# Thread: derivatives of vector functions

1. ## derivatives of vector functions

first bit done stuck on second part:

The equations of motion of a point particle in space are given by

1. x(t) = 2t^3 - 3, y(t)= -3t^3, z(t)= 4t^3 - 1

2. x(t)= asin(wt), y(t) = acos(wt), z(t) = t

In each case, calculate the velocity vector v and the acceleration vector a. Also, calculate the absolute values of these vectors.

Now the bit I can't do is,

what are the curves along which the particle moves?
(find these values for an arbitraryt, with a and w being fixed parameter)

2. Originally Posted by sonia1
first bit done stuck on second part:

The equations of motion of a point particle in space are given by

1. x(t) = 2t^3 - 3, y(t)= -3t^3, z(t)= 4t^3 - 1

2. x(t)= asin(wt), y(t) = acos(wt), z(t) = t

In each case, calculate the velocity vector v and the acceleration vector a. Also, calculate the absolute values of these vectors.

Now the bit I can't do is,

what are the curves along which the particle moves?
(find these values for an arbitraryt, with a and w being fixed parameter)
$x = 2t^3 - 3 \Rightarrow t^3 = \frac{x+3}{2}$ .... (1)

$y = -3t^3 \Rightarrow t^3 = - \frac{y}{3}$ .... (2)

$z = 4t^3 - 1 \Rightarrow t^3 = \frac{z+1}{4}$ .... (3)

From (1), (2) and (3) it follows that the path is a line with cartesian equation $\frac{x+3}{2} = - \frac{y}{3} = \frac{z+1}{4}$

$x = a \sin (\omega t) \Rightarrow x^2 = a^2 \sin^2 (\omega t)$ .... (1)

$y = a \cos (\omega t) \Rightarrow y^2 = a^2 \cos^2 (\omega t)$ .... (2)

(1) + (2): $x^2 + y^2 = a^2$.

You have a spiral with a circular cross-section centred along the z-axis.

3. could you please show me a sketch of what it actually looks like