IBDP Math AI:Topic : AHL 3.12:Vector applications to kinematics-IB style Questions HL Paper 1

Ans:

(a) attempt at chain rule
\(\begin{pmatrix}
v=\frac{dOP}{dt}
\end{pmatrix}\begin{pmatrix}
2t cos t^2\\
-2t sin t^2
\end{pmatrix}\)
(b) attempt at product rule
\(a=\begin{pmatrix}
2 cos t^2 – 4t^2 sin t^2\\
-2 sin t^2 – 4t^2 cos t^2
\end{pmatrix}\)

METHOD 1
let \(S= sin t^2\) and \(C = cos t^2\)
finding cos \(\theta \) using
\(a. \overrightarrow{OP} = 2SC – 4t^2 S^2 – 2SC – 4t^2C^2 = – 4t^2\)
\(|\overrightarrow{OP}| = 1\)
\(|a| = \sqrt{(2C-4t^2S)^2 + (-2S – 4t^2 C)^2}\)
\(=\sqrt{4 + 16t^4}> 4t^2\)
it \(\theta\) is the angle between them, then
\(cos \theta = -\frac{4t^2}{\sqrt{4+16t^4}}\)
so -1 < cos \(\theta\) < 0 therefore the vectors are never parallel

METHOD 2
solve
\(\begin{pmatrix}
2cos t^2 – 4t^2 sin t^2\\ -2 sin t^2 – 4t^2 cos t^2

\end{pmatrix} = k \binom{sin t^2}{cos t^2}\)
then
\((k=) \frac{2cos t^2 – 4t^2 sin t^2}{sin t^2} = \frac{-2 sin t^2 – 4t^2 cos t^2}{cos t^2}\)
\(2cos^2 t^2 – 4t^2 cos t^2 sin t^2 = -2 sin^2 t^2 – 4t^2 cos t^2 sin t^2\)
\(2cos^2 t^2 + 2 sin^2 t^2 =0\)
2 = 0
this is never true so the two vectors are never parallel

MEHTOD 3
embedding vectors in a 3d space and taking the cross product.
\(\begin{pmatrix}
sin t^2\\ cos t^2
\\ 0

\end{pmatrix} \times \begin{pmatrix}
2 cos t^2-4t^2 sin t^2\\-2 sin t^2 – 4t^2 cos t^2
\\ 0

\end{pmatrix} = \begin{pmatrix}
0\\0
\\ -2sin^2t^2 – 4t^2 cos t^2 sin t^2 – 2 cos^2 t^2 + 4t^2 cos t^2 sin t^2
\end{pmatrix}\)
\(=\begin{pmatrix}
0\\0
\\ -2

\end{pmatrix}\)
since the cross product is never zero, the two vectors are never parallel