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Vectors, SPEC2 2024 VCAA 16 MC

Particle 1 has position vector  \({\underset{\sim}{r}}_1(t)=\cos (t) \underset{\sim}{ i }+\sin (t) \underset{\sim}{ j }+\sqrt{\sin (2 t)} \underset{\sim}{ k }\)  and Particle 2 has position vector  \({\underset{\sim}{r}}_2(t)=\sin (t) \underset{\sim}{ i }+\cos (t) \underset{\sim}{ j }+\sqrt{\sin (2 t)} \underset{\sim}{ k }\), where \(t\) is measured in seconds and  \(t \in\left(0, \dfrac{\pi}{2}\right)\).

The number of times the velocity of Particle 1 is perpendicular to the position vector  \({\underset{\sim}{r}}_2(t)\) during the first \(\dfrac{\pi}{2}\) seconds is

  1. \(1\)
  2. \(2\)
  3. \(3\)
  4. \(4\)
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\(A\)

Show Worked Solution

\(\text{By CAS:}\)

\({\underset{\sim}{v}}_1(t)=\dfrac{d}{dt} \left({\underset{\sim}{r}}_1(t)\right)=-\sin (t)\underset{\sim}{i}+\cos (t) \underset{\sim}{j}+\dfrac{\cos (2 t)}{\sqrt{\sin (2 t)}}\, \underset{\sim}{k}\)

\(\text{Find when} \ {\underset{\sim}{v}}_1(t) \perp {\underset{\sim}{r}}_2(t)\)

\(\text{Solve}\ \ {\underset{\sim}{v}}_{1} (t) \cdot {\underset{\sim}{r}}_{2} (t)=0\ \ \text{for}\ \ t \in\left(0, \dfrac{\pi}{2}\right):\)

\({\underset{\sim}{v}}_1(t) \cdot {\underset{\sim}{r}}_2(t)=-\sin ^2(t)+\cos ^2(t)+\cos (2 t)=2 \cos (2 t)\)

\(2 t=\dfrac{\pi}{2} \ \Rightarrow \ t=\dfrac{\pi}{4}\)

\(\Rightarrow A\)

♦ Mean mark 43%.

Vectors, SPEC2 2020 VCAA 1

A particle moves in the  `x\ – y`  plane such that its position in terms of `x` and `y` metres at `t` seconds is given by the parametric equations

`x = 2sin(2t)`

`y = 3cos(t)`

where `t >= 0`

  1. Find the distance, in metres, of the particle from the origin when  `t = pi/6`.   (2 marks)

  2.   i. Express  `(dy)/(dx)`  in terms of `t` and, hence, find the equation of the tangent to the path of the particle at  `t = pi`  seconds.   (3 marks)

  3.  ii. Find the velocity, `underset ~ v`, in `text(ms)^(−1)`, of the particle when  `t = pi`.   (2 marks)

  4. iii. Find the magnitude of the acceleration, in `text(ms)^(−2)`, when  `t = pi`.   (2 marks)

  5. Find the time, in seconds, when the particle first passes through the origin.   (1 mark)

  6. Express the distance, `d` metres, travelled by the particle from  `t = 0`  to  `t = pi/6`  as a definite integral and find this distance correct to three decimal places.   (2 marks)

Show Answers Only
  1. `sqrt39/2\ text(metres)`
  2.   i. `y = −3`
  3.  ii. `4underset~i`
  4. iii. `3\ text(ms)^(−2)`
  5. `pi/2\ \ (text(1st time))`
  6. `1.804\ \ (text(to 3 d.p.))`
Show Worked Solution

a.   `text(At)\ \ t = pi/6,`

`x = 2sin\ pi/3 = sqrt3`

`y = 3cos\ pi/6 = (3sqrt3)/2`

`:.\ text(Distance)` `= sqrt((sqrt3)^2 + ((3sqrt3)/2)^2)`
  `= sqrt39/2\ text(metres)`

 

b.i.   `(dx)/(dt) = 4cos(2t),\ \ (dy)/(dt) = −3sin(t)`

`(dy)/(dx)` `= (dy)/(dt) · (dt)/(dx)`
  `= (−3sin(t))/(4cos(2t))`

 
`text(When)\ t = pi :`

`(dy)/(dx)` `= (−3sin(pi))/(4cos(2pi))=0`

 
`text(Equation of tangent where)\ \ m = 0,\ text(through)\ (0, −3):`

`y = −3`
  

b.ii.   `underset~r(t)` `= 2sin(2t)underset~i + 3cos(t)underset~j`
  `underset~v(t)` `= 4cos(2t)underset~i-3sin(t)underset~j`
  `underset~v(pi)` `= 4cos(2pi)underset~i-3sin(pi)underset~i`
    `= 4underset~i`

