Vector differentiation
The derivative of a vector-valued function is defined similarly to real-valued functions. The first derivative of a position vector function represents velocity, and the second derivative represents acceleration.
A position vector r \mathbf{r} r t t t r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k f f f g g g h h h t t t
The velocity v ( t ) \mathbf{v}(t) v ( t )
v ( t ) = d r d t = f ′ ( t ) i + g ′ ( t ) j + h ′ ( t ) k \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k} v ( t ) = d t d r = f ′ ( t ) i + g ′ ( t ) j + h ′ ( t ) k
The acceleration a ( t ) \mathbf{a}(t) a ( t )
a ( t ) = d 2 r d t 2 = d v d t = f ′ ′ ( t ) i + g ′ ′ ( t ) j + h ′ ′ ( t ) k \mathbf{a}(t) = \frac{d^2\mathbf{r}}{dt^2} = \frac{d\mathbf{v}}{dt} = f''(t)\mathbf{i} + g''(t)\mathbf{j} + h''(t)\mathbf{k} a ( t ) = d t 2 d 2 r = d t d v = f ′′ ( t ) i + g ′′ ( t ) j + h ′′ ( t ) k
Properties of derivatives of vector-valued functions
Suppose u \mathbf{u} u v \mathbf{v} v f f f k k k
d d t ( k u ) = k d u d t \frac{d}{dt}(k\mathbf{u}) = k\frac{d\mathbf{u}}{dt} d t d ( k u ) = k d t d u d d t ( u ± v ) = d u d t ± d v d t \frac{d}{dt}(\mathbf{u} \pm \mathbf{v}) = \frac{d\mathbf{u}}{dt} \pm \frac{d\mathbf{v}}{dt} d t d ( u ± v ) = d t d u ± d t d v d d t ( u ⋅ v ) = u ⋅ d v d t + v ⋅ d u d t \frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} + \mathbf{v} \cdot \frac{d\mathbf{u}}{dt} d t d ( u ⋅ v ) = u ⋅ d t d v + v ⋅ d t d u d d t ( u × v ) = u × d v d t + d u d t × v \frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \mathbf{u} \times \frac{d\mathbf{v}}{dt} + \frac{d\mathbf{u}}{dt} \times \mathbf{v} d t d ( u × v ) = u × d t d v + d t d u × v d d t ( f v ) = d f d t v + f d v d t \frac{d}{dt}(f\mathbf{v}) = \frac{df}{dt}\mathbf{v} + f\frac{d\mathbf{v}}{dt} d t d ( f v ) = d t df v + f d t d v d d t v ( f ( t ) ) = d v d f d f d t \frac{d}{dt}\mathbf{v}(f(t)) = \frac{d\mathbf{v}}{df}\frac{df}{dt} d t d v ( f ( t )) = df d v d t df
Example 1
If f ( t ) = ( e − t − 2 ) i + 6 t j + ( t 2 e − t ) k \mathbf{f}(t) = (e^{-t} - 2)\mathbf{i} + 6t\mathbf{j} + (t^2e^{-t})\mathbf{k} f ( t ) = ( e − t − 2 ) i + 6 t j + ( t 2 e − t ) k f ′ ( t ) \mathbf{f}'(t) f ′ ( t )
Solution:
f ′ ( t ) = d d t ( ( e − t − 2 ) i ) + d d t ( 6 t j ) + d d t ( t 2 e − t k ) \mathbf{f}'(t) = \frac{d}{dt}((e^{-t} - 2)\mathbf{i}) + \frac{d}{dt}(6t\mathbf{j}) + \frac{d}{dt}(t^2e^{-t}\mathbf{k}) f ′ ( t ) = d t d (( e − t − 2 ) i ) + d t d ( 6 t j ) + d t d ( t 2 e − t k )
