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May 28, 2016
Posted by **Melody** on Wednesday, October 14, 2009 at 6:18am.

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 8:53amLooks 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/√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**, Wednesday, October 14, 2009 at 9:00amCan somebody see through the above mess?

Am I on the wrong track? - Maths -
**bobpursley**, Wednesday, October 14, 2009 at 9:36amI 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 12:11pmI 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 + 2kπ = 3t-π/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**, Wednesday, October 14, 2009 at 1:51pmOf 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 2:43pmThat's what happens when you know a lot!

- Maths -
**bobpursley**, Wednesday, October 14, 2009 at 4:14pmNice work.