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Vectors, SPEC2 2024 VCAA 4

A model yacht is sailing on a lake between two buoys. Its path from one buoy to the other, relative to an origin \(O\), is given by

\({\underset{\sim}{r}}_{\text{Y}}(t)=3 \sec (t) \underset{\sim}{i}+2 \tan (t) \underset{\sim}{j}\),  where  \(\dfrac{2 \pi}{3} \leq t \leq \dfrac{4 \pi}{3}\)

Displacement components are measured in metres, and time \(t\) is measured in minutes.

  1. Use a trigonometric identity to show that the Cartesian equation of the path is given by  \(\dfrac{x^2}{9}-\dfrac{y^2}{4}=1\).   (1 mark)
  2. Sketch the path of the yacht on the axes below. Label the endpoints with their coordinates and show the direction of motion.   (2 marks)

    1. Write down an expression, in terms of \(\sec (t)\), for the square of the speed of the yacht at any time, \(t\).   (1 mark)
    2. Find the time, in minutes, when the minimum speed occurs. You do not need to justify that this speed is a minimum.   (1 mark)

    3. State the minimum speed of the yacht in metres per minute.   (1 mark)

    4. State the coordinates of the yacht when the minimum speed occurs.   (1 mark)

    1. Write down a definite integral, in terms of \(t\), that gives the distance travelled by the yacht along the path given by \(\underset{\sim}{ r }(t)\) over the time interval  \(\dfrac{2 \pi}{3} \leq t \leq \dfrac{4 \pi}{3}\).   (1 mark)
    2. Find the distance travelled by the yacht over this time interval.
    3. Give your answer in metres correct to one decimal place.   (1 mark)
  3. The position vector of a drone videoing the yacht, relative to the same origin as the yacht, \(O\), is given by \({\underset{\sim}{ r }}_{ \text{D} }(t)=(2-3 t) \underset{\sim}{ i }+(4 t-1) \underset{\sim}{ j } +(6-t) \underset{\sim}{ k }\),  where  \(0 \leq t \leq 5\).
  4. Displacement components are measured in metres, and time \(t\) is measured in minutes.
  5. What is the shortest distance from the drone to the yacht, as the yacht sails along the path given by  \({\underset{\sim}{ r }}_{\text{Y}}(t)=3 \sec (t) \underset{\sim}{ i }+2 \tan (t) \underset{\sim}{ j }\),  where \(\dfrac{2 \pi}{3} \leq t \leq \dfrac{4 \pi}{3}\) ?
  6. Give your answer in metres, correct to one decimal place.   (2 marks)

Show Answers Only

a.   \(\left(\dfrac{x}{3}\right)^2-\left(\dfrac{y}{2}\right)^2=1\)

b. 
     
 
c.i.  \(\abs{\underset{\sim}{\dot{r}}(t)}^2=13 \sec ^4 t-9 \sec ^2 t\)
c.ii. \(\text{Minimum speed occurs at  \( t=\pi\)  min.}\)

c.iii.  \(\abs{\underset{\sim}{\dot{r}(t)}}=2 \ \text{metres/min}\)

c.iv  \(\text{Yacht has coordinates} \ (-3,0) \ \text {when min speed occurs.}\)
 

d.i   \(\text{Distance }=\displaystyle \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \sqrt{(3 \sec \, t\, \tan\, t)^2+4\left(\sec ^2 t\right)}\,dt\)

d.ii.  \(\text{Distance}=9.4 \ \text{metres}\)

e.   \(D_{\min }=11.1 \ \text{m}\)

Show Worked Solution

a.   \(x\) \(=3 \sec t \) \(\ \Rightarrow  \ \sec t\) \(=\dfrac{x}{3}\)
  \(y\) \(=2 \tan t \) \(\ \Rightarrow  \ \tan t\) \(=\dfrac{y}{2}\)

\(\sec ^2 t+1\) \(=\tan ^2 t\)
\(\left(\dfrac{y}{2}\right)^2+1\) \(=\left(\dfrac{x}{3}\right)^2\)
\(\left(\dfrac{x}{3}\right)^2-\left(\dfrac{y}{2}\right)^2\) \(=1\)

