Derivative Of A Vector Function Vector Function Of A Scalar Variable
Let S ⊂ R, Corresponding to each scalar t ∈ S, there is associated a unique vector r; then r is said to be a vector (vector-valued) function. S is called the domain of r. We express it as r = f(t) where f denotes the law of correspondence.
Let i, j, k be the three mutually perpendicular unit vectors in three-dimensional space. Then the vector function f(t) may be expressed in the form r = f(t) = f1(t)i + f2(t)j + f3(t)k these f1(t), f2(t), and f3 (3) are the real-valued functions and are called the components of r.
Derivative Of A Vector Function Interval
Interval is the subset of R which can be expressed as shown below.
(a, b) = {x/x ∈ R, a < x < b} (a, b] = {x/x ∈ R, a < x ≤ b}
[a, b) = {x/x ∈ R, a ≤ x < b} (a, b) = {x/x ∈ R, a ≤ x ≤ b}
[a, ∞) = {x/x ∈ R, x ≥ a} (a, ∞) = {x/x ∈ R, x > a}
(−∞, a] = {x/x ∈ R, x ≤ a} (−∞, ∞) = {x/x ∈ R}
Derivative Of A Vector FunctionLimit Of A Vector Function
Let f(t) be a vector function over the domain S and a ∈ S. If there exists a vector L such that for each ε > 0, it is possible to find δ > 0 where 0 < | t − a| < δ ⇒ | f(t) − L| < ε then the vector L is called the limit of f(t) as t tends to a.
This is denoted as Ltt →a f(t) = L.
Note. Let Lt t →a A(t) = L and Ltt →a B(t) = M and λ being any constant then (a) Lt t →a [A(t)+ B(t)] = L+ M (b) Ltt →a[λA(t)] = λL (c) Ltt →0[A(t) · B(t)] = LM (d) Ltt →a[A(t) × B(t)] = L × M
Derivative Of A Vector Function Continuity Of Vector Function
Let f be a vector function on an interval I, and a ∈ I. Then f is said to be continuous at a, if Lt t →a
f(t) = f(a)
The function f is said to be continuous on I if f is continuous at each point of
1. Note. If f and g are continuous then f ± g, f · g and f × g are also continuous.
Derivative Of A Vector Function Derivative
Let f be a vector function on an interval I and a ∈ I. Then \(operatorname{Lt}_{t \rightarrow a} \frac{f(t)-f(a)}{t-a}\) if it exists, it is called the derivative of f at ’ a ’ and is denoted by\( f^{\prime}(a) \text { or }\left(\frac{d r}{d t}\right)_{t=a}\)
Also, it is said that f is differentiable at t = a.
Note 1. If f is differentiable at t = a, then it is continuous at t = a.
Note 2. If f is continuous at t = a, then it need not be differentiable at that point.
Note 3. If f is differentiable on an interval I and t ∈ I then the derivative of f at t is denoted by dtf
Note 4. As in calculus of real variables, if the changes in t, f are denoted by δt and δ respectively then we have \(\frac{d I}{d t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\delta f}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}(t+\delta t)-\mathrm{f}(t)}{\delta t}\)
Derivative Of A Vector Function Higher Order Derivatives
Let f be differentiable on an interval I and f0 = df /dt be the derivative of f. If \(L_{t \rightarrow a} \frac{\mathrm{r}^{\prime}(t)-\mathrm{r}^{\prime}(a)}{\mathrm{t}-a}\)exists for each t ∈ I1 ⊂ I, then f’ is said to be differentiable on I1.
Also, f is said to possess second derivative on I1and is denoted by f‘(t) or\( \frac{d^2 d^2}{x^2}\) By induction if f(n−1) is differentiable on In-1⊂ In-2⊂ . . . ⊂ In ⊂ I then f is said to possess nth derivative on In-1 and is denoted by \(f^{(n)}(t) \text { or } \frac{d^n r}{d t^n}\)
Derivative Of A Vector Function Derivative Of Constant Vector
Theorem: Let f be a constant vector function in the interval I and a ∈ I. Then f'(a) = 0
Proof: Let f(t) = c where c is a constant vector
⇒ \(\mathrm{Lt}_{t \rightarrow a} \frac{\mathbf{f}(t)-\mathbf{f}(a)}{t-a}=operatorname{Lt}_{t \rightarrow a} \frac{\mathbf{c}-\mathbf{c}}{t-a}=\mathrm{Lt}_{t \rightarrow a} 0-0 \Rightarrow\left(\frac{d \mathbf{f}}{d t}\right)_{t=a}=\mathrm{f}^{\prime}(a)=0\)
It can be proved that f’(a) = 0 for every a ∈ I.
