Math Help - differentiation - displacement, speed

1. differentiation - displacement, speed

bk2 p90 q24 c d
i don't know how to do c) and d)
question: The displacement x at time t of a particle is given by
x = A sin 2t + B cos 2t
where A and B are constants.
a) if x satisfies the equation
$\frac {d^2 x }{dt^2} + 3 \frac {dx}{dt} - 4x = sin 2t$
find the values of A and B
b) What is the max. displacement of the particle?
c) show that the speed at time t is given by $2 \sqrt {A^2 +B^2 - x^2}$
d) hence find the maximum speed of the particle

my working:
dx/dt = 2A cos 2t - 2B sin 2t
$\frac {d^2 x }{dt^2} = -4A sin 2t - 4B cos 2t$
A = -2/25
B = -3/50
when dx/dt = 0
tan 2t = 4/3
sin 2t = 4/5 or -4/5
cos 2t = 3/5 or -3/5
max. of x = 1/10
don't know how to do c) and d) . thanks

2. Originally Posted by afeasfaerw23231233
bk2 p90 q24 c d
i don't know how to do c) and d)
question: The displacement x at time t of a particle is given by
x = A sin 2t + B cos 2t
where A and B are constants.
a) if x satisfies the equation
$\frac {d^2 x }{dt^2} + 3 \frac {dx}{dt} - 4x = sin 2t$
find the values of A and B
b) What is the max. displacement of the particle?
c) show that the speed at time t is given by $2 \sqrt {A^2 +B^2 - x^2}$
d) hence find the maximum speed of the particle

my working:
dx/dt = 2A cos 2t - 2B sin 2t
$\frac {d^2 x }{dt^2} = -4A sin 2t - 4B cos 2t$
A = -2/25
B = -3/50
when dx/dt = 0
tan 2t = 4/3
sin 2t = 4/5 or -4/5
cos 2t = 3/5 or -3/5
max. of x = 1/10
don't know how to do c) and d) . thanks
d) From (c), the speed will be a maximum when x is a minimum. Can you find the minimum value of x .....?

To get the result in (c):

$v = 2 A \cos (2t) - 2 B \sin (2t) \Rightarrow v^2 = .......... ~$ .... (1)

$x = A \sin (2t) + B \cos (2t) \Rightarrow x^2 = .......... \Rightarrow 4 x^2 = ......... ~$ .... (2)

(1) + (2): $v^2 + 4x^2 = 4B^2 + 4A^2 \Rightarrow ......$