 

b. 
     
 
c.i.   \(\underset{\sim}{r}\) \(=3 \sec (t)\underset{\sim}{i}+2 \tan (t)\underset{\sim}{j}\)
  \(\underset{\sim}{\dot{r}}(t)\) \(=\dfrac{3 \sin t}{\cos ^2 t}\underset{\sim}{i}+\dfrac{2}{\cos ^2 t} \underset{\sim}{j}\)
  \(\abs{\underset{\sim}{\dot{r}}(t)}^2\) \(=\dfrac{9 \sin ^2 t}{\cos ^4 t}+\dfrac{4}{\cos ^4 t}\)
    \(=\dfrac{9-9 \cos ^2 t+4}{\cos ^4 t}\)
    \(=\dfrac{13}{\cos ^4 t}-\dfrac{9}{\cos ^2 t}\)
    \(=13 \sec ^4 t-9 \sec ^2 t\)
♦♦♦ Mean mark c.i. 23%.
 

c.ii. \(\text {Differentiate} \ \ \abs{\underset{\sim}{\dot{r}}(t)}^2: \)

\(\dfrac{d}{d t}\left(\abs{\underset{\sim}{\dot{r}}(t)}^2\right)=\dfrac{52 \sin t}{\cos ^5 t}-\dfrac{18}{\cos ^3 t}=\dfrac{2 \sin t\left(26-9 \sin t\, \cos ^2 t\right)}{\cos ^5 t}\)

\(\text{Find } t \text{ where } \  \sin t=0 \ \text { for } \ \dfrac{2 \pi}{3} \leq t \leq \dfrac{4 \pi}{3}:\)

\(t=\pi\)

\(\text{Minimum speed occurs at  \( t=\pi\)  min.}\)
 

c.iii.  \(\text{Find} \ \ \abs{\underset{\sim}{\dot{r}}(t)}^2 \text{when} \ \ t=\pi:\)

\(\abs{\underset{\sim}{\dot{r}}(t)^2}=\dfrac{13}{\cos ^4 \pi}-\dfrac{9}{\cos ^2 \pi}=13-9=4\)

\(\abs{\underset{\sim}{\dot{r}(t)}}=\sqrt{4}=2 \ \text{metres/min}\)

♦ Mean mark c.iii. 48%.
 
c.iv  \(\text{Find}\ \underset{\sim}{r}(t) \ \text{when} \ \ t=\pi:\)

\(\underset{\sim}{r}(\pi)=3 \sec \pi \underset{\sim}{i}+2 \tan \pi \underset{\sim}{j}=-3 \underset{\sim}{i}\)

\(\text{Yacht has coordinates} \ (-3,0) \ \text {when min speed occurs.}\)
 

d.i   \(\text{Distance }=\displaystyle \int_{\frac{2 \pi}{3}}^{\frac{4 \pi}{3}} \sqrt{(3 \sec \, t\, \tan\, t)^2+4\left(\sec ^2 t\right)}\,dt\)
 

d.ii.  \(\text{Distance}=9.4 \ \text{metres}\)
 

e.    \({\underset{\sim}{r}}_\text{D}(t)=(2-3 t) \underset{\sim}{i}+(4 t-1) \underset{\sim}{j}+(6-t) \underset{\sim}{k} \quad(0 \leqslant t \leqslant 5)\)

\(\underset{\sim}{r}(t)=3 \sec (t) \underset{\sim}{i}+2 \tan (t) \underset{\sim}{j} \quad\left(\dfrac{2 \pi}{3} \leqslant t \leqslant \dfrac{4 \pi}{3}\right)\)

\(D=\sqrt{(3 \sec t-2+3 t)^2+(2 \tan t-4 t+1)^2+(t-6)^2}\)

\(\text{Find} \ \ D_{\min } \  \text{for} \ \ \dfrac{2 \pi}{3} \leq t \leq \dfrac{4 \pi}{3}:\)

\(D_{\min }=11.1 \ \text{m}\)

♦♦♦ Mean mark (e) 28%.