Theorem: Let A and B be two differentiable vector functions of scalar variable t over the domain S then \( \frac{d}{d t}(A \pm B)=\frac{d A}{d t} \pm \frac{d B}{d t}\)
Proof : Let f = \(\mathbf{A} \pm \mathbf{B}, then \mathbf{L t}_{\delta \mathrm{d} \rightarrow 0} \frac{\mathrm{f}(t+\delta t)-\mathrm{f}(t)}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow a} \frac{\mathbf{A}(t+\delta t)-\mathbf{A}(t)}{\delta t} \pm\)
operator name \({Lt}_{\delta t \rightarrow a} \frac{B(t+\delta t)-B(t)}{\delta t}\)
⇒ \(\frac{d \mathbf{A}}{d t} \pm \frac{d \mathbf{B}}{d t} \) ( ∵ A, B are differentiable at t)
∴ f is differentiable at t and ∴ \( \frac{d}{d t}(\mathbf{A} \pm \mathbf{B})=\frac{d \mathbf{A}}{d t} \pm \frac{d \mathbf{B}}{d t}\)
Derivative Of A Vector Function Theorem 1
Let A and B be differentiable vector functions of a scalar variable to ver the domain S, then (1)
⇒ \(\left.\frac{d}{d t}(\mathrm{~A} B)=\frac{d \mathrm{~A}}{d t} \cdot \mathbf{B}+\mathbf{A} \cdot \frac{d B}{d t}(2) \frac{d}{d d}(\mathrm{~A} \times \mathrm{B}) \mathrm{B}\right)=\frac{d A}{d t} \times B+A \times \frac{d B}{d t}\)
Proof (1): ∵ A and B differentiable
⇒ \(\frac{d \mathbf{A}}{d t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathbf{A}(t+\delta t)-\mathbf{A}(t)}{\delta t} \text { and } \frac{d \mathbf{B}}{d t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathbf{B}(t+\delta t)-\mathbf{B}(t)}{\delta t}\)
Let f = A · B
⇒ \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathbf{f}(t+\delta t)-\mathbf{f}(t)}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathbf{A}(t+\delta t) \cdot \mathbf{B}(t+\delta t)-\mathbf{A}(t) \cdot \mathbf{B}(t)}{\delta t}\)
⇒ \(=L_t t_{\delta t-0} \frac{\mathbf{A}(t+\delta t) \cdot[\mathbf{B}(t+\delta t)-\mathbf{B}(t)]}{\delta t}+{ }_{\delta t \rightarrow 0} \frac{[\mathbf{A}(t+\delta t)-\mathbf{A}(t)] \cdot \mathbf{B}(t)}{\delta t} =\mathbf{A} \cdot \frac{d \mathbf{B}}{d t}+\frac{d \mathbf{A}}{d t} \cdot \mathbf{B}\)
(∵A, B are differentiable and continuous at t ).