f ′ ( t ) = − e − t i + 6 j + ( 2 t e − t − t 2 e − t ) k = − e − t i + 6 j + e − t ( 2 t − t 2 ) k \mathbf{f}'(t) = -e^{-t}\mathbf{i} + 6\mathbf{j} + (2te^{-t} - t^2e^{-t})\mathbf{k} = -e^{-t}\mathbf{i} + 6\mathbf{j} + e^{-t}(2t - t^2)\mathbf{k} f ′ ( t ) = − e − t i + 6 j + ( 2 t e − t − t 2 e − t ) k = − e − t i + 6 j + e − t ( 2 t − t 2 ) k
Example 2
If r ( u ) = ( u 2 − 3 ) i + ( 5 cos u ) j + ( 3 sin u ) k \mathbf{r}(u) = (u^2 - 3)\mathbf{i} + (5\cos u)\mathbf{j} + (3\sin u)\mathbf{k} r ( u ) = ( u 2 − 3 ) i + ( 5 cos u ) j + ( 3 sin u ) k d r d u \frac{d\mathbf{r}}{du} d u d r d 2 r d u 2 \frac{d^2\mathbf{r}}{du^2} d u 2 d 2 r
Solution:
d r d u = 2 u i − 5 sin u j + 3 cos u k \frac{d\mathbf{r}}{du} = 2u\mathbf{i} - 5\sin u\mathbf{j} + 3\cos u\mathbf{k} d u d r = 2 u i − 5 sin u j + 3 cos u k
d 2 r d u 2 = 2 i − 5 cos u j − 3 sin u k \frac{d^2\mathbf{r}}{du^2} = 2\mathbf{i} - 5\cos u\mathbf{j} - 3\sin u\mathbf{k} d u 2 d 2 r = 2 i − 5 cos u j − 3 sin u k
Example 3
If f ( t ) = ( cos 4 t − 2 sin t ) i + ( sin 4 t + cos 2 t ) j + ( 5 cos t ) k \mathbf{f}(t) = (\cos 4t - 2\sin t)\mathbf{i} + (\sin 4t + \cos 2t)\mathbf{j} + (5\cos t)\mathbf{k} f ( t ) = ( cos 4 t − 2 sin t ) i + ( sin 4 t + cos 2 t ) j + ( 5 cos t ) k f ′ ( t ) \mathbf{f}'(t) f ′ ( t )
Solution:
f ′ ( t ) = ( − 4 sin 4 t − 2 cos t ) i + ( 4 cos 4 t − 2 sin 2 t ) j + ( − 5 sin t ) k \mathbf{f}'(t) = (-4\sin 4t - 2\cos t)\mathbf{i} + (4\cos 4t - 2\sin 2t)\mathbf{j} + (-5\sin t)\mathbf{k} f ′ ( t ) = ( − 4 sin 4 t − 2 cos t ) i + ( 4 cos 4 t − 2 sin 2 t ) j + ( − 5 sin t ) k
Example 4
If r ( t ) = cos t i + sin t j + 2 t k \mathbf{r}(t) = \cos t\mathbf{i} + \sin t\mathbf{j} + 2t\mathbf{k} r ( t ) = cos t i + sin t j + 2 t k r ( t ) ⋅ r ′ ′ ( t ) \mathbf{r}(t) \cdot \mathbf{r}''(t) r ( t ) ⋅ r ′′ ( t ) r ( t ) × r ′ ′ ( t ) \mathbf{r}(t) \times \mathbf{r}''(t) r ( t ) × r ′′ ( t )
Solution:
r ′ ( t ) = − sin t i + cos t j + 2 k \mathbf{r}'(t) = -\sin t\mathbf{i} + \cos t\mathbf{j} + 2\mathbf{k} r ′ ( t ) = − sin t i + cos t j + 2 k
r ′ ′ ( t ) = − cos t i − sin t j + 0 k \mathbf{r}''(t) = -\cos t\mathbf{i} - \sin t\mathbf{j} + 0\mathbf{k} r ′′ ( t ) = − cos t i − sin t j + 0 k
(a) r ( t ) ⋅ r ′ ′ ( t ) = ( cos t ) ( − cos t ) + ( sin t ) ( − sin t ) + ( 2 t ) ( 0 ) = − cos 2 t − sin 2 t = − 1 \mathbf{r}(t) \cdot \mathbf{r}''(t) = (\cos t)(-\cos t) + (\sin t)(-\sin t) + (2t)(0) = -\cos^2 t - \sin^2 t = -1 r ( t ) ⋅ r ′′ ( t ) = ( cos t ) ( − cos t ) + ( sin t ) ( − sin t ) + ( 2 t ) ( 0 ) = − cos 2 t − sin 2 t = − 1