Vectors, SPEC1 2023 VCAA 10

The position vector of a particle at time \(t\) seconds is given by

\(\underset{\sim}{\text{r}}(t)=\big{(}5-6 \ \sin ^2(t) \big{)} \underset{\sim}{\text{i}}+(1+6 \ \sin (t) \cos (t)) \underset{\sim}{\text{j}}\), where \(t \geq 0\).

  1. Write \(5-6\, \sin ^2(t)\) in the form \(\alpha+\beta\, \cos (2 t)\), where \(\alpha, \beta \in Z^{+}\).  (1 mark)
  1. Show that the Cartesian equation of the path of the particle is \((x-2)^2+(y-1)^2=9.\)  (2 marks)
  1. The particle is at point \(A\) when \(t=0\) and at point \(B\) when \(t=a\), where \(a\) is a positive real constant.
  2. If the distance travelled along the curve from \(A\) to \(B\) is \(\dfrac{3 \pi}{4}\), find \(a\).   (1 mark)
  1. Find all values of \(t\) for which the position vector of the particle, \(\underset{\sim}{\text{r}}(t)\), is perpendicular to its velocity vector, \(\underset{\sim}{\dot{\text{r}}}(t)\).   (2 marks)
Show Answers Only

a.    \(2+3\, \cos (2 t) \)

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

c.    \(a=\dfrac{\pi}{8}\)

d.    \(t =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) \)

Show Worked Solution

a.  \(5-6\, \sin ^2(t)=5-6 \times \dfrac{1}{2}(1-\cos (2 t))\)

\(\ \ \ \quad \quad \quad \quad \quad \quad \begin{aligned}
& =5-3+3\, \cos (2 t) \\
& =2+3\, \cos (2 t)
\end{aligned}\)
 

b.   \( x=2+3\, \cos (2 t) \Rightarrow \dfrac{x-2}{3}=\cos (2 t)\)

\(y=1+6\, \sin (t) \cos (t)=1+3\, \sin (2 t) \Rightarrow \dfrac{y-1}{3}=\sin (2 t)\)

\begin{aligned}
\sin ^2(2 t)+\cos ^2(2 t) &=1 \\
\left(\dfrac{x-2}{3}\right)^2+\left(\dfrac{y-1}{3}\right)^2 & =1 \\
(x-2)^2+(y-1)^2 &=9
\end{aligned}

 
c.  \(\text { Motion is circular, centre }(2,1) \text {, radius }=3\)

\begin{aligned}
\text { Arc length } & = r \theta \\
3 \theta & =\dfrac{3 \pi}{4} \\
\theta & =\dfrac{\pi}{4}
\end{aligned}

\(\therefore a=\dfrac{\pi}{8}\)
 

d.    \(\underset{\sim}{r}=(2+3\, \cos (2 t)) \underset{\sim}{i}+(1+3\, \sin (2 t)) \underset{\sim}{j}\)

\(\underset{\sim}{\dot{r}}=-6\, \sin (2 t) \underset{\sim}{i}+6\, \cos (2 t) \underset{\sim}{j}\)

\(\text { Find } t \text { when } \underset{\sim}{r} \cdot \underset{\sim}{\dot{r}}=0 \text { : }\)

\begin{aligned}
\underset{\sim}{r} \cdot \underset{\sim}{\dot{r}} & =6\left(\begin{array}{l}
2+3\, \cos (2 t) \\
1+3\, \sin (2 t)
\end{array}\right)\left(\begin{array}{l}
-\sin (2 t) \\
\cos (2 t)
\end{array}\right) \\
0 &=-2\,\sin (2 t)-3\, \cos (2 t) \sin (2 t)+\cos (2 t)+3\, \cos (2 t) \sin (2 t)\\
0 &=-2\, \sin (2 t)+\cos (2 t)\\
2\, \sin (2 t) &=\cos (2 t)\\
\tan (2 t) & =\dfrac{1}{2} \\
2 t & =\tan ^{-1}\left(\dfrac{1}{2}\right)+k \pi \\
t & =\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{2}\right)+\dfrac{k \pi}{2}\ \ (\text{where}\ k=0,1,2,…) 
\end{aligned}

Vectors, SPEC1 2021 VCAA 9

Let  `underset~r(t) = (-1 + 4cos(t))underset~i + 2/sqrt3\ sin(t)underset~j`  and  `underset~s(t) = (3 sec(t)-1)underset~i + tan(t)underset~j`  be the position vectors relative to a fixed point `O` of particle `A` and particle `B` respectively for  `0 <= 1 <= c`, where `c` is a positive real constant.

