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Another Calc. 3 problem!!!?
Consider the circular motion r(t) = <3 cos(5t), 3 sin(5t)>. Find its ve-
locity and its acceleration. How do they compare with the position
function at any point in time (i.e. are they parallel, orthogonal or
some other type of special relative position)?
1 Answer
- 9 years ago
To take the derivative of a one-dimensional vector function you take the derivative of the vector's components. This gives r'(t) = v(t) = d/dt <3 cos(5t), 3 sin(5t)> = <d/dt 3 cos(5t), d/dt 3 sin(5t)> = <-15 sin(5t), 15 cos(5t)> and r"(t) = v'(t) = a(t) = d/dt <-15 sin(5t), 15 cos(5t)> = <d/dt -15 sin(5t), d/dt 15 cos(5t)> = <-75 cos(5t), -75 sin(5t)>.
To find if two vectors are perpendicular (or orthogonal), take their dot product. You will find that the position vector is perpendicular (or orthogonal) to the velocity vector.
To find if two vectors are parallel, simply look and see if one is a scalar multiple of the other. You will find that the position vector is parallel to the acceleration vector.