(b) r ( t ) × r ′ ′ ( t ) = ∣ i j k cos t sin t 2 t − cos t − sin t 0 ∣ = ( 0 − ( − 2 t ) ( − sin t ) ) i − ( 0 − ( − 2 t ) ( − cos t ) ) j + ( − cos t sin t − ( − sin t cos t ) ) k = − 2 t sin t i − 2 t cos t j \mathbf{r}(t) \times \mathbf{r}''(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos t & \sin t & 2t \\ -\cos t & -\sin t & 0 \end{vmatrix} = (0 - (-2t)(-\sin t))\mathbf{i} - (0 - (-2t)(-\cos t))\mathbf{j} + (-\cos t \sin t - (-\sin t \cos t))\mathbf{k} = -2t\sin t\mathbf{i} - 2t\cos t\mathbf{j} r ( t ) × r ′′ ( t ) = i cos t − cos t j sin t − sin t k 2 t 0 = ( 0 − ( − 2 t ) ( − sin t )) i − ( 0 − ( − 2 t ) ( − cos t )) j + ( − cos t sin t − ( − sin t cos t )) k = − 2 t sin t i − 2 t cos t j
Example 5
Two vector-valued functions are given by r ( t ) = ln t i − j + 1 t k \mathbf{r}(t) = \ln t\mathbf{i} - \mathbf{j} + \frac{1}{t}\mathbf{k} r ( t ) = ln t i − j + t 1 k u ( t ) = 2 t i − t j + 2 k \mathbf{u}(t) = 2t\mathbf{i} - t\mathbf{j} + 2\mathbf{k} u ( t ) = 2 t i − t j + 2 k
(a) d d t ( r ( t ) ⋅ u ( t ) ) \frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{u}(t)) d t d ( r ( t ) ⋅ u ( t ))
(b) d d t ( u ( t ) × u ′ ( t ) ) \frac{d}{dt}(\mathbf{u}(t) \times \mathbf{u}'(t)) d t d ( u ( t ) × u ′ ( t ))
Solution:
r ′ ( t ) = 1 t i − 1 t 2 k \mathbf{r}'(t) = \frac{1}{t}\mathbf{i} - \frac{1}{t^2}\mathbf{k} r ′ ( t ) = t 1 i − t 2 1 k
u ′ ( t ) = 2 i − j \mathbf{u}'(t) = 2\mathbf{i} - \mathbf{j} u ′ ( t ) = 2 i − j
u ′ ′ ( t ) = 0 \mathbf{u}''(t) = 0 u ′′ ( t ) = 0
(a) r ( t ) ⋅ u ( t ) = 2 t ln t + t + 2 t \mathbf{r}(t) \cdot \mathbf{u}(t) = 2t\ln t + t + \frac{2}{t} r ( t ) ⋅ u ( t ) = 2 t ln t + t + t 2
d d t ( r ( t ) ⋅ u ( t ) ) = 2 ln t + 2 + 1 − 2 t 2 = 2 ln t + 3 − 2 t 2 \frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{u}(t)) = 2\ln t + 2 + 1 - \frac{2}{t^2} = 2\ln t + 3 - \frac{2}{t^2} d t d ( r ( t ) ⋅ u ( t )) = 2 ln t + 2 + 1 − t 2 2 = 2 ln t + 3 − t 2 2
(b) u ( t ) × u ′ ( t ) = ∣ i j k 2 t − t 2 2 − 1 0 ∣ = ( 0 − ( − 2 ) ) i − ( 0 − 4 ) j + ( − 2 t − ( − 2 t ) ) k = 2 i + 4 j + 0 k = 2 i + 4 j \mathbf{u}(t) \times \mathbf{u}'(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2t & -t & 2 \\ 2 & -1 & 0 \end{vmatrix} = (0 - (-2))\mathbf{i} - (0 - 4)\mathbf{j} + (-2t - (-2t))\mathbf{k} = 2\mathbf{i} + 4\mathbf{j} + 0\mathbf{k} = 2\mathbf{i} + 4\mathbf{j} u ( t ) × u ′ ( t ) = i 2 t 2 j − t − 1 k 2 0 = ( 0 − ( − 2 )) i − ( 0 − 4 ) j + ( − 2 t − ( − 2 t )) k = 2 i + 4 j + 0 k = 2 i + 4 j
