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.

]]>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.

]]>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?

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