Homework Sourselet rtsin2ticos2tjpi22tk be the position vector for a point
let r(t)=sin2ti+cos2tj+(((pi)/2)-2t)k be the position vector for a point P=P(t) moving along a curve.
a) find the velocity v and the acceleration a of the point P at the time t=(5(pi)/6).
b) Find the curvature, the unit tangent vector, the principle normal vector, and the binormal to the curve at the point P where t=(5(pi)/6). Determine the tangential and normal components of the acceleration vector at the point P.
a) find the velocity v and the acceleration a of the point P at the time t=(5(pi)/6).
b) Find the curvature, the unit tangent vector, the principle normal vector, and the binormal to the curve at the point P where t=(5(pi)/6). Determine the tangential and normal components of the acceleration vector at the point P.
Solution
(a) Given r(t)=sin2ti+cos2tj+(((pi)/2)-2t)k
We Have on differentiating wrt t
dr/dt = 2cos(2t) i - 2sin(2t) j -2k = Velocity
At t =5/6
Velocity = 2 * cos(5/3)i - 2*sin(5/3)j -2k
= i + 3 j - 2k
Differentiating v wrt time
Acceleration =dv/dt = -4sin(2t) i -4cos(2t) j
At t = 5/6
Acceleration = 23 i -2k
