March 1, 2017

Homework Help: Math

Posted by Melody on Thursday, October 15, 2009 at 2:46am.

Hi, this is a question I posted a couple of days ago and answers below.I am still really confused and unfortunately can't copy a picture into this. I thought it would be y-x but have no idea how to calculate that, but as the pistons move in towards each other, wouldn't the distance have to be 1/2 (y-x)? This is Year 11 maths in Australia. Any further explanations would be really appreciated. Thankyou to Dr Reiny and everyone else who has responded.

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.

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
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.

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