4.15 Motion in Space: Velocity and Acceleration
Given that \(S(t)\) is the arc\(-\)length function. We see that \(\dfrac {dS}{dt}\) is the speed of the particle \(P\) as it moves along the curve.
If \(\overline {r}(t) = x(t)\textbf {i} + y(t) \textbf {j} + z(t)\textbf {k}\hspace {0.2cm}\) is the position vector of a body traveling on a smooth curve in space, then
\[\overline {V} = \frac {d\overline {r}}{dt}= \frac {dx}{dt}\textbf {i} + \frac {dy}{dt}\textbf {j} + \frac {dz}{dt}\textbf {k}\]
is the velocity vector (or simply the velocity) of the particle \(P\) at time \(t\). Clearly, the length of
the velocity vector \(\left |\overline {V}\right | = \left |\overline {r}\right |\) gives the speed of a body at time \(t\).
If the position vector of a body at time \(t\) on a smooth space curve is \(\overline {r}(t) = x(t)\textbf {i} + y(t) \textbf {j} + z(t)\textbf {k}\hspace {0.2cm}\) where \(x, y\) and \(z\) have second derivatives with respect to \(t\), then the acceleration vector of (or simply the acceleration) of \(P\) at time \(t\) is given by \[\overline {a} = \frac {d\overline {V}}{dt} = \frac {d^2 x}{dt^2}\textbf {i} + \frac {d^2 y}{dt^2}\textbf {j} + \frac {d^2 z}{dt^2}\textbf {k}\]
Note that \[K(t) = \frac {\left |\overline {V} \times \overline {a}\right | }{\left |\overline {V}\right |^3 }\]
\[\tau (t) = \frac {\begin {vmatrix} x' & y' & z'\\ x'' & y'' & z''\\ x''' & y''' & z'''\\ \end {vmatrix} }{\left |\overline {V} \times \overline {a}\right |^2 }\]
- 1.
- Find the velocity, acceleration and speed of a particle with position vector \(\overline {r}(t) = \big \langle t^2, e^t, te^t\big \rangle \).
- 2.
- A moving particle starts at an initial position \(\overline {r}(t) = \big \langle 1, 0, 1\big \rangle \) with initial velocity
\(\overline {V}(0) = \textbf {i} - \textbf {j} + \textbf {k}\). Its acceleration vector is \(\overline {a}(t) = 4t\textbf {i} + 6t\textbf {j} + \textbf {k}\). Find its velocity and position at time \(t\).
Solution.
Part 1
\[\overline {V} = \overline {r}'(t) = \big \langle 2t, e^t, e^t + te^t\big \rangle \]
\[\overline {a} = \overline {V}'(t) = \big \langle 2, e^t, e^t+e^t+te^t\big \rangle = \big \langle 2, e^t, 2e^t+te^t\big \rangle \]
\[\text {Speed} = \left |\overline {V}\right | = \sqrt {4t^2 + e^{2t} + e^{2t} + 2te^{2t} + t^2e^{2t}} = \sqrt {4t^2 + e^{2t}\big (2 + 2t + t^2\big )}\]
Part 2
Since \(\overline {a}(t) = \overline {V}'(t) \implies \overline {V}(t) = \displaystyle {\int \overline {a}(t)\hspace {0.2cm}dt}\)
\begin {align*} \overline {V}(t) & = \int \Big ( 4t\textbf {i} + 6t\textbf {j} + \textbf {k}\Big )\hspace {0.1cm}dt\\ & = 2t^2\textbf {i} + 3t^2\textbf {j} + t\textbf {k} + \overline {c}\\\\ \implies \hspace {0.5cm} \overline {V}(0) = \overline {c} & = \textbf {i} - \textbf {j} + \textbf {k} \end {align*}
\[\implies \hspace {0.5cm} \overline {V}(t) = 2t^2\textbf {i} + 3t^2\textbf {j} + t\textbf {k} + \big (\textbf {i} - \textbf {j} + \textbf {k}\big ) = \big (2t^2 + 1\big )\textbf {i} + \big (3t^2-1\big )\textbf {j} + \big (t +1\big )\textbf {k}\]
\begin {align*} \overline {r}(t) & = \int \overline {V}(t)\hspace {0.1cm}dt\\ & = \int \Big [\big (2t^2 + 1\big )\textbf {i} + \big (3t^2-1\big )\textbf {j} + \big (t +1\big )\textbf {k}\Big ]\hspace {0.1cm}dt\\ & = \Big (\frac {2}{3}t^3 + t\Big )\textbf {i} + \big (t^3-t\big )\textbf {j} + \Big (\dfrac {t^2}{2} + t\Big )\textbf {k} + \overline {c}_1 \end {align*}
\[\overline {r}(0) = \overline {c}_1 = \textbf {i}\]
\[\implies \hspace {1cm} \overline {r}(t) = \Big (\frac {2}{3}t^3 + t - i\Big )\textbf {i} + \big (t^3-t\big )\textbf {j} + \Big (\dfrac {t^2}{2} + t\Big )\textbf {k}\]
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