Posted by Melody on .
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

Maths 
Reiny,
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? 
Math experts needed here. 
Reiny,
Can somebody see through the above mess?
Am I on the wrong track? 
Maths 
bobpursley,
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. 
Maths 
MathMate,
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/ 
To MathMate 
Reiny,
Of course!
How simple was that step, eh?
And here I was about to enter the mathematical hinterland.
Thank you! 
To Reiny 
MathMate,
That's what happens when you know a lot!

Maths 
bobpursley,
Nice work.