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Vectors, SPEC2 2025 VCAA 4

The path of a moving particle with position vector

\(\underset{\sim}{r}(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

is shown below for time \(t \geq 0\).

All lengths are in metres and time is measured in seconds.
 

  1. Write down the coordinates of the particle's starting point.   (1 mark)

  2. On the graph above, draw an arrow from the point \((9,0)\) to indicate the direction of motion of the particle.   (1 mark)

  3. Find the value of \(t\) for which the particle will first return to its starting point.   (1 mark)

  4. Show that the speed of the particle, in m s\(^{-1}\), at time \(t\) can be expressed as  \(\sqrt{125-100 \cos \left(\dfrac{3 t}{2}\right)}\).   (3 marks)

  5. What is the maximum speed of the particle in m s\(^{-1}\)?   (1 mark)

  6. On the graph above, trace the path of the particle for  \(t \in[0, \pi]\).   (1 mark)

  7. Find the length of the path traced in part f, giving your answer in metres, correct to one decimal place.   (2 marks)

Show Answers Only

a.    \(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(t=4 \pi \ \text{seconds}\)

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max}}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

Show Worked Solution

a.    \(\text{At} \ \ t=0:\)

\(r(t)=(5 \cos 0-4 \cos 0)\underset{\sim}{i} + (5 \sin 0-4 \sin 0) {\underset{\sim}{j}}=\underset{\sim}{i}\)

\(\text{Initial coordinates:}\ (1,0)\)
 

b.    
     

c.   \(\text{Solve for}\ t \ \text{(by CAS):}\)

\(5 \cos (t)-4 \cos \left(\dfrac{5t}{2}\right)=1\)

\(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)=0\)

\(t=4 \pi \ \text{seconds}\)

Mean mark (c) 53%.
 

d.    \(r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}\)

\(\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}\)

\(\text {Speed}=\abs{\dot{r}(t)}:\)

\(\text{Speed}^2\) \(=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2\)
  \(=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)\)
       \(\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)\)
  \(=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)\)
  \(=125-100 \cos \left(\dfrac{5 t}{2}-t\right)\)
\(\text{speed}\) \(=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}\)

 

e.    \(\text{Speed}_{\text {max }} \ \text {occurs when} \ \ \cos \left(\frac{3 t}{2}\right)=-1\)

\(\text{Speed}_{\text {max}}=\sqrt{125+100}=15 \ \text{ms}^{-1}\)
 

f.    \(\text{At} \ \ t=\pi, \quad r(\pi)=-5\underset{\sim}{i}-4\underset{\sim}{j}\)

\(\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).\)
 


 

g.    \(\text{Length of curve (by CAS):}\)

\(\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}\)

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}