 

b.iii.  `underset~a(t) = −8sin(2t)underset~i-3cos(t)underset~j`

♦ Mean mark (b)(iii) 48%.
`underset~a(pi)` `= −8sin(2pi)underset~i-3cos(pi)underset~j`
  `= 3underset~j`
`|underset~a(pi)|` `= sqrt(0^2 + 3^2)`
  `= 3\ text(ms)^(−2)`

 

c.   `text(Find)\ t\ text(when)\ \ x = 2sin(2t) = 0\ \ text(and)\ \ y = 3cos(t) = 0:`

`t = pi/2\ \ (text(1st time))`

 

d.    `text(Distance)` `= int_0^(pi/6) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(pi/6) sqrt((4cos(2t))^2 + (−3sin(t))^2)\ dt`
    `~~ 1.804\ \ (text(to 3 d.p.))`

Vectors, SPEC1 2012 VCAA 9

The position of a particle at time  `t`  is given by

`underset ~r (t) = (2 sqrt (t^2 + 2) - t^2) underset ~i + (2 sqrt (t^2 + 2) + 2t) underset ~j,\ \ t >= 0.`

  1. Find the velocity of the particle at time  `t.`   (1 mark)

  2. Find the speed of the particle at time  `t = 1`  in the form  `(a sqrt b)/c`, where `a, b` and `c` are positive integers.   (2 marks)

  3. Show that at time  `t = 1,\ \ (dy)/(dx) = (1 + sqrt 3)/(1 - sqrt 3).`   (2 marks)

  4. Find the angle in terms of `pi`, between the vector  `-sqrt 3 underset ~i + underset ~j`  and the vector  `underset ~r (t)`  at time  `t = 0.`   (2 marks)

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  1. `((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`
  2. `(4 sqrt 6)/3`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
  4. `(7 pi)/12`
Show Worked Solution

a.  `underset ~r (t) = (2 sqrt (t^2 + 2) – t^2) underset ~i + (2 sqrt (t^2 + 2) +2t) underset ~j`

  `underset ~v(t)` `=dot underset ~r(t)`
    `= (2(2t)(1/2)(t^2 + 2)^(-1/2) – 2t) underset ~i + (2(2t)(1/2)(t^2 + 2)^(-1/2) + 2)underset ~j`
    `= ((2t)/sqrt(t^2 + 2) – 2t) underset ~i + ((2t)/sqrt(t^2 + 2) + 2) underset ~j`

 

b.    `underset ~v(1)` `= (2/sqrt(1 + 2) – 2)underset ~i + (2/sqrt(1 + 2) + 2) underset ~j`
    `= (2/sqrt 3 – 2) underset ~i + (2/sqrt 3 + 2)underset ~j`
  `|\ underset ~v(1)\ |` `= sqrt((2/sqrt 3 – 2)^2 + (2/sqrt 3 + 2)^2)`
    `= sqrt(4/3 – 8/sqrt 3 + 4 + 4/3 + 8/sqrt 3 + 4)`
    `= sqrt(8/3 + 8)`
    `= sqrt(32/3)`
    `= (4sqrt2)/sqrt3`
    `= (4 sqrt 6)/3`

♦ Mean mark part (c) 41%.

c.    `(dy)/(dt)` `= (dy)/(dx) xx (dx)/(dt)`
  `(dy)/(dx)` `=((dy)/(dt))/((dx)/(dt))`
  `:. (dy)/(dx)|_(t=1)` `= (2/sqrt 3 + 2)/(2/sqrt 3 – 2)`
    `= ((2 + 2 sqrt 3)/sqrt 3) xx (sqrt 3/(2 – 2 sqrt 3))`
    `= (2 + 2 sqrt 3)/(2 – 2 sqrt 3)`
    `= (1 + sqrt 3)/(1 – sqrt 3)`

 

d.   `text(At)\ \ t=0:` 

♦♦ Mean mark part (d) 32%.

`underset ~r(0) = 2 sqrt 2 underset ~i + 2 sqrt 2 underset ~j`

`theta_1` `= tan^(-1)((2 sqrt 2)/(2 sqrt 2)) = pi/4`
`theta_2` `= tan^(-1)(1/sqrt 3)=pi/6`

 
`text(Let)\ \ theta=\ text(angle between the vectors:)`

`theta` `= pi – theta_1 – theta_2`
  `= pi – pi/4 – pi/6`
  `= (12 pi – 3 pi – 2 pi)/12`
  `= (7 pi)/12`

Vectors, SPEC2 2018 VCAA 4

Two yachts, `A` and `B`, are competing in a race and their position vectors on a certain section of the race after time  `t`  hours are given by

`underset ~ r_A (t) = (t + 1) underset ~i + (t^2 + 2t) underset ~j \ and \ underset ~r_B (t) = t^2 underset ~i + (t^2 + 3) underset ~j, \ t >= 0`

where displacement components are measured in kilometres from a given reference buoy at origin `O`.