    1. Show that the cartesian equation of the path of particle `A` is  `((x + 1)^2)/16 + (3y^2)/4 = 1`.   (1 mark)

    2. Show that the cartesian equation of the path of particle `A` in the first quadrant can be written as  `y = sqrt3/6 sqrt(-x^2-2x + 15)`.   (1 mark)

    1. Show that the particles `A` and `B` will collide.   (1 mark)

    2. Hence, find the coordinates of the point of collision of the two particles.   (1 mark)

    1. Show that  `d/(dx)(8arcsin ((x + 1)/4) + ((x + 1)sqrt(-x^2 -2x + 15))/2) = sqrt(-x^2-2x + 15)`.   (2 marks)


    2.    

      Hence, find the area bounded by the graph of  `y = sqrt3/6 sqrt(-x^2-2x + 15)`,  the `x`-axis and the lines  `x = 1`  and  `x = 2sqrt3-1`,  as shown in the diagram above. Give your answer in the form  `(asqrt3pi)/b`, where `a` and `b` are positive integers.   (2 marks)

Show Answers Only
    1. `text(See Worked Solutions)`
    2. `text(See Worked Solutions)`
    1. `text(See Worked Solutions)`
    2. `-1 + 2sqrt3, 1/sqrt3`
    1. `text(See Worked Solutions)`
    2. `(2sqrt3pi)/9`
Show Worked Solution

a.i.   `text(Particle A)`

`underset~r(t) = (-1 + 4cos(t))underset~i + 2/sqrt3\ sin(t)underset~j`

`x` `= -1 + 4cos(t)`
`x + 1` `= 4cos(t)`
`cos(t)` `= (x + 1)/4`
`y` `= 2/sqrt3\ sin(t)`
`sin(t)` `= (sqrt3 y)/2`

 
`text(Using)\ \ cos^2(t) + sin^2(t) = 1`

`((x + 1)^2)/16 + (3y^2)/4 = 1`

 

a.ii.  `((x + 1)^2)/16 + (3y^2)/4 = 1`

♦ Mean mark part (a)(ii) 41%.
`(x + 1)^2 + 12y^2` `= 16`
`12y^2` `= 16-x^2-2x-1`
`y^2` `= 1/12(15-x^2-2x)`
`y` `= ±sqrt(1/12 (-x^2-2x + 15))`

 
`text(In the 1st quadrant,)\ \ y > 0`

`:. y` `= 1/sqrt12 sqrt(-x^2-2x + 15)`
  `= 1/(2sqrt3) xx sqrt3/sqrt3 sqrt(-x^2-2x + 15)`
  `= sqrt3/6 sqrt(-x^2-2x + 15)`

 

b.i.   `text(If particles collide, find)\ \t\ text(that satisfies)`

`-1 + 4cos(t)` `= 3sec(t)-1\ \ text(and)`
`2/sqrt3 sin(t)` `= tan(t)`

 
`text(Equate)\ underset~j\ text(components:)`

`2/sqrt3 sin(t)` `= (sin(t))/(cos(t))`
`cos(t)` `= sqrt3/2`
`t` `= pi/6`

 
`text(Check)\ underset~i\ text(components at)\ \ t= pi/6 :`

`-1 + 4cos(pi/6)` `= 3 sec(pi/6)-1`
`-1 + 4 · sqrt3/2` `= 3· 2/sqrt3-1`
`2sqrt3` `= 2sqrt3`

 
`:.\ text(Particles collide.)`

 

b.ii.   `text(Collision occurs at)\ \ r(pi/6)`

`r(pi/6)` `= (-1 + 4cos\ pi/6, 2/sqrt3 sin\ pi/6)`
  `= (-1 + 2sqrt3, 1/sqrt3)`
♦ Mean mark part (c)(i) 37%.

