he motion of a set of particles moving along the x-axis is governed by the differential equation; dt/dx t^3-x^3,where x(t) denotes the position at time t of the particle. Show that the acceleration of a particle is given by d^2x/dt^2=3t^2-3t^33x^2 +3x^5 .

Question

experimentx · Answer

'dt/dx t^3-x^3' is not an equation ... it's an expression, can you revise the question??

anonymous · Answer

its dt/dx= t^3-x^3 sorry!!

experimentx · Answer

dx/dt =  1/(t^3-x^3)
$$ \frac{d^2x}{dt^2} = \frac{-1}{(t^3 - x^3)^2} (3t^2 - 3x^2 \frac{dx}{dt})
\
=\frac{-1}{(t^3 - x^3)^2} (3t^2 - 3x^2 (t^3 - x^3))$$
simplify ... simplify

experimentx · Answer

I think you meant dx/dt= t^3-x^3 rather than dt/dx= t^3-x^3

anonymous · Answer

yes I did mean dx/dt! but y is dx/dt= 1/(t^3-X^3)^2

experimentx · Answer

Oh ... just cut that -1/(t^3-x^3)^2 part ... you will have your answer!!

anonymous · Answer

i just dont understand how you get 3t^2-3x^2?

anonymous · Answer

(3t^2−3x^2dxdt) this part

experimentx · Answer

yes ...and put dx/dt = (t^3 - x^3)

anonymous · Answer

ok, but where is this part coming from (3t^2−3x^2dxdt)

anonymous · Answer

sorry just a little confused

experimentx · Answer

differentiation ... 
dt^3/dt  - dx^3/dt

anonymous · Answer

wouldnt that give you 3t^2

experimentx · Answer

yes ... and the other would give your 3x^2 dx/dt

anonymous · Answer

sorry i have been away from calculus for a while, how does dx^3/dt = 3x^2? wouldn't it be equal to 0?

experimentx · Answer

if x is not a function of t, then it would be 0, but since x is a function of t, this is not zero!!

experimentx · Answer

i mean x is x(t) i.e. x depends on t ... if x were independent of t then it would have been zero

anonymous · Answer

oh ok i understand! Thank you!!

experimentx · Answer

yw