Explanations & Calculations
Once you know the position vector as a function of time (t), the velocity & the acceleration vectors can be found by derivating the position vector once & twice respectively. Useful operator is d ( u . v ) d t = d u d t , v + u . d v d t \qquad\qquad
\begin{aligned}
\small \frac{d(u.v)}{dt}&=\small \frac{du}{dt},v+u.\frac{dv}{dt}
\end{aligned} d t d ( u . v ) = d t d u , v + u . d t d v
And once you know any two vectors, the included angle can be found by performing the dot product of vectors. And I hope you know how to perform the dot product for 2 vectors [ex: ( p i + q j ) \small (p\it{i}+q\it{j}) ( p i + q j ) ( r i + s j ) \small (r\it{i}+s\it{j}) ( r i + s j ) a ‾ . b ‾ = ∣ a ‾ ∣ ∣ b ‾ ∣ cos θ cos θ = a ‾ . b ‾ ∣ a ‾ ∣ ∣ b ‾ ∣ ⋯ ⋯ ( 1 ) \qquad\qquad
\begin{aligned}
\small \underline{a} .\underline{b}&=\small |\underline{a}||\underline{b}|\cos\theta\\
\small \cos\theta&=\small \frac{\underline{a}.\underline{b}}{|\underline{a}||\underline{b}|}\cdots\cdots(1)
\end{aligned} a . b cos θ = ∣ a ∣∣ b ∣ cos θ = ∣ a ∣∣ b ∣ a . b ⋯⋯ ( 1 )
Then following what is mentioned above in the calculations ahead, needed quantities can be found. r ‾ = e − t cos θ . i + e − t sin θ . j ⋯ ⋯ ( g i v e n ) v ‾ = d r d t = ( − e − t cos θ − e − t sin θ ) i + ( − e − t sin θ + e − t cos θ ) j = − e − t ( cos θ + sin θ ) i − e − t ( sin θ − cos θ ) j T h e n r ‾ . v ‾ = − e − 2 t ( cos 2 θ + sin θ cos θ ) − e − 2 t ( sin 2 θ − sin θ cos θ ) = − e − 2 t ( cos 2 θ + sin θ cos θ + sin 2 θ − sin θ cos θ ) = − e − 2 t ( 1 ) = − e − 2 t ∣ r ‾ ∣ = ( e − t cos θ ) 2 + ( e − t sin θ ) 2 = ( e − t ) 2 × ( cos 2 θ + sin 2 θ ) = e − t ∣ v ‾ ∣ = [ − e − t ( cos θ + sin θ ) ] 2 + [ − e − t ( sin θ − cos θ ) ] 2 = ( − e − t ) 2 × [ ( cos θ + sin θ ) 2 + ( sin θ − cos θ ) 2 ] = e − t . cos 2 θ + sin 2 θ + 2 sin θ cos θ + sin 2 θ + cos 2 θ − 2 sin θ cos θ = e − t . 1 + 1 = 2 e − t cos θ = − e 2 t e − t × 2 e − t = − 1 2 θ = 3 π 4 ⋯ ⋯ ( ∵ π ≤ θ ≤ 0 ) \qquad\qquad
\begin{aligned}
\small \underline{r}&=\small e^{-t}\cos\theta.\it{i}+e^{-t}\sin\theta.\it{j}\cdots\cdots(given)\\\\
\small \underline{v}&=\small \frac{dr}{dt}\\
\small&=\small (-e^{-t}\cos\theta-e^{-t}\sin\theta)\it{i}+(-e^{-t}\sin\theta+e^{-t}\cos\theta)\it{j}\\
&=\small -e^{-t}(\cos\theta+\sin\theta)\it{i}-e^{-t}(\sin\theta-\cos\theta)\it{j}\\\\
Then\\
\small \underline{r}.\underline{v}&=\small -e^{-2t}(\cos^2\theta+\sin\theta \cos\theta )-e^{-2t}(\sin^2\theta-\sin\theta\cos\theta)\\
&=\small -e^{-2t}(\cos^2\theta+\sin\theta\cos\theta+\sin^2\theta-\sin\theta\cos\theta)\\
&=\small -e^{-2t}(1)\\
&=\small -e^{-2t}\\\\
\small |\underline{r}|&=\small \sqrt{(e^{-t}\cos\theta)^2+(e^{-t}\sin\theta)^2}\\
&=\small \sqrt{(e^{-t})^2\times(\cos^2\theta+\sin^2\theta)}\\
&=\small e^{-t}\\\\
\small |\underline{v}|&=\small \sqrt{[-e^{-t}(\cos\theta+\sin\theta)]^2+[-e^{-t}(\sin\theta-\cos\theta)]^2}\\
