Difficult math problem - am i on the right track?

<p>a searchlight is located at point A 40 feet from a wall. The searchlight revolves counterclockwise at a rate of π/30 radians per second. At any point B on the wall, the strength of the light L, is inversely proportional to the square of the distance d from A; that is, at any point on the wall L = k/d^2 . At the closest point P, L = 10,000 lumens.</p>

<p>a) Find the constant of proportionality k.
b) Express L as a function of θ , the angle formed by AP and AB.
c) How fast (in lumens/second) is the strength of the light changing when θ =π/4? Is it
increasing or decreasing? Justify your answer.
d) Find the value of θ between θ =0 and θ =π/2 after which L<1000 lumens.</p>

<p>for a:
(40)^2 + (10,000)^2 = d^2
L = k/d^2
10,000 = k/10,000.08
k = 100,000,800</p>

<p>for b:
L = [0.5 (AP x AB)] x sin(theta)
L = [0.5 (40 x 10,000)] x sin(theta)
L = 200,000 x sin(theta)</p>

<p>for c:
Who knows?</p>

<p>for d:
999.999 <= 200,000 x sin(theta)
theta <= 0.005</p>

<p>Hm, i think i got everything wrong, but i just want to know if I'm on the right track. The problem seems too easy the way i solved it so that's why i think I'm wrong. Thanks guys.</p>

<p>P.S. <= means "less than or equal too", haha</p>

<p>People should really start reading the top of the forum which alongside “IMPORTANT” clearly state:
<a href=“http://talk.collegeconfidential.com/high-school-life/452903-no-homework-help-posts-allowed-cc.html[/url]”>http://talk.collegeconfidential.com/high-school-life/452903-no-homework-help-posts-allowed-cc.html&lt;/a&gt;&lt;/p&gt;

<p>So sorry.</p>

<p>There’s no mention whether this is homework or not I think.</p>

<p>Ugh, i knew some a-hole would post something like that…</p>

<p>@meadow I’ve done the work, couldn’t just tell me if i was right or wrong?</p>

<p>Blah sorry. I had a post like this before, with no mention of homework. Got an infraction and 1 point for “homework help request”. Sorry if it was somewhat offensive.</p>

<p>it’s no problem dude, thanks for the lookout =)</p>