d d t ( u ( t ) × u ′ ( t ) ) = 2 i + 0 j + 0 k = 2 i \frac{d}{dt}(\mathbf{u}(t) \times \mathbf{u}'(t)) = 2\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = 2\mathbf{i} d t d ( u ( t ) × u ′ ( t )) = 2 i + 0 j + 0 k = 2 i
Example 6
A particle is moving so that its position after time t t t r ( s ) = ( 3 cos 2 s ) i + ( 2 sin 2 s ) j + e − s k \mathbf{r}(s) = (3\cos 2s)\mathbf{i} + (2\sin 2s)\mathbf{j} + e^{-s}\mathbf{k} r ( s ) = ( 3 cos 2 s ) i + ( 2 sin 2 s ) j + e − s k s = 4 t s = 4t s = 4 t t t t
Solution:
Using the chain rule, d r d t = d r d s d s d t \frac{d\mathbf{r}}{dt} = \frac{d\mathbf{r}}{ds} \frac{ds}{dt} d t d r = d s d r d t d s
d r d s = ( − 6 sin 2 s ) i + ( 4 cos 2 s ) j − e − s k \frac{d\mathbf{r}}{ds} = (-6\sin 2s)\mathbf{i} + (4\cos 2s)\mathbf{j} - e^{-s}\mathbf{k} d s d r = ( − 6 sin 2 s ) i + ( 4 cos 2 s ) j − e − s k
d s d t = 4 \frac{ds}{dt} = 4 d t d s = 4
d r d t = 4 ( ( − 6 sin 2 s ) i + ( 4 cos 2 s ) j − e − s k ) = ( − 24 sin 8 t ) i + ( 16 cos 8 t ) j − 4 e − 4 t k \frac{d\mathbf{r}}{dt} = 4((-6\sin 2s)\mathbf{i} + (4\cos 2s)\mathbf{j} - e^{-s}\mathbf{k}) = (-24\sin 8t)\mathbf{i} + (16\cos 8t)\mathbf{j} - 4e^{-4t}\mathbf{k} d t d r = 4 (( − 6 sin 2 s ) i + ( 4 cos 2 s ) j − e − s k ) = ( − 24 sin 8 t ) i + ( 16 cos 8 t ) j − 4 e − 4 t k
Example 7
The position vector of a plane above the ground is r ( t ) = ( t 3 + 3 ) i + ( t 2 − 2 ) j + t k \mathbf{r}(t) = (t^3 + 3)\mathbf{i} + (t^2 - 2)\mathbf{j} + t\mathbf{k} r ( t ) = ( t 3 + 3 ) i + ( t 2 − 2 ) j + t k t = 2 t = 2 t = 2
Solution:
v ( t ) = d r d t = 3 t 2 i + 2 t j + k \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 3t^2\mathbf{i} + 2t\mathbf{j} + \mathbf{k} v ( t ) = d t d r = 3 t 2 i + 2 t j + k
a ( t ) = d v d t = 6 t i + 2 j \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 6t\mathbf{i} + 2\mathbf{j} a ( t ) = d t d v = 6 t i + 2 j
At t = 2 t = 2 t = 2
v ( 2 ) = 3 ( 2 2 ) i + 2 ( 2 ) j + k = 12 i + 4 j + k \mathbf{v}(2) = 3(2^2)\mathbf{i} + 2(2)\mathbf{j} + \mathbf{k} = 12\mathbf{i} + 4\mathbf{j} + \mathbf{k} v ( 2 ) = 3 ( 2 2 ) i + 2 ( 2 ) j + k = 12 i + 4 j + k
a ( 2 ) = 6 ( 2 ) i + 2 j = 12 i + 2 j \mathbf{a}(2) = 6(2)\mathbf{i} + 2\mathbf{j} = 12\mathbf{i} + 2\mathbf{j} a ( 2 ) = 6 ( 2 ) i + 2 j = 12 i + 2 j
Vector integration
