An object travels the path given by the parametric equations x=ln2t + t^3 and y=(e^t) - 2 sint. At t=2.1 sec, what is the acceleration of the particle?

Answer 1

x=ln2t + t^3
dx/dt = 1/t + 3t^2

y=(e^t) - 2 sint
dy/dt = e^t - 2cost
v = dy/dx = (dy/dt) / (dx/dt) = (e^t - 2cost)/((1/t + 3t^2)
a = dv/dt = [(1/t + 3t^2)(e^t + 2sint) - (e^t - 2cost)(-1/t^2 + 6t)]/(1/t + 3t^2)^2
phewww
now let t = 2.1, I am sure you have a calculator, make sure you are set to radians

Answer 2

v = dy/dx = dy/dt / dx/dt = (e^t-2cost)/(1/t + 3t^2)
a = dv/dx = dv/dt / dx/dt
= [(e^t+2sint)(1/t + 3t^2) - (6t - 1/2)(e^t-2cost)]/(1/t + 3t^2)^3

This is just a stupid exercise -- needlessly complicated.