&=\small \sqrt{(-e^{-t})^2\times[(\cos\theta+\sin\theta)^2+(\sin\theta-\cos\theta)^2 ]}\\
&=\small e^{-t}.\sqrt{\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta+\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta}\\
&=\small e^{-t}.\sqrt{1+1}\\
&=\small \sqrt2 e^{-t}\\\\
\small \cos\theta&=\small \frac{-e^{2t}}{e^{-t}\times\sqrt2e^{-t}}\\
&=\small -\frac{1}{\sqrt2}\\
\small \theta&=\small \frac{3\pi}{4}\cdots\cdots(\because\,\pi\leq \theta\leq0)
\end{aligned} r v T h e n r . v ∣ r ∣ ∣ v ∣ cos θ θ = e − t cos θ . i + e − t sin θ . j ⋯⋯ ( given ) = d t d r = ( − e − t cos θ − e − t sin θ ) i + ( − e − t sin θ + e − t cos θ ) j = − e − t ( cos θ + sin θ ) i − e − t ( sin θ − cos θ ) j = − e − 2 t ( cos 2 θ + sin θ cos θ ) − e − 2 t ( sin 2 θ − sin θ cos θ ) = − e − 2 t ( cos 2 θ + sin θ cos θ + sin 2 θ − sin θ cos θ ) = − e − 2 t ( 1 ) = − e − 2 t = ( e − t cos θ ) 2 + ( e − t sin θ ) 2 = ( e − t ) 2 × ( cos 2 θ + sin 2 θ ) = e − t = [ − e − t ( cos θ + sin θ ) ] 2 + [ − e − t ( sin θ − cos θ ) ] 2 = ( − e − t ) 2 × [( cos θ + sin θ ) 2 + ( sin θ − cos θ ) 2 ] = e − t . cos 2 θ + sin 2 θ + 2 sin θ cos θ + sin 2 θ + cos 2 θ − 2 sin θ cos θ = e − t . 1 + 1 = 2 e − t = e − t × 2 e − t − e 2 t = − 2 1 = 4 3 π ⋯⋯ ( ∵ π ≤ θ ≤ 0 )
Therefore, for the first case, the two vectors make this angle between them.
Then for the acceleration, a ‾ = d 2 r d t 2 = d v d t = ( e − t ( cos θ + sin θ ) − e − t ( − sin θ + cos θ ) ) i − ( − e − t ( sin θ − cos θ ) + e − t ( cos θ + sin θ ) ) j = 2 e − t sin θ . i − 2 e − t cos θ . j T h e n a ‾ . r ‾ = ( 2 e − t sin θ × e − t cos θ ) + ( − 2 e − t cos θ × e − t sin θ ) = 2 e − 2 t sin θ cos θ − 2 e − 2 t sin θ cos θ = 0 \qquad\qquad
\begin{aligned}
\small \underline{a}&=\small \frac{d^2r}{dt^2}= \frac{dv}{dt}\\
&=\small (e^{-t}(\cos\theta+\sin\theta)-e^{-t}(-\sin\theta+\cos\theta))\it{i}-(-e^{-t}(\sin\theta-\cos\theta)+e^{-t}(\cos\theta+\sin\theta))\it{j}\\
&=\small 2e^{-t}\sin\theta .\it{i}-2 e^{-t}\cos\theta.\it{j}\\\\
Then\\
\small \underline{a}.\underline{r}&=\small (2e^{-t}\sin\theta\times e^{-t}\cos\theta)+(-2e^{-t}\cos\theta\times e^{-t}\sin\theta)\\
&=\small 2e^{-2t}\sin\theta\cos\theta-2e^{-2t}\sin\theta\cos\theta\\
&=\small 0
\end{aligned} a T h e n a . r = d t 2 d 2 r = d t d v = ( e − t ( cos θ + sin θ ) − e − t ( − sin θ + cos θ )) i − ( − e − t ( sin θ − cos θ ) + e − t ( cos θ + sin θ )) j = 2 e − t sin θ . i − 2 e − t cos θ . j = ( 2 e − t sin θ × e − t cos θ ) + ( − 2 e − t cos θ × e − t sin θ ) = 2 e − 2 t sin θ cos θ − 2 e − 2 t sin θ cos θ = 0
When the dot product becomes zero for two vectors, that means (according to [1],) the cosine of the contained angle is zero hence they are 90 degrees apart which places them in an orthogonal position. Then the acceleration vector is perpendicular to the position vector.