The anti-derivative of a vector function is a family of vector-valued functions. If r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} r ( t ) = f ( t ) i + g ( t ) j + h ( t ) k
∫ r ( t ) d t = ( ∫ f ( t ) d t ) i + ( ∫ g ( t ) d t ) j + ( ∫ h ( t ) d t ) k = F ( t ) i + G ( t ) j + H ( t ) k + C \int \mathbf{r}(t) dt = \left(\int f(t) dt\right)\mathbf{i} + \left(\int g(t) dt\right)\mathbf{j} + \left(\int h(t) dt\right)\mathbf{k} = F(t)\mathbf{i} + G(t)\mathbf{j} + H(t)\mathbf{k} + \mathbf{C} ∫ r ( t ) d t = ( ∫ f ( t ) d t ) i + ( ∫ g ( t ) d t ) j + ( ∫ h ( t ) d t ) k = F ( t ) i + G ( t ) j + H ( t ) k + C
where F ( t ) F(t) F ( t ) G ( t ) G(t) G ( t ) H ( t ) H(t) H ( t ) f ( t ) f(t) f ( t ) g ( t ) g(t) g ( t ) h ( t ) h(t) h ( t ) C = C 1 i + C 2 j + C 3 k \mathbf{C} = C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k} C = C 1 i + C 2 j + C 3 k
Example 1
Find ∫ ( 2 t i + 3 j − 3 t 2 k ) d t \int (2t\mathbf{i} + 3\mathbf{j} - 3t^2\mathbf{k}) dt ∫ ( 2 t i + 3 j − 3 t 2 k ) d t
Solution:
∫ ( 2 t i + 3 j − 3 t 2 k ) d t = ( ∫ 2 t d t ) i + ( ∫ 3 d t ) j − ( ∫ 3 t 2 d t ) k \int (2t\mathbf{i} + 3\mathbf{j} - 3t^2\mathbf{k}) dt = \left(\int 2t dt\right)\mathbf{i} + \left(\int 3 dt\right)\mathbf{j} - \left(\int 3t^2 dt\right)\mathbf{k} ∫ ( 2 t i + 3 j − 3 t 2 k ) d t = ( ∫ 2 t d t ) i + ( ∫ 3 d t ) j − ( ∫ 3 t 2 d t ) k
= t 2 i + 3 t j − t 3 k + C = t^2\mathbf{i} + 3t\mathbf{j} - t^3\mathbf{k} + \mathbf{C} = t 2 i + 3 t j − t 3 k + C
Example 2
Find the anti-derivative of v ′ ( t ) = cos 2 t i − 2 sin t j + 1 1 + t 2 k \mathbf{v}'(t) = \cos 2t\mathbf{i} - 2\sin t\mathbf{j} + \frac{1}{1+t^2}\mathbf{k} v ′ ( t ) = cos 2 t i − 2 sin t j + 1 + t 2 1 k v ( 0 ) = 3 i − 2 j + k \mathbf{v}(0) = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} v ( 0 ) = 3 i − 2 j + k
Solution:
v ( t ) = ∫ v ′ ( t ) d t = ( ∫ cos 2 t d t ) i − ( ∫ 2 sin t d t ) j + ( ∫ 1 1 + t 2 d t ) k \mathbf{v}(t) = \int \mathbf{v}'(t) dt = \left(\int \cos 2t dt\right)\mathbf{i} - \left(\int 2\sin t dt\right)\mathbf{j} + \left(\int \frac{1}{1+t^2} dt\right)\mathbf{k} v ( t ) = ∫ v ′ ( t ) d t = ( ∫ cos 2 t d t ) i − ( ∫ 2 sin t d t ) j + ( ∫ 1 + t 2 1 d t ) k
= 1 2 sin 2 t i + 2 cos t j + arctan t k + C = \frac{1}{2}\sin 2t\mathbf{i} + 2\cos t\mathbf{j} + \arctan t\mathbf{k} + \mathbf{C} = 2 1 sin 2 t i + 2 cos t j + arctan t k + C
Using the initial condition v ( 0 ) = 3 i − 2 j + k \mathbf{v}(0) = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} v ( 0 ) = 3 i − 2 j + k
