解微積分兩題（２）

(2) Since [(q-1)^3 + (q+1)^3]
= (q^3 – 3q^2 + 3q –1) + (q^3 + 3q^2 + 3q +1)
= 2 q^3 + 6 q
= 2q (q^2 +3),
thus p = [q^2 +3] / [(q-1)^3 + (q+1)^3]
= [q^2 +3] /2q (q^2 +3)
= 1/ (2q)

Hence, dp/dq = - 1/(2q^2)
and d2p /dq2 = 1/q^3.

(4) (a) velocity v = t^2 – 4t +3.
When v=0, we have t^2 – 4t +3 =0,
(t-3)(t-1)=0,
t=1 or t=3.
Acceleration a=dv/dt = 2t –4.
When t=1, a =dv/dt= 2(1)-4 = -2
When t=3, a= dv/dt= 2(3)-4 =2.

(b) when t=0, v=3. From the above, when t=1, v=0 ;
as such, v > 0 when 0 < = t < 1 and the particle is moving forward during the period.
When t=1, v=0 it means that the particle is steady and not moving backward nor forward at that moment.
When t=1, a=-2, it means that the particle is starting moving backward. Since v= t^2 - 4t + 3 = (t-2)^2 -1. v will decrease from 0 to -1 which will happen when t =2. And v will remain to be negative until v resumes 0 which will happen when t =3. As such, the particle is moving backward from 1 < t < 3.
When t =3, v=0 . It means that the particle is steady and not moving backward nor forward at that moment.
When t = 3, a = dv/dt =2 >0, it means that v will become >0 at the instant following t=3. Hence the particle becomes moving forward again. And v = t^2-4t+3
= (t-2)^2 –1. It means that for t>3, v>0. As such, the particle is moving forward when t>=3.

(c) acceleration a = dv/dt = 2t-4.
For 0 < = t < 2, a < 0 and hence the velocity is decreasing.
For 2= t, a =0 and hence the velocity is steady for that instant.
For t>2, a >0 and hence the velocity is increasing.