Problem with angles and differentiation

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TW Cantor
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Homework Statement



A "big screen" AB of height L is placed at a height h above a point C on the ground as shown (in the attachment). A person wishes to sit on the ground at a point D in order to watch a video on the big screen. She wishes to sit at a horizontal distance x away from C which gives her the best view. This means that she wishes to sit at a value of x which maximises the viewing angle . If L = 16.2m and h = 7.7m, find this value of x.




Homework Equations






The Attempt at a Solution



i know i have to get an equation which i can then differentiate but I am not sure how to get this equation and what i would then differentiate with respect to.
 

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umm i got cos(θ)=((x^2)+184.03)/((sqrt((x^2)+571.21)*(sqrt((x^2)+59.29))

is that the sort of thing i should get getting for cos in terms of x?
 
haha obviously not the right way then. i tried to find the lengths of AD and BD in terms of x, and then used the cosine rule to get theta in terms of x?
 
sorry does that mean inverse cos of the whole thing?
 
no! stop making it complicated! :biggrin:

you want the largest θ …

that'll be the smallest cosθ …

put y = cosθ (if that makes you happier) …

now y is a function of x, so use dy/dx to find a minimum for y (= cosθ) :wink:
 
ok, so i say that y=((x^2)+184.03)/((sqrt((x^2)+571.21)*(sqrt((x^2)+59.29))
i then say dy/dx=0 and use that to calculate a value for x?
 
this method worked :-) i got x=13.57, thanks a lot for your help :-) :-)
 
TW Cantor said:
… i then say dy/dx=0 and use that to calculate a value for x?

(have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

Yes.

Except, now that I look at it again, it would be easier to use y2 instead of y (to avoid those nasty square-roots ). :wink:
 
yeah it did come out as quite a nasty differential :-( but i got the right answer so its fine :-)