3 i − 2 j + k = 1 2 sin 0 i + 2 cos 0 j + arctan 0 k + C 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} = \frac{1}{2}\sin 0\mathbf{i} + 2\cos 0\mathbf{j} + \arctan 0\mathbf{k} + \mathbf{C} 3 i − 2 j + k = 2 1 sin 0 i + 2 cos 0 j + arctan 0 k + C
3 i − 2 j + k = 2 j + C 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} = 2\mathbf{j} + \mathbf{C} 3 i − 2 j + k = 2 j + C
C = 3 i − 4 j + k \mathbf{C} = 3\mathbf{i} - 4\mathbf{j} + \mathbf{k} C = 3 i − 4 j + k
Therefore, v ( t ) = ( 1 2 sin 2 t + 3 ) i + ( 2 cos t − 4 ) j + ( arctan t + 1 ) k \mathbf{v}(t) = \left(\frac{1}{2}\sin 2t + 3\right)\mathbf{i} + (2\cos t - 4)\mathbf{j} + (\arctan t + 1)\mathbf{k} v ( t ) = ( 2 1 sin 2 t + 3 ) i + ( 2 cos t − 4 ) j + ( arctan t + 1 ) k
Example 3
Evaluate ∫ 1 2 ( t 2 i + 3 t j + 1 t 4 k ) d t \int_1^2 (t^2\mathbf{i} + 3t\mathbf{j} + \frac{1}{t^4}\mathbf{k}) dt ∫ 1 2 ( t 2 i + 3 t j + t 4 1 k ) d t
Solution:
∫ 1 2 ( t 2 i + 3 t j + t − 4 k ) d t = [ t 3 3 i + 3 t 2 2 j − 1 3 t 3 k ] 1 2 \int_1^2 (t^2\mathbf{i} + 3t\mathbf{j} + t^{-4}\mathbf{k}) dt = \left[\frac{t^3}{3}\mathbf{i} + \frac{3t^2}{2}\mathbf{j} - \frac{1}{3t^3}\mathbf{k}\right]_1^2 ∫ 1 2 ( t 2 i + 3 t j + t − 4 k ) d t = [ 3 t 3 i + 2 3 t 2 j − 3 t 3 1 k ] 1 2
= ( 8 3 i + 6 j − 1 24 k ) − ( 1 3 i + 3 2 j − 1 3 k ) = 7 3 i + 9 2 j + 7 24 k = \left(\frac{8}{3}\mathbf{i} + 6\mathbf{j} - \frac{1}{24}\mathbf{k}\right) - \left(\frac{1}{3}\mathbf{i} + \frac{3}{2}\mathbf{j} - \frac{1}{3}\mathbf{k}\right) = \frac{7}{3}\mathbf{i} + \frac{9}{2}\mathbf{j} + \frac{7}{24}\mathbf{k} = ( 3 8 i + 6 j − 24 1 k ) − ( 3 1 i + 2 3 j − 3 1 k ) = 3 7 i + 2 9 j + 24 7 k
Example 4
An engineer found that the velocity v \mathbf{v} v t t t v ( t ) = 2 t 2 i + 6 e − 4 t j + 8 cos 4 t k \mathbf{v}(t) = 2t^2\mathbf{i} + 6e^{-4t}\mathbf{j} + 8\cos 4t\mathbf{k} v ( t ) = 2 t 2 i + 6 e − 4 t j + 8 cos 4 t k t t t
Solution:
r ( t ) = ∫ v ( t ) d t = ( ∫ 2 t 2 d t ) i + ( ∫ 6 e − 4 t d t ) j + ( ∫ 8 cos 4 t d t ) k \mathbf{r}(t) = \int \mathbf{v}(t) dt = \left(\int 2t^2 dt\right)\mathbf{i} + \left(\int 6e^{-4t} dt\right)\mathbf{j} + \left(\int 8\cos 4t dt\right)\mathbf{k} r ( t ) = ∫ v ( t ) d t = ( ∫ 2 t 2 d t ) i + ( ∫ 6 e − 4 t d t ) j + ( ∫ 8 cos 4 t d t ) k
= 2 3 t 3 i − 3 2 e − 4 t j + 2 sin 4 t k + C = \frac{2}{3}t^3\mathbf{i} - \frac{3}{2}e^{-4t}\mathbf{j} + 2\sin 4t\mathbf{k} + \mathbf{C} = 3 2 t 3 i − 2 3 e − 4 t j + 2 sin 4 t k + C
Example 5
