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Calculus, EXT1 C3 2023 HSC 3 MC

The diagram shows the direction field of a differential equation. A particular solution to the differential equation passes through \((-2,1)\).

Where does the solution that passes through \((-2,1)\) cross the \(y\)-axis?

  1. \(y=1.12\)
  2. \(y=1.34\)
  3. \(y=1.56\)
  4. \(y=1.78\)
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\(C\)

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\(\text{Following gradients → cross y-axis slightly above 1.5}\)

\(\Rightarrow C\)

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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`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, EXT1 C3 2021 SPEC2 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) = x+2y`
  4. `(dy)/(dx) = y - 2x`
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`D`

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`text(By elimination:)`

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

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

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

Calculus, EXT1 C3 2020 SPEC2 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.
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`B`

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

`(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, EXT1 C3 2019 SPEC2 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) = 1/(cos(y - x))`
  4. `(dy)/(dx) = 1/(sin(y - x))`
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`B`

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`text(By elimination:)`

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

`:.\ text(Eliminate A and D.)`
 

`text{At (1, 0),  0 < gradient < 1}`

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

`:.\ text(Eliminate C)`

`=>B`

Calculus, EXT1 C3 2017 SPEC1 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` `= 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, EXT1 C3 2015 SPEC2 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)`

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`D`

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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, EXT1 C3 2018 SPEC2 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)`

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`A`

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

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

`=>  A`

Calculus, EXT1 C3 2014 SPEC2 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`

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`C`

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`text(Consider quadrant 2,)`

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

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

 
`=> C`

Calculus, EXT1 C3 2013 SPEC2 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) = −x/y`

D.   `(dy)/(dx) = x/y`

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`D`

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`text(By inspection:)`

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

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

`:.\ text(Eliminate A, and B)`
 

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

`:.\ text(Eliminate C)`

`=> D`

Calculus, EXT1 C3 2012 SPEC2 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, EXT1 C3 2011 SPEC2 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, EXT1 C3 2017 SPEC2 9 MC

The gradient of the tangent to a curve at any point  `P(x, y)`  is half the gradient of the line segment joining `P` and the point  `Q(-1, 1)`.

The coordinates of points on the curve satisfy the differential equation

A.   `(dy)/(dx) = (y + 1)/(2(x - 1))`

B.   `(dy)/(dx) = (2(y - 1))/(x + 1)`

C.   `(dy)/(dx) = (x - 1)/(2(y + 1))`

D.   `(dy)/(dx) = (y - 1)/(2(x + 1))`

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`D`

Show Worked Solution
`m_text(tang)` `= 1/2 m_(PQ)`
`m_(PQ)` `= (y – 1)/(x – (-1))`
  `= (y – 1)/(x + 2)`

 
`:. m_text(tang) = (dy)/(dx) = (y – 1)/(2(x + 1))`

`=>   D`