Derivative of a Vector Function Solved Problems
Example. 1 . If r = a cos ti + a sin tj + at tan θk find \( \left|\frac{d r}{d t} \times \frac{d^2}{d t^2}\right| \text { and } \mid \frac{d r}{d t} \frac{d^2 t}{d t^2} \frac{d^2}{d t^2}\)
Solution.
Given
r = a cos ti + a sin tj + at tan θk ∴ dtr = −a sin ti + a cos tj + a tan θk
⇒ \(\frac{d^2 r}{d t^2}=-a \cos t \mathbf{i}-a \sin t \mathbf{j}+0 . \quad \cdots \frac{d^3 r}{d t^2}=a \sin t \mathbf{i}-a \cos t \mathbf{j}\)
∴ \(\frac{d r}{d t} \times \frac{d^2 r}{d t^2}\)
= \(\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\-a \sin t & a \cos t & a \tan \theta \\-a \cos t & -a \sin t & 0
\end{array}\right|=a^2 \sin t \tan \theta \mathrm{i}-a^2 \cos t \tan \theta \mathbf{j}+a^2 \mathbf{k}\)
∴ \(\left|\frac{\mathbf{d r}}{d t} \times \frac{d^2 r}{d t^2}\right|\)
= \(a^2 \sqrt{\left(\sin ^2 t \tan ^2 \theta+\cos ^2 t \tan ^2 \theta+1\right)}=a^2 \sqrt{\left(\tan ^2 \theta+1\right)}\)
= \(d^2 \sec \theta \text { And }\left[\frac{d t}{d t} \frac{d^2 t}{d t^2} \frac{d^2 c}{d t^2}\right]= \)
⇒\(\left(\frac{d r}{d t} \times \frac{d^2 r}{d t^2}\right) \cdot \frac{d^2 r}{d t^3}\)
= (a2 sin t tan θi − a2 cos t tan θj + a2k) · (a sin ti− a cos tk) = a3 sin2 t tan θ+ a3 cos2 t tan θ = a3 tan θ
Derivative Of A Vector Function With Scalar Variable
Example. 2. If A = t2i−tj+(2t+1)k and B = (2t−3)i+j−tk find
(1) (A × B)0
(2) (| A + B|)0 · atit = 1
Solution.
Given A = t2i − tj + (2t + 1)k
⇒ \(\mathbf{A}^{\prime}=\frac{d \mathbf{A}}{d t}=2 t \mathbf{i}-\mathbf{j}+2 \mathbf{k} \text { and } \mathbf{B}=(2 t-3) \mathbf{i}+\mathbf{j}-t \mathbf{k}\)
⇒ \(\mathbf{B}^{\prime}=\frac{d \mathbf{B}}{d t}=2 \mathbf{i}-\mathbf{k} \text { at } t=1, \mathbf{A}=\mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{B}=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)
⇒ \(\frac{d \mathrm{~A}}{d t}=2 \mathrm{t}-\mathrm{j}+2 \mathrm{k}, \frac{d \mathrm{~B}}{d t}=2 \mathrm{i}-\mathrm{k}\)
∴ A × B’ = \(\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & -1 & 3 \\2 & 0 & -1\end{array}\right|\)
= \(\mathbf{i}+7 \mathbf{j}+2 \mathbf{k} \quad \mathbf{A}^{\prime} \times \mathbf{B}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\2 & -1 & 2 \\-1 & 1 & -1\end{array}\right|= \)
−i + k
∴ (A × B)’ = (i + 7j + 2k) + (−i + k) = 7j + 3k (5)
A + B = t2 i − tj + (2t + 1)k + (2t − 3)i + j − tk
A + B = (t2 + 2t − 3) i + (1 − t)j + (t + 1)k
|A + B| = \(\sqrt{\left[\left(t^2+2 t-3\right)^2+(1-t)^2+(t+1)^2\right]}=\sqrt{\left(t^4+4 t^3-12 t+11\right)}\)
∴ \((|\mathbf{A}+\mathbf{B}|)^{\prime}=\frac{4 t^3+12 t-12}{2 \sqrt{\left(t^4+4 t^3-12 t+11\right)}} \quad ∴ \text { At } t=1,(|\mathbf{A}+\mathbf{B}|)^{\prime}=1\)
Example. 3. If r is a vector function, of the scalar t, whose modulus is r and a, b are constant vectors, differentiate the following with respect to t. \(\begin{array}{ll}\text { (1) } \frac{r}{r^2}+\frac{r b}{a \cdot r} & \text { (2) } r^3 r+a \times r\end{array}\)
Solution.