  1. Find the cartesian equation of the path for each yacht.   (2 marks)

  2. Show that the two yachts will not collide if they follow these paths.   (2 marks)

  3. Find the coordinates of the point where the paths of the two yachts cross. Give your coordinates correct to three decimal places.   (2 marks)

One of the rules for the race is that the yachts are not allowed to be within 0.2 km of each other. If this occurs there is a time penalty for the yacht that is travelling faster.

  1. For what values of `t` is yacht `A` travelling faster than yacht `B`?   (2 marks)

  2. If yacht `A` does not alter its course, for what period of time will yacht `A` be within 0.2 km of yacht `B`? Give your answer in minutes, correct to one decimal place.   (2 marks)

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  1. `y_A = x_A^2-1, x_A > 1; qquad y_B = x_B + 3, x_B > 0`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `(x, y) ~~ (2.562, 5.562)`
  4. `0 < t < 5/2`
  5. `4.1\ text(minutes)`
Show Worked Solution

a.  `x_A = t + 1,\ \ y_A = = t^2 + 2t`

`y_A` `= t^2 + 2 t + 1-1`
  `= (t + 1)^2-1`
`:. y_A` `= x_A^2-1`

 
`x_B = t^2,`

`y_B` `= t^2+3`
`:.y_B` `= x_B+3`

 

b.  `text(Same)\ \ xtext(-coordinate occurs when:)`

`t+1` `= t^2`
`t` `= (1+sqrt5)/2,\ \ \ (t>0)`

 
`text(When)\ \ t = (1+sqrt5)/2,`

`y_A=(3sqrt5+5)/2`

`y_B= (sqrt5+9)/2  !=y_A`
 

`:.\ text(No collision)`

 

c.   `x + 3` `= x^2-1`
  `x^2-x-4` `= 0`
  `x` `= (1 + sqrt 17)/2,\ \ \ (t>0 \ =>\ x>1)`
  `y` `= (7 + sqrt 17)/2`
  `:. (x, y)` `= ((1 + sqrt 17)/2, (7 + sqrt 17)/2)`
    `=(2.562, 5.562)`

 

d.    `underset ~dot r_A (t)` `= underset ~i + (2t + 2)underset ~j`
  `|underset ~dot r_A (t)|` `= sqrt(1 + (2t + 2)^2)`
    `= sqrt(4t^2 + 8t + 5)`
     
  `underset ~dot r_B (t)` `= 2t underset ~i + 2t underset ~j`
  `|underset ~dot r_B (t)|` `= sqrt (4t^2 + 4t^2)`
    `= sqrt(8t^2)`

  
 `text(Yacht A is travelling faster when:)`

♦ Mean mark part (d) 41%.

`sqrt(4t^2 + 8t + 5)` `> sqrt (8t^2)`
`4t^2 + 8t + 5` `> 8t^2`
`4t^2-8t-5` `< 0`

 
`:. 0 < t < 5/2`

 

e.    `underset ~ r_B-underset ~r_A` `= (t^2-(t + 1)) underset ~ i + (t^2 + 3-(t^2 + 2t)) underset ~j`
    `= (t^2-t-1) underset ~i + (-2t + 3) underset ~j`

 

`d = |underset ~r_B-underset~r_A|= sqrt ((t^2-t-1)^2 + (3-2t)^2)`
 

`text(Find)\ \ t\ \ text(such that:)`

♦ Mean mark part (e) 37%.

`sqrt ((t^2-t-1)^2 + (3-2t)^2) < 0.2`

`=> 1.52883 < t < 1.59734`

 
`:.\ text(Period of time yachts are within 0.2 km)`

`~~ 1.59734-1.52883`

`~~0.068506\ text(hours)`

`~~ 4.1\ text(minutes)`

Vectors, SPEC2-NHT 2018 VCAA 13 MC

Let  `underset~i`  and  `underset~j`  be unit vectors in the east and north directions respectively

At time `t, t >= 0`, the position of particle `A` is given by  `underset~r_A = (t^2 - 5t + 6) underset~i + (5t - 8) underset~j`  and the position of particle `B` is given by  `underset~r_B = (3 - t) underset~i + (t^2 - t) underset~j`.

Particle `A` will be directly east of particle `B` when `t` equals

  1. 1
  2. 2
  3. 1 and 2
  4. 2 and 4
  5. 4
Show Answers Only

`E`

Show Worked Solution

`text(Find)\ t\ text(when particles align:)`

`5t – 8` `= t^2 – t`
`0` `= t^2 – 6t – 8`
`0` `= (t – 4)(t – 2)`

 
`t = 2, quad t = 4`
 

`x_A(2)` `= 4 – 10 + 6 = 0`
`x_B(2)` `= 1 > x_A (2) \ ->\  B\ text(is East of)\ A`
`x_A(4)` `= 16 – 20 + 6 = 2`
`x_B(4)` `= -1 < x_A (2) \ ->\  A\ text(is East of)\ B`

 

`=>  E`