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December 8, 2016
Posted by **Melody** on Thursday, October 15, 2009 at 2:46am.

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(3t-pi/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, Wednesday, October 14, 2009 at 8:53am

Looks like a rather nasty question

I see the distance (d) as

d = 3sin(2t)+3 + 2sin(3t-pi/4)+8.5

= 3sin(2t)+ 2sin(3t-pi/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(3t-pi/4)

= 0 for a max/min of d

So we have to solve

6cos(2t) - 6cos(3t-pi/4) = 0 or

cos(2t) - cos(3t-pi/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/ã2)(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, Wednesday, October 14, 2009 at 9:00am

Can somebody see through the above mess?

Am I on the wrong track?

Maths - bobpursley, Wednesday, October 14, 2009 at 9:36am

I see the distance between the pistons as y-x. It is so hard to see without a diagram.

d= 2sin(3t-pi/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, Wednesday, October 14, 2009 at 12:11pm

I think I have made some headway.

I started from Reiny's result

cos(2t) - cos(3t-pi/4) = 0

which I validated as correct.

By equating the two cosines, I conclude that:

2t + 2kƒÎ = 3t-ƒÎ/4

after adding 2kƒÎ for generality, which gives

t = ƒÎ/4 + 2kƒÎ

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 + 2kƒÎ = 3t - %pi;/4

which gives

t=(1/5)(ƒÎ/4+2kƒÎ)

Numerical solutions are: ... -6.12, -4.86, -3.61, -1.1, 0.157, 1.41, 2.67, 3.92, 5.18, 6.44 ...

To MathMate - Reiny, Wednesday, October 14, 2009 at 1:51pm

Of course!

How simple was that step, eh?

And here I was about to enter the mathematical hinterland.

Thank you!

To Reiny - MathMate, Wednesday, October 14, 2009 at 2:43pm

That's what happens when you know a lot!

Maths - bobpursley, Wednesday, October 14, 2009 at 4:14pm

Nice work.

- Math -
**MathMate**, Thursday, October 15, 2009 at 8:55amMelody, the difference is indeed y-x, which is a short form for f(t) = y(t)-x(t).

Since they both vary with time in a periodic manner, it is important to find out when the f(t) attains the maximum, and what the value of the maximum is.

This graph shows exactly the function

f(t) = y(t) - x(t) as a function of time:

http://img62.imageshack.us/i/1255515491func.png/

You will notice peaks and valleys at different times t.

To find the maximum (and minimum), Dr. Reiny has evaluated the derivative of f(t) with respect to t and equated f'(t) to zero.

The solutions where t is a maximum or minimum have been given as:

t1 = ƒÎ/4 + 2kƒÎ

t2 = (1/5)(ƒÎ/4+2kƒÎ)

and k is an integer constant.

By substituting different integral values of k into the above solutions, we obtain values of t for which the function f'(t)=0, or where f(t) is a maximum or a minimum.

These values have been evaluated as:

t1 = -5.5, 0.78, 7.07, ... and

t2 = -6.12, -4.86, -3.61, -1.1, 0.157, 1.41, 2.67, 3.92, 5.18, 6.44...

respectively.

Now if you would go back to the graph of f(t) and you can pick out the minimum values of about -9.545 at t=2.67, corresponding to (1/5)*(π/4+4π).

You can evaluate y(t)-x(t) at different locations where f'(t) = 0 to make sure you get the global maximum. The minimum distance can be read from the graph and the exact time when it occurs can also be calculated from the solutions t1 and t2 above.

I hope that explains what you need to know, but do not hesitate if you need more details.

Final note: it would be much clearer if you study the graph of the function y(t)-x(t) which is at:

http://img62.imageshack.us/i/1255515491func.png/

The graph of the derivative is shown in :

http://img379.imageshack.us/i/1255515491deriv.png/ - Math -
**MathMate**, Thursday, October 15, 2009 at 8:58amNote:

the minimum distance occurs at

t=(1/5)(1/4+6)π

where

f'(t)=0 and

f(t)=0.5 - Math -
**MathMate**, Thursday, October 15, 2009 at 9:04amAlso, please check the signs, I believe y(t)-x(t) is always positive. The graph shown is x(t)-y(t). My apologies.