Question

CRAZY CALCULUS PROBLEM: a searchlight is located at point A, 40 ft from a wall. The search light revolves counterclockwise at a rate of (pi/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.

Answer 1

I will now draw a picture!

Answer 2

a.) Find the constant of proportionality k. <--- does that mean just solve for k????

Answer 3

you have L = 10000 for d=40
just plug it in L=(k/d^2)

Answer 4

Yep, just solve for \(k\).

Answer 5

i am sure there are more parts of this problem yet to come...
(because this isn't calculus yet :P)

Answer 6

THat makes sense. Oh, the calculus will come soon enough. x) b.) express L as a function of theta, the angle formed by the lines AP and AB...

Answer 7

can you find 'd' in terms of theta first ??
using trigonometry ?
hint : cos ratio

Answer 8

lot easier to find cos theta from that right triangle...
cos theta =... ?

Answer 9

or sec theta will also do

Answer 10

sOrry,my computer froze. hold on, i'm trying to process xD

Answer 11

i am sure you know how to find cos theta, right ?
also no calculus yet :P i am curious about it being crazy!

Answer 12

Gah! Leave me alone! haa jk DOn't go! Cos(theta)=40/d!

Answer 13

so, d= 40/cos theta or d= 40 sec theta
just plug this in L = k/d^2
and done with b)!

Answer 14

not to mention that you already know 'k' from part a)

Answer 15

i won't leave until i make you solve this.... and i am not jk :P

Answer 16

lol thanks! So... does it reduce to 10,000cos(theta)?

Answer 17

can you please post all the parts ? so that we can think on further steps....
first what you got 'k' from part a) ?

Answer 18

did you forget to square cos theta ?
because i get
L = 10000 cos^2 theta

Answer 19

Do you mean all the parts to the problem? c.) How fast (in Lumens/Second) is the strength of the light changing when theta=(pi)/4? Justify ura nswere! (ANd yeah, i forgot to square -_-)

Answer 20

now comes the calculus.
can you differentiate,
L = 10000 cos^2 theta ?

Answer 21

(with respect to t)

Answer 22

shouldnt it be 400000Cos^2(theta). did u forget to square D in part a? cuz i did lol

Answer 23

:O
wait, let me recalculate

Answer 24

^_^

Answer 25

i still get L=10000 cos^2 theta
k=16*10^6

Answer 26

d^2 =40^2 = 1600

Answer 27

yeah, so that is d^2, then it's multiplied by L. or am I doing this wrong?

Answer 28

1000 = k/ 1600
k = 16*10^6
L = 16*10^6 / d^2
d= 40 sec theta
L = 16*10^6 / (40^2 sec^2 theta )
L = (16*10^6 / 1600 ) cos^2 theta
L =10000 cos^2 theta.

Answer 29

Oops ^_^ lol! Sorry! I promise im not always this dumb! I'm one of the only 3 A's in my calculus class!!!! lol ok, go on xD

Answer 30

any doubts so far ?
can you differentiate,
L = 10000 cos^2 theta with respect to theta ?
its ok :)

Answer 31

You want the derivative with respect to time.

Answer 32

oh yes :P

Answer 33

20000Cos(theta)*(-sin(theta))

Answer 34

But you need it with respect to theta for the chain rule\[
\frac{dL}{dt}=\frac{dL}{d\theta}\frac{d\theta}{dt}
\]

Answer 35

that is dL/ d theta = 20000Cos(theta)*(-sin(theta))
can you simplify a bit ?
and yes, you need dL/dt, so i suggest you divide both sides by dt

Answer 36

"The search light revolves counterclockwise at a rate of (pi/30) radians per second. "

This is a hint about \(d\theta/dt\)

Answer 37

Simplify? Uhm... i dont know.

Answer 38

\(2 \sin \theta \cos \theta = \sin 2\theta \)
but not necessary.

Answer 39

so, we have
\(dL = -10000 \sin 2\theta \: d\theta \\dL/dt = -10000 \sin 2\theta \: (d\theta/dt) \)
dL/ dt is what u need to find, and theta, dtheta/dt is given.

Answer 40

ok. So how do I find dL/dt?

Answer 41

theta, dtheta/dt is given.can you figure out those ?

Answer 42

no :(

Answer 43

what does this give you ?
The search light revolves counterclockwise at a rate of (pi/30) radians per second''

Answer 44

and part D directly gives you '..... when theta=(pi)/4'

Answer 45

I'm not sure what I'm suppoased to be doing 0.0

Answer 46

The search light revolves counterclockwise at a rate of (pi/30) radians per second----------> dθ/dt = pi/30,
θ = pi/4, given in part d) of question,
plug those in
dL/dt=−10000sin2θ(dθ/dt)
and you get dL/dt

Answer 47

OHHHHHHHHHHHHHHHHHHHHHH! BLAH! Lol ok! Hold on xD

Answer 48

-1047.2?

Answer 49

yes, that correct....or you can keep that as -1000pi/3

Answer 50

Ok, so the strength of the light is increasing at pi/4? and why is it negative?

Answer 51

when theta = pi/4 strength of light is changing at the rate of -1000pi/3 lumens/sec means decreasing @ 1000pi/3 lumens/sec (negative implies decreasing, positive implies increasing)
means in one second, strength of light decreases by 1000pi/3 lumens

Answer 52

I ment increasing xD lol Idk why i wrote decreasing thank you so much! i have no idea what time it is where u are but. GOOD NIGHT xD

Answer 53

good night and sweet dreams to you ^_^
(its 1.07 pm here :) )

Answer 54

welcome so much ! ^_^

Answer 55

wahhh! 11:39. Goku has saved the day again