The second derivative question

[disfused_3289]

Homework Statement

Given (d^2)x/ (dt^2) = 2 for all t>= 0 and dx/dt= -5 and x= 4 when t= 0 find t when

a) dx/ dy= 0
b) x= 0

2. The attempt at a solution

a) Given (d^2)x/ (dt^2) = 2, then
dx/dt= 2x + c = -5

given dx/ dt= 0 then 2x + c= 0

c= -2.5 using 2x + c= 0 and 2x + c = -5

using dx/dt= 2x + c, t= x^2 + cx + d
where 0= 16 - 2.5 x 4 + d (using x= 4 when t= 0 )
where d= -6

Thus,

t= x^2 - 2.5x - 6

when dx/dt= 0 then x= 1.25 in 2x + c= 0 (c= -2.5)

subsituting x= 1.25 into x^2 - 2.5x - 6 gives t= -121/16,

however the answer seems to differ giving t= 2.5 when dx/dt= 0
How is this so?

b) not to sure as confused about a).

 

[neutrino]

Homework Statement

Given (d^2)x/ (dt^2) = 2 for all t>= 0 and dx/dt= -5 and x= 4 when t= 0 find t when

a) dx/ dy= 0

Perhaps you meant dx/dt=0?

2. The attempt at a solution

a) Given (d^2)x/ (dt^2) = 2, then
dx/dt= 2x + c = -5

x and its derivatives are a function of t here, so from the second-order equation, you'll end-up at
$$\frac{dx}{dt} = 2t+C$$

given dx/ dt= 0

Was it? The question tells you what $\frac{dx}{dt}$ is at t=0. Use this condition to find the constant of integration. With one more integration, you can find x as a function of t. The rest is just algebra.

c= -2.5 using 2x + c= 0 and 2x + c = -5

using dx/dt= 2x + c, t= x^2 + cx + d
where 0= 16 - 2.5 x 4 + d (using x= 4 when t= 0 )
where d= -6

Thus,

t= x^2 - 2.5x - 6

when dx/dt= 0 then x= 1.25 in 2x + c= 0 (c= -2.5)

subsituting x= 1.25 into x^2 - 2.5x - 6 gives t= -121/16,

however the answer seems to differ giving t= 2.5 when dx/dt= 0
How is this so?

b) not to sure as confused about a).

Part b should become clear once you rework part a.

 

[J.D.]

Given (d^2)x/ (dt^2) = 2 for all t>= 0 and dx/dt= -5 and x= 4 when t= 0.

I think what you mean is:

Given $$\frac{d^2 x}{dt^2} = 2 , \quad \forall t \geq 0$$ and $$\frac{dx}{dt}(0) = -5$$ and $$x(0) = 4$$. You will then find that c is not -2.5.

And I now see that neutrino already told you that...