Maths
posted by Melody .
Two pistons A & B move backwards and forwards in a cyclinder. Distance x cm of the right hand end of piston A from point 0 at time t seconds is x=3sin(2t)+3 and the distance y cm of the left hand end of piston B by formula y=2sin(3tpi/4)+8.5
The pistons are set in motion at t=0
How do I work out the minimun distance between the pistons and the time it occurs? I am really lost on this, can you please help.
Thanks

Looks like a rather nasty question
I see the distance (d) as
d = 3sin(2t)+3 + 2sin(3tpi/4)+8.5
= 3sin(2t)+ 2sin(3tpi/4)+ 11.5
I first ran this through a primitive graphing program I have, and it showed the above to have a period of 2pi, and a minimum value roughly between t = 1.5 and t = 1.7
to get exact answers is a messy Calculus problem
dd/dt = 6cos(2t)  6cos(3tpi/4)
= 0 for a max/min of d
So we have to solve
6cos(2t)  6cos(3tpi/4) = 0 or
cos(2t)  cos(3tpi/4) = 0
the trouble is that we need to have our trig functions contain the same period.
we know cos 2t = 2cos^2(t)  1
and cos(3t  pi/4) = cos3tcospi/4 + sin3tsinpi/4
= (1/√2)(cos3t + sin3t)
now there are expansion formulas for
cos 3x and sin 3x in terms of cos x, but I have a feeling that the question couldn't possibly be that messy.
Are you using graphig calculators?
What level is this? 
Can somebody see through the above mess?
Am I on the wrong track? 
I see the distance between the pistons as yx. It is so hard to see without a diagram.
d= 2sin(3tpi/4)+8.5 3sin(2t)+3
Then proceed as in Dr Reiny's solution.
In past days, we would have put this on the analogue computer and ran a graph for few seconds. 
I think I have made some headway.
I started from Reiny's result
cos(2t)  cos(3tpi/4) = 0
which I validated as correct.
By equating the two cosines, I conclude that:
2t + 2kπ = 3tπ/4
after adding 2kπ for generality, which gives
t = π/4 + 2kπ
Numerical solutions are: ... 5.5, 0.78, 7.07, ...
Since cosine is an even function, one of the arguments of cosine could be negative, therefore
2t + 2kπ = 3t  %pi;/4
which gives
t=(1/5)(π/4+2kπ)
Numerical solutions are: ... 6.12, 4.86, 3.61, 1.1, 0.157, 1.41, 2.67, 3.92, 5.18, 6.44 ...
These results can be verified in the graphed links below:
for the function:
http://img62.imageshack.us/i/1255515491func.png/
for the derivative:
http://img379.imageshack.us/i/1255515491deriv.png/ 
Of course!
How simple was that step, eh?
And here I was about to enter the mathematical hinterland.
Thank you! 
That's what happens when you know a lot!

Nice work.