Answer
The amplitude is $\pm \sqrt{13}$ and the period is $2\pi $.
Work Step by Step
The given equation is provided:
$d=2\cos t+3\sin t$
Multiply and divide the given equation by $\sqrt{{{2}^{2}}+{{3}^{2}}}=\sqrt{13}$. Then we get
$d=\left( \frac{2}{\sqrt{13}}\cos t+\frac{3}{\sqrt{13}}\sin t \right)\sqrt{13}$
The $\cos $ is computed by dividing the base and hypotenuse and sine is computed by dividing the perpendicular and hypotenuse.
It is given that $\tan \theta =\frac{3}{2}$.
Hence, assume a right triangle with the perpendicular $=3k$ and base $=2k$ for some constant $k$ and let the hypotenuse be ‘H’.
By using Pythagoras theorem:
$\begin{align}
& {{\left( 2k \right)}^{2}}+{{\left( 3k \right)}^{2}}={{H}^{2}} \\
& H=\left( \sqrt{{{2}^{2}}+{{3}^{2}}} \right)k \\
& H=\sqrt{13}k
\end{align}$
Then, we further solve:
$\begin{align}
& \theta =\frac{3k}{\sqrt{13}k} \\
& =\frac{3}{\sqrt{13}}\text{ }
\end{align}$
Now, the value of cos is:
$\begin{align}
& \cos \theta =\frac{2k}{\sqrt{13}k} \\
& =\frac{2}{\sqrt{13}}
\end{align}$
Put the values:
$\begin{align}
& d\left( t \right)=\left( cos\theta \cdot \cos t+\sin \theta \cdot \sin t \right)\sqrt{13} \\
& =\left( \cos t\cdot \cos \theta +\sin t\cdot \sin \theta \right)\sqrt{13} \\
& =\cos \left( t-\theta \right)\sqrt{13}\text{ }
\end{align}$
Further solve the equation:
$\begin{align}
& d\left( t \right)=2\cos t+3\sin t \\
& =\sqrt{13}\cos \left( t-\theta \right)\text{ }
\end{align}$
By applying the aforementioned formula:
$\left( \cos A\cos B+\sin A\sin B=\cos \left( A-B \right) \right)$
Hence, it is proved.
The amplitude will be $\sqrt{13}\left( \text{maximum or minimum of cos}\left( t-\theta \right) \right)$:
$\begin{align}
& \text{Amplitude}=\sqrt{13}\left( \pm 1 \right) \\
& =\pm \sqrt{13}
\end{align}$
The period of the ball's motion will be $\text{cos}\left( t-\theta \right)$ $=2\pi $.
