Given the position of the particle, find when is its speed minimized.

Question

If the position of a particle is given by

$$r(t) = \left\langle t^2, t^2-2t, 1-t^2 \right\rangle$$

when is its speed minimized?

Answer 1

As Lord said, work out the speed formula. Here is a start: $$s=\sqrt{([t^2]')^2+([2t-2]')^2+[1-t^2]')^2}.$$ (apostrophe notation means "derivative") This will result in a quadratic polynomial inside the square root. The vertex of this quadratic polynomial is what you need. Now for the sake of learning, give it a try and share your findings with us. We can tell you if you did it right...

Answer 2

Hint:

The velocity vector is $$ \vec v(t)=\frac{d}{dt} {\vec r} (t)=(2t,2t-2,-2t) $$

minimize the modulus of this vector.

Answer 3

$r'(t)=<2t,2t-2,-2t>$

Consider $f(t)=||r'(t)||^2=12 t^2-8 t+4$

$f'(t)=24t-8$ and $f'(t)=0$ for $t=\dfrac{1}{3}$s