2011-03-20, 01:04 PM
The example in the book uses position vector R = (cos t + t sin t)i + (sin t - t cos t)j.
Then v = dR/dt = (t cos t)i + (t sin t)j, then of course a = dv/dt = (cos t - t sin t)i + (t cos t + sin t)j.
|v| is then the square root of the sum (t cos t)^2 + (t sin t)^2) = square root of t^2(cos^2 t + sin^2 t) = t, using cos^2 t + sin^2 t = 1.
Then it says that a[SIZE="1"]t[/SIZE] is equal to the derivative of of |v|, which is d(t)/dt = 1
I guess this is because the tangential component is the one that has an effect on velocity so it would be the second derivative of position?
Then v = dR/dt = (t cos t)i + (t sin t)j, then of course a = dv/dt = (cos t - t sin t)i + (t cos t + sin t)j.
|v| is then the square root of the sum (t cos t)^2 + (t sin t)^2) = square root of t^2(cos^2 t + sin^2 t) = t, using cos^2 t + sin^2 t = 1.
Then it says that a[SIZE="1"]t[/SIZE] is equal to the derivative of of |v|, which is d(t)/dt = 1
I guess this is because the tangential component is the one that has an effect on velocity so it would be the second derivative of position?