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Calculus, SPEC2-NHT 2019 VCAA 7 MC

The gradient of the line that is perpendicular to the graph of a relation at any point  `P(x, y)`  is half the gradient of the line joining `P` and the point  `Q(−1,1)`.

The relation satisfies the differential equation

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

Show Worked Solution

`m_(PQ) = (y – 1)/(x + 1) \ => \ m_text(perp) = 1/2((y – 1)/(x + 1))`

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

 
`=>\ E`

Calculus, SPEC1 2019 VCAA 10

Find  `(dy)/(dx)`  at the point  `((sqrtpi)/sqrt6,(sqrtpi)/sqrt3)` for the curve defined by the relation  `sin(x^2) + cos(y^2) = (3sqrt2)/pi\ xy`.

Give your answer in the form  `(pi - asqrtb)/(sqrta(pi + sqrtb))`, where  `a`, `b ∈ ZZ^+`.  (5 marks)

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`(pi – 2sqrt3)/(sqrt2(pi + sqrt3))`

Show Worked Solution

`2x · cos(x^2) – 2y · sin(y^2) · (dy)/(dx) = (3sqrt2)/pi(y + x · (dy)/(dx))`

`text(Substitute)\ (sqrtpi/sqrt6, sqrtpi/sqrt3)\ text(into equation:)`

`(2sqrtpi)/sqrt6 cos(pi/6) – (2sqrtpi)/sqrt3 sin(pi/3) · (dy)/(dx)` `= (3sqrt2)/pi (sqrtpi/sqrt3 + sqrtpi/sqrt6 · (dy)/(dx))`
`(2sqrtpi)/sqrt6 · sqrt3/2 – (2sqrtpi)/sqrt3 · sqrt3/2 · (dy)/(dx)` `= sqrt18/pi (sqrtpi/sqrt3 + sqrtpi/sqrt6 · (dy)/(dx))`
`sqrtpi/sqrt2 – sqrtpi · (dy)/(dx)` `= sqrt6/sqrtpi + sqrt3/sqrtpi · (dy)/(dx)`
`(dy)/(dx)(sqrt3/sqrtpi + sqrtpi)` `= sqrtpi/sqrt2 – sqrt6/sqrtpi`
`(dy)/(dx)((sqrt3 + pi)/sqrtpi)` `= (pi – sqrt12)/(sqrt2 sqrtpi)`
`(dy)/(dx)` `= (pi – sqrt12)/(sqrt2 sqrtpi) xx sqrtpi/(sqrt3 + pi)`
  `= (pi – 2sqrt3)/(sqrt2(pi + sqrt3))`

Calculus, SPEC2-NHT 2018 VCAA 5

A horizontal beam is supported at its endpoints, which are 2 m apart. The deflection `y` metres of the beam measured downwards at a distance `x` metres from the support at the origin `O` is given by the differential equation  `80 (d^2y)/(dx^2) = 3x-4`.
 


 

  1. Given that both the inclination, `(dy)/(dx)`, and the deflection, `y`, of the beam from the horizontal at  `x = 2`  are zero, use the differential equation above to show that  `80 y = 1/2 x^3-2x^2 + 2x`.   (2 marks)

  2. Find the angle of inclination of the beam to the horizontal at the origin `O`. Give your answer as a positive acute angle in degrees, correct to one decimal place.   (2 marks)

  3. Find the value of `x`, in metres, where the maximum deflection occurs, and find the maximum deflection, in metres.   (3 marks)

  4. Find the maximum angle of inclination of the beam to the horizontal in the part of the beam where  `x >= 1`. Give your answer as a positive acute angle in degrees, correct to one decimal place.   (2 marks)

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  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `1.4^@`
  3. `x = 2/3; quad 1/135\ text(m)`
  4. `0.5^@`
Show Worked Solution
a.   `80 * int (d^2y)/(dx^2)\ dx` `= int 3x-4\ dx`
  `80* (dy)/(dx)` `= (3x^2)/2-4x + c_0`

 
`text(When)\ \ x=2,\ \ dy/dx=0:`

`80 xx 0` `= (3xx2^2)/2-4xx2 + c_0`
`c_0` `= 2`

 

`80* (dy)/(dx)` `= (3x^2)/2-4x + 2`
`80y` `= int (3x^2)/2-4x + 2\ dx`
  `= 1/2 x^3-2x^2+2x +c_1`

 

`text(When)\ \ x=2,\ \ y=0:`

`0` `= 1/2 2^3-2 xx 2^2 + 2 xx 2 + c_1`
`c_1` `= 0`

`:. 80y = 1/2x^3-2x^2 + 2x`

 

b.   `text(Find)\ \ dy/dx\ \ text(when)\ \ x=0:`

  `80 *(dy)/(dx)` `= 0-0 + 2`
  `(dy)/(dx)` `= 1/40`
  `:. theta` `= tan^(-1) (1/40) ~~ 1.4^@`

 

c.   `text(Find)\ \ x\ \ text(when)\ \ dy/dx=0:`

`x=2\ \ text(or)\ \ 2/3`

`=>y_text(max)\ \ text(occurs at)\ \ x=2/3\ \ (y=0\ \ text(at)\ \ x=2)`

 

`80*y_max` `= 1/2(2/3)^3-2(2/3)^2 + 2(2/3)`
`:. y_max` `= 1/80 xx 16/27`
  `= 1/135\ text(metres)`

 

d.   `text(Max inclination when convexity changes)\ => (d^2y)/(dx^2) =0`

`80*(d^2y)/(dx^2) = 3x-4 = 0`

`=> x = 4/3`

`text(Find)\ \ dy/dx\ \ text(when)\ \ x = 4/3 :`

`(dy)/(dx)` `= 1/80 (3/2(4/3)^2-4(4/3) + 2)`
  `= (-1)/120`

 

`theta` `= tan^(-1) (|-1/120|)`
  `~~ 0.5^@`

Calculus, SPEC2-NHT 2017 VCAA 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) = (2(x - 1))/(y + 1)`

E.   `(dy)/(dx) = (y - 1)/(2(x + 1))`

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`E`

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))`

`=>   E`