Whats is the meaning of the derivative of a vector function?

$\begingroup$

Assuming we have a continuous and well behaved vector function $$ R(t) = \langle f(t), g(t), h(t) \rangle $$ then its derivative at an arbitrary point a is $R'(a)$, which can be computed (I'm not using the definition not to clutter the post) as $$ R'(x) = \langle f'(a), g'(a), h'(a)\rangle $$ This gives the vector tangent to the vector function curve, at point a.

But what does the function defined by R'(t) describes? I have tried to think about it and search it on the books that I have, but I have found no answer.

$\endgroup$

3 Answers

$\begingroup$

A physical perspective: if $R(t)$ refers to the coordinates of an object in $3$-dimensional space at time $t$ (so that the curve is a trajectory through space), then $R'(t)$ gives the velocity vector of the object.

In particular, the direction of $R'(t)$ is the instantaneous direction of travel at time $t$, and the length of $R'(t)$ is the speed of travel at time $t$.

$\endgroup$ 3 $\begingroup$

The derivative of a function $f:\mathbb{R}^n\mapsto\mathbb{R}^m$ at a point $x_0$ is the unique linear function $A$ that satisfies $$ f(x)=f(x_0)+A(x-x_0)+o(|x_0-x|), $$ where the little o means that there is some correction term that is small in terms of $|x_0-x|$. Put in words, $A$ is the best affine approximation to $f$ at the point $x_0$.

$\endgroup$ 1 $\begingroup$

The derivative of a vector function is defined as, “the measure of the change of the vector function value (output value) per unit change in its argument value (input value) when change in argument value approaches to zero”. e.g If r is position vector of a particle which changes with time, then its derivative w.r.t to time is (dr(t))/dt and is given by (dr(t))/dt= lim┬(Δt→0)⁡〖(Δr(t))/Δt〗
= lim┬(Δt→0)⁡〖(r(t+Δt)-r(t))/Δt〗 Derivatives are fundamental tools of calculus. For example, the derivative of the position vector of a moving object with respect to time is the object's velocity: this measures how quickly the position of the object changes when time advances. i.e V = (dr(t))/dt

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like