A billboard d = 85 feet wide is perpendicular to a straight road and is 40 feet from the road (see figure). Find the point on the road at which the angle θ subtended by the billboard is a maximum. (Round your answer to two decimal places.) x = ft
tan(a) = 40/x tan(a + o) = (40+85)/x = 125/x
Now I'm stuck :P.
take the arctan of bot sides... so that you can write an equation for \(\theta\) \[\theta=???\]
\(\theta\) as a function of x
well rifght now I have tan(a+o).
I gotta get rid of the tan(a) huh..
\[\large a= \tan^{-1} \left( 40 \over x \right)\] \[\large a+\theta = \tan^{-1} \left( 125\over x \right)\] \[\large\theta =(a+\theta)-a\]
so hmmm...
arctan(125/x) -arctan(40/x)?
yeah
hmmm
I'm still confused :P.
what part
what exactly we are doing lol :P. Maybe as we go along I'll get it..
Just no sure where this is going to end up...
\[\Large \theta =\tan^{-1}\left( 125 \over x \right)-\tan^{-1}\left( 40 \over x \right)\] so... \[\Large {d \theta \over dx} = \dots?\]
seems like it's going ot be some UGLYness...
iknow ... groan
125/(x^2 * 15625?)
err +
- (40/(x^2+1600)?
lol I guess there was more to this.... \[\frac{1210(x^2-50000)}{(x^2+1600) (x^2) (1562500)}\]
the fawk..
i'm not good at that i used wolfram alpha you flipped the signs ... it should be -125/.... +40/...
so did I lol... Wolfman FTW!
Okies so I'm not retypijng that, lol link plz?
http://www.wolframalpha.com/input/?i=differentiate+arctan%28125%2Fx%29-arctan%2840%2Fx%29 http://www.wolframalpha.com/input/pod.jsp?id=MSP48681a22hb622d3hg3b500001446aa99b6f3bb56&s=45&button=1
and now we solve for x which I'm not going to attempt :P.
actually... I think it'as sqrt(5000?)
yeah that works :P 70.71
yes
now onto the next question, we must hurry!
:P
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