Expert: Paul Klarreich - 2/8/2006

Question
In the figure ( http://img.photobucket.com/albums/v211/silverpyro/11.jpg ),
BD is an advertisement painted on a vertical wall BDC of a building. BD =
20m, DC = 10m. An observer at A, x metres from the wall, finds the angle
subtended by the advertisement at his eyes to be (theta).

(theta will be subsituted with a @)

Find the value of d@/dx at x=50. Hence, estimate the increase in the
distance between the observer and the wall if the angle subtended is to be
decreased by 1-dregree from that observed at x=50 (your answer should be
correct to the nearest 1/10 metre)


Work Done

I found d@/dx to be

  20(300-x^2)
-----------------
x^4+1000x^2+90000

and when x=50, d@/dx = 0.1897.

I don't understand what the question is asking...can you clarify? Is it
asking how much x would have to increase for @ to be @-1..? If so, how
would one go about doing that =|

Thank you =)

Answer
Hi, Mona,

Your Question: In the figure ( http://img.photobucket.com/albums/v211/silverpyro/11.jpg ),
BD is an advertisement painted on a vertical wall BDC of a building. BD = 20m, DC = 10m. An observer at A, x metres from the wall, finds the angle subtended by the advertisement at his eyes to be (theta).

(theta will be subsituted with a @)

Find the value of d@/dx at x=50. Hence, estimate the increase in the distance between the observer and the wall if the angle subtended is to be decreased by 1-dregree from that observed at x=50 (your answer should be correct to the nearest 1/10 metre)

Work Done

I found d@/dx to be

20(300-x^2)
-----------------
x^4+1000x^2+90000

and when x=50, d@/dx = 0.1897.

I don't understand what the question is asking...can you clarify? Is it asking how much x would have to increase for @ to be @-1..? If so, how would one go about doing that =|

Thank you =)
===========================
Two items:

1. About these problems in general:  You are doing something called 'linearization.'  Sounds terribly complicated, but just a lot of letters meaning estimating  delta-x  based on delta-@.  So what you do is:

A. Find your  d@/dx = f'(x), although you might use implicit differentiation.

B. Write   d@ = f(x) dx

C. Put in your x = 50, or whatever.

D. Put in  d@ = -1 degree, converted to radians, of course.

E. Solve for  dx.

So, if you found that when x=50, d@/dx = 0.1897, assuming it is correct, all you need to do is write:

d@ = 0.1897 dx, then put d@ = 1 degree(in radians) and solve for dx.
--------------------

2. About this particular example.  Your dt/dx looks correct. (I will write 't' now, not '@', for a while.  The @ sign looks ugly.)

We need an expression for theta:  In your picture (I got it with no trouble.) here's what I did, which is probably the same as you:

Write  t2 = angle BAC,  t1 = angle DAC.  (You mentioned D in your description, but didn't put it in the diagram.  No problem.)

Then t = t2 - t1, and
                      tan t2 - tan t1
tan t = tan(t2 - t1) = ----------------
                      1 + tan t2 tan t1

Now  tan t2 = 30/x, and tan t1 = 10/x, so
       30/x - 10/x      20x
tan t = ----------- = ---------
       1 + 300/x^2   x^2 + 300

t = arctan(.../...)

dt           1             6000 - 20x^2  
-- = --------------------- -------------  =
dx   1 + [20x/(x^2+300)]^2 (x^2 + 300)^2

  6000 - 20x^2
---------------------- =
(x^2 + 300)^2 + 400x^2

     6000 - 20x^2
----------------------------- =
x^4 + 600x^2 + 90000 + 400x^2

     6000 - 20x^2
----------------------
x^4 + 1000x^2 + 90000

Now multiply across by dx, to have (going back to your @ symbol):

       6000 - 20x^2
d@ = --------------------- dt
    x^4 + 1000x^2 + 90000

Now you put your  d@ = 1 degree(in radians) and x = 50 and you are home, solving for dx.