Derivative Of A Vector Function Vector With Constant Magnitude
Theorem:1 The necessary and sufficient condition that f(t) is a vector of constant magnitude is f · df/dt= 0.
Proof : (1) The condition is necessary
Let f(t) be a vector of constant magnitude. Then f(t). f(t) = |f(t)|2 = const Differentiating w.r. to t we get
⇒ \(f \cdot \frac{d f}{d t}+\frac{d f}{d t} \cdot f=0 \Rightarrow f \cdot \frac{d f}{d t}=0\)
(2) Condition is sufficient \(\text { Let } f \cdot \frac{d f}{d t}=0 \text { then } \frac{1}{2} \frac{d}{d t}(f . f)=0 \Rightarrow \text { f.f }=\text { const. }\)
Continuous Magnitude Vector Function Properties
Derivative of a Vector Function Vectors With Direction Constant
Theorem: The necessary and sufficient condition for f(t) to have constant direction is \(\mathrm{f} \times \frac{\mathrm{df}}{\mathrm{dt}}=0 \text { Proof : Let } \mathrm{f}(t)=f(t) \mathbf{F}(t) \)
Where f(t) denotes the magnitude of f(t) and F(t), is a vector function with unit magnitude for every value of t :
Now, \(\mathbf{f}=f \mathbf{F} \quad \Rightarrow \quad \frac{d \mathbf{f}}{d t}=f \frac{d \mathbf{F}}{d t}+\frac{d f}{d t} \mathbf{F} \)
∴ \(\mathbf{f} \times \frac{d \mathbf{F}}{d t}=f \mathbf{F} \times\left(f \frac{d \mathbf{f t}}{d t}+\frac{d f}{d t} \mathbf{F}\right)=f^2 \mathbf{F} \times \frac{d \mathbf{F}}{d t}\)
∵ F × F = 0
(1) The condition is necessary
Let the direction of f be constant. Then F = const ⇒ \(\frac{d F}{d t}=0 \Rightarrow f \times \frac{d f}{d t}=0\)
(2) The condition is sufficient
Let \(\mathrm{f} \times \frac{d \mathrm{f}}{d t}=0 \text { i.e. } f^2 \mathrm{~F} \times \frac{d \mathrm{~F}}{d t}=0 \Rightarrow \mathbf{F} \times \frac{d \mathrm{~F}}{d t}=0 \quad[because f \neq 0]\)
Also F being of constant magnitude \(\mathrm{F} \cdot \frac{d \mathrm{~F}}{d t}=0\)
Hence from (1) and (2) we have \(\mathrm{F} \cdot \frac{d \mathrm{~F}}{d t}=0\) ⇒ F = constant i.e. the direction of f remains the same.
Derivative of a Vector Function Composite Vector Function
Theorem: Let s be a scalar function defined over the domain S and differentiable at t ∈ S. If f is a vector function differentiable at s(t) in the range of functions then the composite function f(s) is differentiable at t and \(f[s(t)]^{\prime}=\mathrm{f}^{\prime}[s(t)] s^{\prime}(t) \text { or } \frac{d f}{d t}=\frac{d f}{d x} \frac{d s}{d t}\)
Proof: Since f(s) and s(t) are differentiable, we have \(\frac{d \Delta}{d t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\delta s}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\Delta(t+\delta t)-s(t)}{\delta t} \text { and } \frac{d f}{d s}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\delta \mathrm{f}}{\delta s}=\mathrm{L}_{\delta t \rightarrow 0} \frac{\mathrm{r}(t+\delta t)-\mathrm{f}(t)}{\delta t}\)
Now \({Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}[s(t+\delta t)]-\mathrm{f}[s(t)]}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}[s(t+\delta t)]-\mathrm{f}[a(t)]}{\lambda(t+\delta t)-s(t)} \cdot \frac{\Delta(t+\delta t)-\Delta(t)}{\delta t} .\)
= \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}(s+\delta t)-\mathrm{f}(s)}{\delta t}\)
⇒ \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{s(t+\delta t)-s(t)}{d t}=\frac{d r}{d s} \cdot \frac{d a}{d t}\)
Thus f[s(t)] is differentiable at t and \(\frac{d f}{d t}=\frac{d f}{d s}-\frac{d s}{d t}\)
Understanding Continuous Magnitude In Vector Functions
Derivative of a Vector Function Velocity And Acceleration Of A Particle
If r is the position vector of a point P in space and if r(t) = x(t)i+ y(t)j +z(t)k is a function of time t, then \(\frac{d \mathbf{r}}{d t}\) will represent the velocity of the point P at any time and it is directed by v.
v = \(\frac{d \mathbf{r}}{d t}\) = velocity of P at any time
Similarly,\( \frac{d v}{d t}=\frac{d^2 r}{d t^2}\) represents the acceleration of the particle at any time and is denoted by a
a = \( \frac{d v}{d t}=\frac{d^2 r}{d t^2}\) = accelaration of P at any time.