(1) \(\frac{d}{d t}\left[\frac{r}{r^2}+\frac{r \mathrm{~b}}{\mathrm{a}-\mathrm{r}}\right]=\frac{d}{d t}\left(\frac{r}{r^2}\right)+\mathrm{b} \frac{d}{d t}\left(\frac{r}{\mathrm{a} \cdot \mathrm{r}}\right)\)
⇒ \(\left.=\frac{r^2 \frac{d r}{d t}-\mathbf{r}(2 r) \frac{d r}{d t}}{r^4}+\frac{\mathbf{b}\left[(\mathbf{a} \cdot \mathbf{r}) \frac{d r}{d t}-r\left(\mathbf{a} \cdot \frac{d r}{d t}\right)\right]}{\left(\mathbf{a} \cdot \mathbf{r}^2\right.}\right)\)
⇒ \(=\frac{r^2 \mathbf{r}^{\prime}-2 r \mathbf{r}^{\prime}}{r^4}+\mathbf{b}\left[\frac{1}{(\mathbf{a} \cdot \mathbf{r})} \cdot \frac{d r}{d t}-\frac{r}{(\mathbf{a} \cdot \mathbf{r})^2}\left(\mathbf{a} \cdot \frac{d \mathbf{r}}{d t}\right)\right]=\)
⇒ \(=\frac{\mathbf{r}^{\prime}}{r^2}-\frac{2 r^{\prime} \mathbf{r}}{r^3}+\frac{\mathbf{b} r^{\prime}}{(\mathbf{a} \cdot \mathbf{r})}-\frac{\mathbf{b} r\left(\mathbf{a} \cdot \mathbf{r}^{\prime}\right)}{(\mathbf{a} \cdot \mathbf{r})^2}\)
(2) \(\frac{d}{d t}\left[r^3 \mathbf{r}+\mathbf{a} \times \mathbf{r}^{\prime}\right]=\frac{d}{d t}\left[r^3 \mathbf{r}\right]+\frac{d}{d t}\left(\mathbf{a} \times \mathbf{r}^{\prime}\right)\)
⇒ \(=r^3 \frac{d \mathbf{r}}{d t}+3 r^2 \frac{d t}{d t} \mathbf{r}+\mathbf{a} \times \frac{d(\mathbf{r})}{d t}=r^3 \mathbf{r}^{\prime}+3 r^2 r^{\prime} \mathbf{r}+\mathbf{a} \times \mathbf{r}^{\prime \prime}\)
Continuity Of Vector Functions Explained
Example. 4. If r = aeλt + be−λt where a and b are constant vectors show that\(\frac{d^2 \mathrm{r}}{d t^2}=\lambda^2 \mathrm{r}\)
Solution.
Given, r = aeλt + be−λt
Diffirentiating w.r.t. t, \(\frac{d r}{d t}=\lambda\left(a e^{\lambda t}-b e^{-\lambda t}\right) \cdots\)
⇒ \(\frac{d^2 r}{d t^2}=\lambda^2\left(a e^{i t}+b e^{-\lambda t}\right)=\lambda^2 r\)
Derivative of a Vector Function Solved Problems
Example. 5. Show that if r = acos wt + b sin wt, where a, b are constant vectors and w is a constant scalar, then \(\frac{d^2 r}{d d^2}=-w^2 \mathbf{r} \text { and } \mathbf{r} \times \frac{d r}{d t}=w(\mathbf{a} \times \mathbf{b}) \)
Solution.
Given r = acos wt + b sin wt, a, b are constant vectors and w is a constant scalar.