A particle travels so that its acceleration is a ( t ) = 2 sin 2 t i − 3 cos t j + k \mathbf{a}(t) = 2\sin 2t\mathbf{i} - 3\cos t\mathbf{j} + \mathbf{k} a ( t ) = 2 sin 2 t i − 3 cos t j + k t = 0 t = 0 t = 0 ( 2 , 1 , 0 ) (2, 1, 0) ( 2 , 1 , 0 ) 4 i + 2 j 4\mathbf{i} + 2\mathbf{j} 4 i + 2 j
(a) The velocity of the particle at any time t t t
(b) The displacement of the particle at any time t t t
Solution:
(a) v ( t ) = ∫ a ( t ) d t = ( ∫ 2 sin 2 t d t ) i − ( ∫ 3 cos t d t ) j + ( ∫ d t ) k \mathbf{v}(t) = \int \mathbf{a}(t) dt = \left(\int 2\sin 2t dt\right)\mathbf{i} - \left(\int 3\cos t dt\right)\mathbf{j} + \left(\int dt\right)\mathbf{k} v ( t ) = ∫ a ( t ) d t = ( ∫ 2 sin 2 t d t ) i − ( ∫ 3 cos t d t ) j + ( ∫ d t ) k
= − cos 2 t i − 3 sin t j + t k + C 1 = -\cos 2t\mathbf{i} - 3\sin t\mathbf{j} + t\mathbf{k} + \mathbf{C}_1 = − cos 2 t i − 3 sin t j + t k + C 1
At t = 0 t=0 t = 0 v ( 0 ) = 4 i + 2 j \mathbf{v}(0) = 4\mathbf{i} + 2\mathbf{j} v ( 0 ) = 4 i + 2 j 4 i + 2 j = − i + C 1 4\mathbf{i} + 2\mathbf{j} = -\mathbf{i} + \mathbf{C}_1 4 i + 2 j = − i + C 1 C 1 = 5 i + 2 j \mathbf{C}_1 = 5\mathbf{i} + 2\mathbf{j} C 1 = 5 i + 2 j
v ( t ) = ( 5 − cos 2 t ) i + ( 2 − 3 sin t ) j + t k \mathbf{v}(t) = (5 - \cos 2t)\mathbf{i} + (2 - 3\sin t)\mathbf{j} + t\mathbf{k} v ( t ) = ( 5 − cos 2 t ) i + ( 2 − 3 sin t ) j + t k
(b) r ( t ) = ∫ v ( t ) d t = ( ∫ ( 5 − cos 2 t ) d t ) i + ( ∫ ( 2 − 3 sin t ) d t ) j + ( ∫ t d t ) k \mathbf{r}(t) = \int \mathbf{v}(t) dt = \left(\int (5 - \cos 2t) dt\right)\mathbf{i} + \left(\int (2 - 3\sin t) dt\right)\mathbf{j} + \left(\int t dt\right)\mathbf{k} r ( t ) = ∫ v ( t ) d t = ( ∫ ( 5 − cos 2 t ) d t ) i + ( ∫ ( 2 − 3 sin t ) d t ) j + ( ∫ t d t ) k
= ( 5 t − 1 2 sin 2 t ) i + ( 2 t + 3 cos t ) j + 1 2 t 2 k + C 2 = (5t - \frac{1}{2}\sin 2t)\mathbf{i} + (2t + 3\cos t)\mathbf{j} + \frac{1}{2}t^2\mathbf{k} + \mathbf{C}_2 = ( 5 t − 2 1 sin 2 t ) i + ( 2 t + 3 cos t ) j + 2 1 t 2 k + C 2
At t = 0 t=0 t = 0 r ( 0 ) = 2 i + j \mathbf{r}(0) = 2\mathbf{i} + \mathbf{j} r ( 0 ) = 2 i + j 2 i + j = 3 j + C 2 2\mathbf{i} + \mathbf{j} = 3\mathbf{j} + \mathbf{C}_2 2 i + j = 3 j + C 2 C 2 = 2 i − 2 j \mathbf{C}_2 = 2\mathbf{i} - 2\mathbf{j} C 2 = 2 i − 2 j
r ( t ) = ( 5 t − 1 2 sin 2 t + 2 ) i + ( 2 t + 3 cos t − 2 ) j + 1 2 t 2 k \mathbf{r}(t) = (5t - \frac{1}{2}\sin 2t + 2)\mathbf{i} + (2t + 3\cos t - 2)\mathbf{j} + \frac{1}{2}t^2\mathbf{k} r ( t ) = ( 5 t − 2 1 sin 2 t + 2 ) i + ( 2 t + 3 cos t − 2 ) j + 2 1 t 2 k