∴ f is differentiable at t and\( \frac{d}{d t}(\mathrm{~A} \cdot \mathrm{B})=\mathrm{A} \cdot \frac{d \mathrm{~B}}{d t}+\frac{d \mathrm{~A}}{d t} \cdot \mathbf{B}\)
Proof (2): Let g = A × B
⇒ \(therefore \mathrm{Lt}_{d t \rightarrow 0} \frac{\mathrm{g}(t+d t)-\mathrm{g}(t)}{d t}=\frac{\mathrm{A}(t+d t) \times \mathrm{B}(t+\delta t)-\mathrm{A}(t) \cdot \mathrm{B}(t)}{\delta t}\)
⇒ operator name \({Lt}_{\delta t \rightarrow 0} \frac{A(t+\delta t) \times[B(t+\delta t)-B(t)]}{\delta t}\)
⇒ \(+operatorname{Lt}_{\delta t \rightarrow 0} \frac{[A(t+\delta t)-A(t)] \times B(t)}{\delta t}=\mathbf{A} \times \frac{d B}{d t}+\frac{d \mathbf{A}}{d t} \times \mathbf{B}\)
( ∵ A, B are differentiable and continuous at t)
∴ f is differentiable at t and\( \frac{d}{d t}(\mathrm{~A} \times \mathrm{B})=\mathbf{A} \times \frac{d \mathrm{~B}}{d t}+\frac{d \mathbf{A}}{d t} \times \mathbf{B} \)
Derivative Of A Vector Function Theorem 2
Let A, B, and C be three differentiable vector functions of scalar variable t over a domain S. Then
⇒ \(\text { (1) } \frac{d}{d t}(A B C)=\left[\frac{d A}{d t} B C\right]+\left[A \frac{d B}{d t} C\right]+\left[A B \frac{d C}{d t}\right] \)
⇒ \(\text { (2) } \frac{d}{d t}\{A \times(B \times C)\}=\frac{d A}{d t} \times(B \times C)+A \times\left(\frac{d B}{d t} \times C\right)+A \times\left(B \times \frac{d C}{d t}\right)\)
Proof : (1) \(\frac{d}{d t}[\mathbf{A B C}]=\frac{d}{d t}[\mathbf{A} \cdot(\mathbf{B} \times \mathbf{C})]=\frac{d \mathbf{A}}{d t} \cdot(\mathbf{B} \times \mathbf{C})+\mathbf{A} \cdot \frac{d}{d t}(\mathbf{B} \times \mathbf{C}) \)
⇒ \(\left[\frac{d \mathrm{~A}}{d t} \mathrm{BC}\right]+\mathbf{A} \cdot\left(\frac{d \mathrm{~B}}{d t} \times \mathbf{C}+\mathbf{B} \times \frac{d \mathbf{C}}{d t}\right)=\left[\frac{d \mathrm{~A}}{d t} \mathbf{B C}\right]+\mathbf{A} \cdot\left(\frac{d \mathrm{~B}}{d t} \times \mathbf{C}\right)+\mathbf{A} \cdot\left(\mathbf{B} \times \frac{d \mathbf{C}}{d t}\right) \)
⇒ \(\left[\frac{d \mathrm{~A}}{d t} \mathbf{B C}\right]+\left[\mathbf{A} \frac{d \mathbf{B}}{d t} \mathbf{C}\right]+\left[\mathbf{A B} \frac{d \mathbf{C}}{d t}\right] \)
⇒ \( \text { (2) } \frac{d}{d t}\{\mathbf{A} \times(\mathbf{B} \times \mathbf{C})\}=\frac{d \mathbf{A}}{d t} \times(\mathbf{B} \times \mathbf{C})+\mathbf{A} \times \frac{d}{d t}(\mathbf{B} \times \mathbf{C})\)
⇒ \( =\frac{d \mathbf{A}}{d t} \times(\mathbf{B} \times \mathbf{C})+\mathbf{A} \times\left(\frac{d \mathbf{B}}{d t} \times \mathbf{C}+\mathbf{B} \times \frac{d \mathbf{C}}{d t}\right)\)
⇒ \(\frac{d \mathbf{A}}{d t} \times(\mathbf{B} \times \mathbf{C})+\mathbf{A} \times\left(\frac{d \mathbf{B}}{d t} \times \mathbf{C}\right)+\mathbf{A} \times\left(\mathbf{B} \times \frac{d \mathbf{C}}{d t}\right)\)
Derivative Of A Vector Function Theorem 3
Let f be a differentiable vector function and φ a scalar differentiable function on a common domain S. Then φf is differentiable on S and\( \frac{d}{d t}(\phi f)=\phi \frac{d f}{d t}+\frac{d \phi}{d t} f\)
Proof: \({Lt}_{\delta t \rightarrow 0} \frac{\phi(t+\delta t) \mathrm{P}(t+\delta t)-\phi(t) \mathrm{r}(t)}{\delta t}\)
