# find speed and angle of particle from position parametric equation

• Nov 22nd 2009, 03:57 PM
superdude
find speed and angle of particle from position parametric equation
The position at time t of a particle P is given by $r(t)=\sqrt{2}ti+\frac{1}{2}t^2j+\ln(t)k$ for t>0. Find the speed and the angle the particle came through the xy-plane.

when I thought about his question I took the derivative r'(t) but why do I need to take the distance |r'(t)| ? Does this have to do with the difference between speed and velocity?

For the second part the work is $\arccos \frac{r'(1) \cdot k}{|r'(1)|\dot |k|}$ I know that the particle crosses the xy plane at t=1 so I get that part, but I'm lost as to how there's a triangle happening.

so the numerator has a dot product where the denominator has regular multiplication
• Nov 29th 2009, 11:07 AM
Media_Man
Speed and Velocity
Yes. You are correct. In fact, the precise definition of the velocity of a particle whose position is given by the vector equation r(t) is another vector r'(t). Speed is defined as the magnitude of velocity, |r'(t)|.

For the second part, the vector r'(1) represents the trajectory of the particle at the instant of its position r(1). Since k is the normal of the plane of interest, the "triangle" of which you speak is really just the shape formed by the two vectors r'(1) and k crossing at an angle.