 

c.i.   `d/dx (8sin^(-1) ((x + 1)/4) + ((x + 1)sqrt(-x^2 -2x + 15))/2)`

`= 8/(sqrt(1-((x + 1)/4)^2)) xx 1/4 + ((x + 1))/2 xx (-2x-2)/(2sqrt(-x^2-2x + 15)) + sqrt(-x^2-2x + 15) xx 1/2`

`= 8/(sqrt(16-(x + 1)^2))-((x + 1)^2)/(2sqrt(-x^2-2x + 15)) + sqrt(-x^2 -2x + 15)/2`

`= 16/(2sqrt(-x^2-2x + 15))-((x + 1)^2)/(2sqrt(-x^2-2x + 15))+ (-x^2-2x + 15)/(2sqrt(-x^2-2x + 15))`

`= (16-x^2-2x-1-x^2-2x + 15)/(2sqrt(-x^2-2x + 15))`

`= (2(-x^2-2x + 15))/(2sqrt(-x^2-2x + 15))`

`= sqrt(-x^2-2x + 15)`

♦ Mean mark part (c)(ii) 45%.

 

c.ii.    `text(Area)` `= int_1^(2sqrt3-1) sqrt3/6 sqrt(-x^2-2x + 15)\ dx`
    `= sqrt3/6 int_1^(2sqrt3-1) sqrt(-x^2-2x + 15)\ dx`
    `= sqrt3/6 [8sin^(-1)((x + 1)/4) + ((x + 1)sqrt(-x^2-2x + 15))/2]_1^(2sqrt3-1)`
    `= sqrt3/6 [(8sin^(-1)(sqrt3/2) + 2sqrt3 sqrt(-(2sqrt3 -1)^2-2(2sqrt3-1) + 15)/2)-(8sin^(-1)(1/2) + (2sqrt(-1-2 + 15))/2)]`
    `= sqrt3/6 [(8pi)/3 + sqrt3 sqrt(-12 + 4sqrt3-1-4sqrt3 + 2 + 15)-((8pi)/6 + sqrt12)]`
    `= sqrt3/6 ((8pi)/3 + 2sqrt3-(8pi)/6-2sqrt3)`
    `= sqrt3/6 ((8pi)/6)`
    `= (2sqrt3pi)/9`

Vectors, SPEC2 2020 VCAA 15 MC

Two forces, `underset~F_(text(A)) = 4 underset~i - 2 underset~j`  and  `underset~F_(text(B)) = 2 underset~i - 5 underset~j`, act on a particle of mass 3 kg. The particle is initially at rest at position  `underset~i + underset~j`. All force components are measured in newtons and displacements are measured in metres.

The cartesian equation of the path of the particle is

  1. `y = x/2`
  2. `y = x/2 - 1/2`
  3. `y = ((x + 1)^2)/2 + 1`
  4. `y = ((x - 1)^2)/1 + 1`
  5. `y = x/2 + 1/2`
Show Answers Only

`E`

Show Worked Solution

`text(Net Force)\ (underset~F) = 4underset~i – 2underset~j +  2underset~i + 5underset~j = 6underset~i + 3underset~j`

`underset~a = underset~F/m = 2underset~i + underset~j`

♦ Mean mark 38%.

`underset~v = int_0^t 2underset~i + underset~j\ dt = 2tunderset~i + tunderset~j`

`underset~r` `= int_0^t 2tunderset~i + tunderset~j\ dt + (underset~i + underset~j)`
  `= t^2 underset~i + (t^2)/2 underset~j + underset~i + underset~j`
  `= (t^2 + 1)underset~i + ((t^2)/2 + 1)underset~j`

 
`x = t^2 + 1 \ => \ t^2 = x – 1`

`y` `= (t^2)/2 + 1`
  `= ((x – 1))/2 + 1`
  `= x/2 + 1/2`

 
`=>E`

Vectors, SPEC2 2011 VCAA 13 MC

The position of a particle at time  `t`  is given by  `underset~r(t) = (sqrt(t - 2))underset~i + (2t)underset~j`  for  `t >= 2`.

