Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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irspow
2006-02-21 12:45:05

No bother johny, I am glad that we aren't all crazy around here.  With all of the objections I brought up during that thread, I am glad to be now vindicated.  Thanks for letting me know.

johny
2006-02-21 01:04:19

Thanks alot Irspow!!!!! Sorry i couldnt reply before as my PC wasnt working........and i asked a maths teacher about the Windscreen washer problem which u said wasnt correct, well got to say u were right as the Question did contain inaccuracies!!!! Sorry for bothering u soo much.

irspow
2006-02-16 10:04:38

This is just a type of composite function, but here goes.

10sin(πt/15)(π/15);

2π/3 (sin(πt/15))

edit*

I'm sorry, I guess that was a bit vague.

You take the derivative of the first part, multiply that times the second part.  Add that to the first part times the derivative of the second part.  Okay, not such a good explanation.  Look;

d/dt -10 = 0, because it is a constant.

d/dt cos(πt/15) = -sin(πt/15) (π/15), the second part was a composite function.

Now you have, if you can follow the bumbling above;

(0)(cos[πt/15]  +  -10(-sin[πt/15])(π/15);

0 + 10sin(πt/15)(π/15) = (2π/3)sin(πt/15)

Sorry, if that is still a bad explanation, I am not a teacher at all.

johny
2006-02-16 09:45:36

Ok, i get it now, and yes i am using radians!!!!! and just one last question......how would i differentiate this equation which would finally give me an equation for Velocity of the burner at any time t (as velocity is rate of change of displacement)???

irspow
2006-02-16 09:17:01

If I am reading this right;

r = -10cos(πt/15)

At t = 0,  r = -10

At t = 15, r = 10

Since T = 30, maximum displacement occurs at T/2 or t = 15

dr = r(15) - r(0) = 20

edit*

I am assuming that you are using radians because of the 2π/T identity.

johny
2006-02-16 08:46:48

Sorry guys recently i have been asking too many questions but i have been struggling with such topics!!!!! So whenever anyone finds time, could hopefully post a solution to the following question:

A small high-intensity gas burner runs on a narrow rail, forwards and backwards underneath plastic ski brush components.

Each components is heated for one cycle of the burner, and this lasts for 30 seconds (the period T)

The displacement of the burner is given by the equation (shown in the pic).

So ignoring the thickness of the burner block, what is the minimum length of narrow rail needed?