[Math4u] Re: Calculus Help

Hi:

If you draw right triangle, LOY, with legs of length, 1 and 2), you have lighthouse (L), 'opposite point' (O), and your position (Y) as vertices.  Now, choose a point, P, on segment(OY) and let x = length of segment OP.  Then x = tan(theta), where theta = measure of angle PLO.  Moreover, dx/dt = d/dt[tan(theta)] or, dx/dt = sec^2(theta) * d/dt(theta).  Because the light makes one revolution per 15 sec, d/dt(theta) = 8pi radians/min.  When the light reaches point Y, sec^2(theta) = (sqrt(5))^2 = 5.  Therefore, dx/dt = 5 * 8pi = 40pi mi/min.

Regards,

Rich B.

--- In Math4u@yahoogroups.com, "luvmath03" <luvmath03@...> wrote:
>
> I have just a couple of questions that are weirding me out. I think
> I should be able to get them but the unit analysis is confusing me
> more. I am just not really able to form the ininial equations and
> such.
>
> Question 1.
> A lighthouse is 1 mile off shore, due west. You are on the shore 2
> miles north of the point opposite the lighthouse. The light rotates;
> if you see the flash every 15 seconds, how fast is the beam moving
> when it passes you?
>
> I have an answer of 2/2.625 miles/sec but I don't feel its write
> since I feel like I need to use the chain rule (section we are
> in)....How would I set up an equation to relate Theta and x miles?
> when I differentiate I have d(theta)/dt and I would solve for d
> (x)/dt.
>
>
> Question 2.
> A trough is triangular in cross section, an isosceles triangle with
> sides of 12 inches and a top of 10 inches. The trough is 40 inches
> long. How fast is the depth change if you are pumping 1 cubic
> foot/min into the trough. I know the answer will depend upon the
> height of the liquid at an instantaneous time (t). I think I kind of
> have this one but just want to see another thought.
>
> Question 3.
> I am really lost here.
> A light is attached to a 13 in radius bicycle tire, at a point 8
> inches from the center. If Harvey rides the bike at 15 mph, how fast
> is the light moving up and down at its fastest.
>
> I know the light is at a point that is horizontal to the road for the
> fastest up and down point.
>
> Thanks in advance for assistance.
>
> William
>

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