The cartesian equation of the path of the particle is

A.   `y = 2x^2 + 4`, `x >= 2`
B.   `y = 2x^2 + 2`, `x >= 2`
C.   `y = 2x^2 + 4`, `x >= 0`
D.   `y = sqrt((x - 4)/2)`, `x >= 4`
E.   `y = 2x^2 + 2`, `x >= 0`

 

Show Answers Only

`C`

Show Worked Solution

`tilder(t) = (sqrt(t – 2)tildei + (2t)tildej)`

`x = sqrt(t – 2)`

`text(Given)\ \ t >= 2\ \ =>\ \ x >=0`

`y = 2t\ \ =>\ \ t = y/2`

`:. x` `= sqrt(y/2 – 2)`
`x^2` `= y/2 – 2`
`y/2` `= x^2 + 2`
`y` `= 2x^2 + 4`

 
`=> C`

Vectors, SPEC1 2013 VCAA 7

The position vector  `underset ~r (t)`  of a particle moving relative to an origin `O` at time `t` seconds is given by

`underset ~r(t) = 4 sec (t) underset ~i + 2 tan (t) underset ~j,\ t in [0, pi/2)`

where the components are measured in metres.

  1. Show that the cartesian equation of the path of the particle is  `x^2/16-y^2/4 = 1.`   (2 marks)

  2. Sketch the path of the particle on the axes below, labelling any asymptotes with their equations.   (2 marks)

     

        
     VCAA 2013 spec 7b
     

  3. Find the speed of the particle, in `text(ms)^-1`, when `t = pi/4.`   (2 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
  3. `4 sqrt 3\ \ text(ms)^-1`
Show Worked Solution
a.   `x` `= 4sec(t)`
  `x/4` `= sec(t)`
  `y` `= 2tan(t)`
  `y/2` `= tan(t)`

  
`text(Using)\ \ tan^2 (t) + 1 = sec^2(t),`

`(x^2)/16 +1 ` `= y^2/4`
`:. (x^2)/16-(y^2)/4` `=1`

 

b.   `y^2/4 = x^2/16 -1\ \ =>\ \ y=+- sqrt(x^2/4 -4)`

♦ Mean mark part (b) 45%.

`lim_(x->oo) y = +- x/2`

 

♦ Mean mark part (c) 39%.

c.    `overset·underset~r(t)` `= d/(dt)(4(cos(t))^(−1))underset~i + d/(dt)(2tan(t)) underset~j`
    `= 4(−1)(−sin(t))(cos(t))^(−2)underset~i + 2sec^2(t)underset~j`
    `= 4sin(t)sec^2(t)underset~i + 2sec^2(t)underset~i`

 

`|overset·underset~r(pi/4)|` `= sqrt(16sin^2(pi/4)sec^4(pi/4) + 4sec^4(pi/4))`
  `= sqrt(16(1/sqrt2)^2(sqrt2)^4 + 4(sqrt2)^4)`
  `= sqrt(16(1/2)(4) + 4(4))`
  `= sqrt(48)`
  `= 4sqrt3\ \ text(ms)^(-1)`

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)

Show Answers Only
  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, SPEC1 2014 VCAA 2

The position vector of a particle at time  `t >= 0`  is given by

`underset ~r (t) = (t-2) underset ~ i + (t^2-4t + 1) underset ~j`

  1.  Show that the cartesian equation of the path followed by the particle is  `y = x^2-3`.  (1 mark)

  2.  Sketch the path followed by the particle on the axes below, labelling all important features.  (2 marks)


 

 

  1.  Find the speed of the particle when  `t = 1`.  (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(See Worked Solutions)`
  3. `sqrt 5`

Show Worked Solution

a.    `x` `= t-2\ \ =>\ \ t=x+2`
`y` `= t^2-4t + 1`
  `= (x + 2)^2-4 (x + 2) + 1`
`y` `= x^2 + 4x + 4-4x-8 + 1`
  `= x^2-3`

 

b.   `t >= 0`

`x = t-2\ \ =>\ \ x >= -2`

`y(-2)` `= (-2)^2-3`
  `= 1`
`y(0)` `= -3`

 

