 2015 VCAA 13 MC
 

The direction field for a certain differential equation is shown above.

The solution curve to the differential equation that passes through the point  `(–2.5, 1.5)`  could also pass through

A.   `(0, 2)`

B.   `(1, 2)`

C.   `(3, 1)`

D.   `(3, –0.5)`

E.   `(–0.5, 2)`

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`C`

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`text{Draw a graph that goes through (–2.5, 1.5) such that all}`

♦ Mean mark 47%.

`text{gradient curve lines are tangential:}`

`(3, –0.5)`

`=> D`

Calculus, SPEC1 2017 VCAA 8

A slope field representing the differential equation  `dy/dx = −x/(1 + y^2)`  is shown below.

  1. Sketch the solution curve of the differential equation corresponding to the condition  `y(−1) = 1`  on the slope field above and, hence, estimate the positive value of `x` when  `y = 0`. Give your answer correct to one decimal place.  (2 marks)
  2. Solve the differential equation  `(dy)/(dx) = (−x)/(1 + y^2)`  with the condition  `y(−1) = 1`. Express your answer in the form  `ay^3 + by + cx^2 + d = 0`, where `a`, `b`, `c` and `d` are integers.  (2 marks)
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  1.  

  2. `2y^3 + 6y + 3x^2 – 11 = 0`
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a.   

♦♦ Mean mark part (a) 32%.
MARKER’S COMMENT: Solution curve should follow slope ticks and not cross them.

 

b.    `(1 + y^2)(dy)/(dx)` `= −x`
  `int 1 + y^2 dy` `= −int x\ dx`
  `y + (y^3)/3` `= −(x^2)/2 + C, C ∈ R`

 
`text(Substituting)\ (-1,1):`

`1 + (1^3)/3` `= −((−1)^2)/2 + C`
`1 + 1/3` `= −1/2 + C`
`:. C` `= 1/2 + 4/3`
  `= 11/6`

 

`y + 1/3y^3` `= −1/2x^2 + 11/6`
`6y + 2y^3` `= −3x^2 + 11`

 

`:. 2y^3 + 6y + 3x^2 – 11 = 0`

Calculus, SPEC2 2018 VCAA 10 MC

The differential equation that best represents the direction field above is

A.  `(dy)/(dx) = (2x + y)/(y - 2x)`

B.  `(dy)/(dx) = (x + 2y)/(2x - y)`

C.  `(dy)/(dx) = (2x - y)/(x + 2y)`

D.  `(dy)/(dx) = (x - 2y)/(y - 2x)`

E.  `(dy)/(dx) = (2x + y)/(2y - x)` 

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`A`

Show Worked Solution

`text(When)\ \ x=0, \ m=1`

`text(When)\ \ y=0, \ m=-1`

`=>  A`