r = a e^T where T is theta
yes omega = w = dT/dt where w is constant angular velocity
so T, theta = w t and r = a e^wt
well, here is a hard way (do not have text handy)
now x = r cos T = r cos (wt) and y = r sin T = r sin (wt)
x = ae^wt cos (wt) and y = a e^wt sin(wt)
dx/dt = -aw e^wt sin wt +aw e^wt cos wt = aw e^wt (coswt-sinwt)
d^2x/dt^2 =-aw^2e^wt (sinwt+coswt)+aw^2 e^wt (coswt-sinwt)
d^2x/dt^2 = a w^2 e^wt(-2sinwt) = -2 a w^2 e^wt sin wt
dy/dt = a w e^wt cos wt + a w e^wt sin wt = a w e^wt (coswt+sinwt)
d^2y/dt^2 = a w^2 e^wt(-sinwt+coswt) +a w^2e^wt(coswt+sinwt)
d^2y/dt^2 = +2 a w^2 e^wt cos wt (who would have guessed:)
magnitude of acceleration^2 = [2 a w^2 e^wt]^2 [sin^2+cos^2]
so magnitude of acceleration = 2 a w e^w t = 2 r
so we have a position vector
Z= ae^wt cos (wt)i + a e^wt sin(wt) j
and an acceleration vector
Acc = -2 a w^2 e^wt sin wt i + +2 a w^2 e^wt cos wt j
the dot product is the component in the direction from the origin, r direction times the magnitude of r which is a e^wt (so it cancels)
now you can use the magnitude and the r component to get the theta component
A point p describes a curve r= ae^(theta) with a constant angular velocityabout its pole Find the radial and transverse acceleration of point P
Should we take some value for angular velocity and solve this or?
2 answers
so magnitude of acceleration = 2 a w^2 e^w t = 2 w^2 r
whoops, left w^2 out
whoops, left w^2 out