Question

For any time t≥0, if the position of a particle in the xy-plane is given by x=t^2+1 and y=ln(2t+3), then the acceleration vector is

Answer 1

isnt it the derivative of the unit tangent vector?

Answer 2

r(t) = (x(t), y(t))

r'(t) = (x'(t), y'(t)) , get the unit form of that and derive again ....

Answer 3

I thought it was a parametrization problem

Answer 4

parametric equations are just a vector from the origin pointing to the x,y point at its current location

Answer 5

if your simply trying to get dy/dx, then get (dy/dt)/(dx/dt) and see if its simpler to play with

Answer 6

so i get x and y and derive it

Answer 7

y=ln(2t+3)
x=t^2+1


y'=2/(2t+3)
x'=2t

dy/dx = (2t^2+3t)^(-1)

Answer 8

the key here is this; what we have here is not dy/dx perse so just deriving again is not going to work out so well. what we actually want for acceleration is:\[\frac{\frac{d}{dt}\frac{dy}{dx}}{dx/dt}\]
if memory serves

Answer 9

okay then so, to find the acceleration vector, V(x), one needs to first derive x and y right

Answer 10

yes

Answer 11

so it would be in this form, (x', y')

Answer 12

yep, thats the tangent vector ... we want to turn that into a unit vector and derive again

Answer 13

thanks i think i can take it from here

Answer 14

good :)