D. w.r.t t we get dr/dt = w(−asin wt + b cos wt)
⇒ \(\frac{d^2 \mathbf{r}}{d t^2}=\lambda^2\left(a e^{i t}+b e^{-\lambda t}\right)=\lambda^2 \mathbf{r}\)
⇒ \(\mathbf{r} \times \frac{d \mathbf{r}}{d t}=(\mathbf{a} \cos w t+\mathbf{b} \sin w t) \times w(-\mathbf{a} \sin w t+\mathbf{b} \cos w t)\)
= w (a × b cos2 wt − b × asin2 wt)
= w(a × b) (cos2 wt + sin2 wt) = w(a × b)
Example. 6. A particle moves so that its position vector is given by r = (cos ωt)i+(sin ωt)j, where ω is a constant scalar.
(1) Find the velocity of the particle and show that it is perpendicular to r.
(2) Find the acceleration and show that it is directed opposite to the direction of r.
(3) Show that the magnitude of the acceleration is proportional to the magnitude of r.
Solution.
Given r = (cos ωt)i + (sin ωt)j is the position vector of the particle.
D. W.r to t, we get dtr = [(− sin ωt)i + (cos ωt)j]ω
v = velocity = ω(− sin ωti + cos ωtj)
(1) r. v = ω[(cos ωti + sin ωtj) · (− sin ωti + cos νtj)
= ω(− sin ωt cos ωt + sin ωt cos ωt) = 0
(2) \(\frac{d^2 \mathrm{r}}{d t^2}=\omega^2(-\cos \omega t \mathrm{i}-\sin \omega t \mathrm{j})=-\omega^2 \bar{r}\)
acceleration = a = −ω²r, which has that it is the direction of opposite the direction of r.
(3) |a| = w²|r|
Example. 7. Suppose a particle P moves along a curve where parametric equations are x = t − sin t, y = 1 − cos t, z = 4sin(t/2), where t is time, (1) find the velocity and acceleration at time t = 0 (2) Find the magnitude of velocity and acceleration t = 0
Solution.
The position vector of the particle P is r = xi + yj + zk = (t − sin t)i + (1 − cos t)j + (4sin(t/2))k
D. w. r to t \(\frac{d \mathbf{r}}{d t}=(1-\cos t) \mathbf{i}+(\sin t) \mathbf{j}+\left(2 \cos \frac{t}{2}\right) \mathbf{k} \text { and } \frac{d^2 \mathbf{r}}{d t^2}=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+\left(-\sin \frac{t}{2}\right) \mathbf{k}\)
(1) The velocity v and acceleration a of P are, \(\mathbf{v}=\frac{d r}{d t}=(1-\cos t) \mathbf{i}+(\sin t) \mathbf{j}+\left(2 \cos \frac{t}{2}\right) \mathbf{k} ; \quad \mathbf{a}=\frac{d^2 \mathbf{r}}{d t^2}=(\sin t) \mathbf{i}+\cos t \mathbf{j}+ \)
(− sin t/2)k (b) at, t = 0, v = i + 2k, a = j
|v| = √1 + 4 = √5. | a| = 1
Examples Of Vector Function Scalar Variable Derivatives
Example. 8. A particle moves along a curve whose parametric equations are x = e −t, y = 2cos3t, z = 2sin3t, where t is the time, determine its velocity and acceleration at any time.
Solution.
Given
A particle moves along a curve whose parametric equations are x = e −t, y = 2cos3t, z = 2sin3t, where t is the time
The position vector of the particle is r = xi + yj + zk = e −ti + (2cos3t)j + (2sin3t)k
D.W.R. to t
⇒ \(\frac{d \mathbf{r}}{d t}=\text { Velocity }=\mathbf{v}=-e^{-t^2} \mathbf{i}+(-6 \sin 3 t) \mathbf{j}+(6 \cos 3 t) \mathbf{k}\)
⇒ \(\mathbf{a}=\text { acceleration }=\frac{d^2 \mathbf{r}}{d t^2}=e^{-t} \mathbf{i}+(-18 \sin 3 t) \mathbf{j}+(-18 \cos 3 t) \mathbf{k} \)