Position Vector
If a particle has coordinates \(x(t),y(t),z(t)\), then its position vector is
\[ \mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}. \]
Velocity and Acceleration
Vector Functions and Motion Along a Curve
Find velocity and acceleration vectors by differentiating the position vector. These IOE vector calculus problems use the standard line by line method expected in Engineering Mathematics II.
UnitVector Calculus
TopicVector Differentiation
Question Type2 Mark Problems
Theory Required for These Problems
Velocity Vector
Velocity is the first derivative of the position vector:
\[ \mathbf{v}(t)=\frac{d\mathbf{r}}{dt} =\frac{dx}{dt}\mathbf{i} +\frac{dy}{dt}\mathbf{j} +\frac{dz}{dt}\mathbf{k}. \]
Acceleration Vector
Acceleration is the derivative of velocity or the second derivative of position:
\[ \mathbf{a}(t)=\frac{d\mathbf{v}}{dt} =\frac{d^2\mathbf{r}}{dt^2}. \]
Helical Motion
When \(x\) and \(y\) contain sine and cosine while \(z\) increases linearly, the particle moves along a helix. The velocity vector is tangent to the path.
Note: Differentiate first and substitute the given value of \(t\) afterwards.
Method used: Form the position vector, differentiate for velocity, differentiate again for acceleration, and then substitute the given time.
Continue through the IOE Engineering Mathematics II Digital Workbook. This velocity and acceleration page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
21
2 Marks
Velocity and Acceleration of a Particle
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 3(a) • 2 Marks
A particle moves along the curve
\[
x=3\cos t,\qquad y=3\sin t,\qquad z=6t.
\]
Find its velocity and acceleration at \(t=\frac{\pi}{3}\).
Solution
Differentiation of the position vector
The position vector is
\[ \mathbf{r}(t)=3\cos t\,\mathbf{i}+3\sin t\,\mathbf{j}+6t\,\mathbf{k}. \]Velocity is
\[ \mathbf{v}(t)=\frac{d\mathbf{r}}{dt} =-3\sin t\,\mathbf{i}+3\cos t\,\mathbf{j}+6\mathbf{k}. \]Acceleration is
\[ \mathbf{a}(t)=\frac{d\mathbf{v}}{dt} =-3\cos t\,\mathbf{i}-3\sin t\,\mathbf{j}. \]At \(t=\frac{\pi}{3}\),
\[ \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}, \qquad \cos\frac{\pi}{3}=\frac{1}{2}. \]Therefore,
\[ \mathbf{v}\left(\frac{\pi}{3}\right) =-\frac{3\sqrt{3}}{2}\mathbf{i} +\frac{3}{2}\mathbf{j}+6\mathbf{k}. \]Also,
\[ \mathbf{a}\left(\frac{\pi}{3}\right) =-\frac{3}{2}\mathbf{i} -\frac{3\sqrt{3}}{2}\mathbf{j}. \]\[
\mathbf{v}=-\frac{3\sqrt{3}}{2}\mathbf{i}+\frac{3}{2}\mathbf{j}+6\mathbf{k}
\]
\[
\mathbf{a}=-\frac{3}{2}\mathbf{i}-\frac{3\sqrt{3}}{2}\mathbf{j}
\]
22
2 Marks
Velocity and Acceleration of a Particle
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 3(b) • 2 Marks
A particle moves along the curve
\[
x=\sqrt{2}\cos t,\qquad y=\sqrt{2}\sin t,\qquad z=4t.
\]
Find its velocity and acceleration at \(t=\frac{\pi}{4}\).
Solution
Differentiation of the position vector
The position vector is
\[ \mathbf{r}(t)=\sqrt{2}\cos t\,\mathbf{i}+\sqrt{2}\sin t\,\mathbf{j}+4t\,\mathbf{k}. \]Velocity is
\[ \mathbf{v}(t)=\frac{d\mathbf{r}}{dt} =-\sqrt{2}\sin t\,\mathbf{i}+\sqrt{2}\cos t\,\mathbf{j}+4\mathbf{k}. \]Acceleration is
\[ \mathbf{a}(t)=\frac{d\mathbf{v}}{dt} =-\sqrt{2}\cos t\,\mathbf{i}-\sqrt{2}\sin t\,\mathbf{j}. \]At \(t=\frac{\pi}{4}\),
\[ \sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \qquad \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}. \]Therefore,
\[ \mathbf{v}\left(\frac{\pi}{4}\right) =-\mathbf{i}+\mathbf{j}+4\mathbf{k}. \]Also,
\[ \mathbf{a}\left(\frac{\pi}{4}\right) =-\mathbf{i}-\mathbf{j}. \]\[
\mathbf{v}=-\mathbf{i}+\mathbf{j}+4\mathbf{k}
\]
\[
\mathbf{a}=-\mathbf{i}-\mathbf{j}
\]