⇒ \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\phi(t+\delta t)\{\mathbf{f}(t+\delta t)-\mathbf{f}(t)\}}{\delta t}+\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\{\phi(t+\delta t)-\phi(t)\} \mathbf{f}(t)}{\delta t} \)
⇒ \(\phi(t) \frac{d}{d t}[\mathbf{f}(t)]+\frac{d}{d t}[\phi(t)] \mathbf{f}(t)\)
∴ φf is differentiable at t and \( \frac{d}{d t}(\phi \mathrm{f})=\phi \frac{d \mathrm{f}}{d t}+\frac{d \phi}{d t} \mathrm{f}\)
Note. If f is a constant vector then \(\frac{d f}{d t}=0 \Rightarrow \frac{d}{d t}(\phi \mathrm{f})=0+\frac{d \phi}{d t} \mathrm{f}=\frac{d \phi}{d t} \mathrm{f} \)
Derivative Of A Vector Function Theorem 4
If f = f1(t)i + f2(t)j + f3(t)k, where f1(t), f2(t) and f3(t) are the cartesian components of the vector f, then \(\frac{d f}{d t}=\frac{d f}{d t} i+\frac{d f_2}{d t} j+\frac{d f_a}{d t} k \)
Proof : Given f = f1(t)i + f2(t)j + f3(t)k
⇒ \(\frac{d \mathbf{f}}{d t}=\frac{d}{d t}\left(f_1 \mathbf{i}+f_2 \mathbf{j}+f_3 \mathbf{k}\right)=\frac{d}{d t}\left(f_1 \mathbf{i}\right)+\frac{d}{d t}\left(f_2 \mathbf{j}\right)+\frac{d}{d t}\left(f_3 \mathbf{k}\right) \quad=\frac{d f_1}{d t} \mathbf{i}+\frac{d f_2}{d t} \mathbf{j}+\frac{d f_3}{d t} \mathbf{k} \)
Derivative Of A Vector Function Definition
If f1, f2, and f3 are constant functions then f = f1 i + f2j + f3k is called a constant vector function.
Derivative Of A Vector Function Theorem 5
The necessary and sufficient condition that f(t) may be a constant vector function is \(\frac{d f}{d t}=0 \text {. }\)
Proof : (1) The condition is necessary i.e. to prove \(\frac{d f}{d t}=0 \text {. }\) if f(t) is a constant vector function. Let f(t) = C
Then \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{r}(t+\delta t)-\mathrm{f}(t)}{\Delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{C}-\mathrm{C}}{s t}=0 \Rightarrow \frac{d \mathrm{ft}}{d t}=0 \)
(2) The condition is sufficient
To prove f(t) is a constant vector if \(\frac{d f}{d t}=0 \text {. }\)
Let f(t) = f1(t)i + f2(t)j + f3(t)k
⇒ Therefore\(\quad \frac{d f^{\prime}}{d t}=0 \Rightarrow \frac{d f_1}{d t} \mathbf{i}+\frac{d f_2}{d t} \mathbf{j}+\frac{d f_1}{d t} \mathbf{k}=0 \Rightarrow \frac{d f_t}{d t}=0, \frac{d f_2}{d t}=0, \frac{d f_2}{d t}=0\)
⇒ f1, f2, and f3 are constant functions ⇒ f(t) is a constant vector function
Derivative Of A Vector Function Theorem 6
If A is a differentiable vector function of a scalar t over a domain S then
⇒ \(\frac{d}{d t}\left(\mathrm{~A}^2\right)=2 \mathrm{~A} \cdot \frac{d \mathrm{~A}}{d t} \)
Proof: We know that A2 = A. A
⇒ Therefore\(\quad \frac{d}{d t}\left(\mathbf{A}^2\right)=\frac{d}{d t}(\mathbf{A} \cdot \mathbf{A})=\mathbf{A} \cdot \frac{d \mathbf{A}}{d t}+\frac{d \mathbf{A}}{d t} \cdot \mathbf{A}=2 \mathbf{A} \cdot \frac{d \mathbf{A}}{d t} \)
Corollary: Let r be a vector function over a domain S and r denote the value |r(t)|, then\( \mathbf{r} \cdot \frac{d r}{d t}=r \frac{d r}{d r}\)
Proof : Since r2 = r · r = r2
⇒ Therefore \(\quad \frac{d}{d t}(\mathbf{r} \cdot \mathbf{r})=\frac{d}{d t}\left(r^2\right) \quad 2 \mathbf{r} \cdot \frac{d r}{d t}=2 r \frac{d r}{d t} \Rightarrow \bar{r} \cdot \frac{d \vec{r}}{d t}=r \frac{d r}{d t} \)