`0` `= x^2-3`
`x^2` `= 3`
`x` `= +- sqrt 3`

 

c.    `underset ~ dot r (t)` `= underset ~ dot i + (2t-4) underset ~j`
  `underset ~ dot r (1)` `= underset ~ dot i + (2-4) underset ~j`
    `= underset ~ dot i-2 underset ~j`
  `|underset ~ dot r(1)|` `= sqrt (1^2 + (-2)^2)`
    `= sqrt 5`

Calculus, SPEC1 2018 VCAA 9

A curve is specified parametrically by  `underset ~r(t) = sec(t) underset ~i + sqrt 2/2 tan(t) underset ~j, \ t in R`.

  1.  Show that the cartesian equation of the curve is  `x^2-2y^2 = 1`.   (2 marks)

  2.  Find the `x`-coordinates of the points of intersection of the curve  `x^2-2y^2 = 1`  and the line  `y = x-1`.   (1 mark)

  3.  Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the `x`-axis.   (2 marks)

Show Answers Only

  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `x = 1 or x = 3`
  3. `(2 pi)/3`

Show Worked Solution

a.     `x = sec(t), qquad y = sqrt 2/2 tan(t)`

`x^2 = sec^2(t), qquad y^2 = 1/2 tan^2(t)`

`x^2 = sec^2(t), qquad 2y^2 = tan^2(t)`

`1 + tan^2(t)` `= sec^2(t)`
`1 + 2y^2` `= x^2`
`:.x^2-2y^2` `=1\ \ text(.. as required)`

 

b.    `x^2-2(x-1)^2` `= 1`
  `x^2-2(x^2-2x + 1)` `=1`
  `x^2-2x^2 + 4x-2` `=1`
  `-x^2 + 4x-2-1` `=0`
  `x^2-4x + 3` `=0`
  `(x-3) (x-1)` `=0`

 
`:. x = 1 or x = 3`

♦♦ Mean mark 30%.

 

c.   `x^2-{:2y_1:}^2` `=1`
  `{:2y_1:}^2` `=x^2-1`
  `{:y_1:}^2` `= (x^2-1)/2`
  `{:y_2:}^2` `= (x-1)^2`

 

`V` `=pi int_1^3 {:y_1:}^2-{:y_2:}^2 \ dx`
  `= pi int_1^3 (x^2-1)/2-(x-1)^2\ dx`
  `= pi [x^3/6-x/2-(x-1)^3/3]_1^3`
  `= pi [(3^3/6-3/2-2^3/3)-(1^3/6-1/2-0)]`
  `= pi (9/2-3/2-8/3-1/6 + 1/2)`
  `= pi (7/2-8/3-1/6)`
  `= pi ((21-16-1)/6)`
  `= (2 pi)/3`

Vectors, SPEC2 2017 VCAA 12 MC

Let   `underset~r(t) = (1 - sqrt(a)sin(t))underset~i + (1 - 1/b cos(t))underset~j`  for  `t >= 0`  and  `a, b ∈ R^−`  be the path of a particle moving in the cartesian plane.

The path of the particle will always be a circle if

  1. `ab^2 = 1`
  2. `a^2b = 1`
  3. `ab^2 != 1`
  4. `ab = 1`
  5. `a^2b != 1`
Show Answers Only

`A`

Show Worked Solution

♦ Mean mark 49%.

`x` `= 1 – sqrta sin(t)`
`1 – x` `= sqrta sin(t)`
`sin(t)` `= (1 – x)/(sqrta)`
`sin^2(t)` `= ((1 – x)^2)/a`

 

`y` `= 1 – 1/b cos(t)`
`1 – y`  `= 1/b cos(t)`
`b(1 – y)` `= cos(t)`
`b^2(1 – y)^2` `= cos^2(t)`

 

` ((1 – x)^2)/a + b^2(1 – y)^2` `= sin^2(t) + cos^2(1)`
`((1 – x)^2)/a + b^2(1 – y)^2` `= 1`

 
`text(Equation describes a circle when:)`

 `1/a = b^2\ \ =>\ \ ab^2 = 1`

`=> A`