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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}