Question

The bottom of a large theater screen is 3ft above your eye level and the top of the acreen is 10ft above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of 3ft/s while looking at the screen. What is the change of the viewing angle when you are 30ft from the wall on which the screen hangs assuming the floor is horizontal ?

Answer 1

@mayankdevnani

Answer 2

i had a diagram but it erased it :{

anyway if you draw it out you should notice 2 triangles
using the tangent relationship

Answer 3

\[\tan A = \frac{3}{x}\]
\[\tan B = \frac{10}{x}\]
\[\tan \theta = \tan (B-A) = \frac{\tan B - \tan A}{1 + \tan A \tan B}\]

Answer 4

\[\tan \theta = \frac{7x}{x^2 +30}\]
now use relation
\[\frac{d \theta}{dt} = \frac{d \theta}{dx}*\frac{dx}{dt}\]
dx/dt is given as 3

Answer 5

Ok I get that will try and see :)

Answer 6

ok :)

Answer 7

Where u get the formula for tan(b-a)

Answer 8

?? :)

Answer 9

http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

Answer 10

Where does the 30,ft come in

Answer 11

that is the "x" value
you plug it in at the very end for x after you have dtheta/dt

Answer 12

How do I set up once I have the 7x/x^2+30

Answer 13

you have to differentiate.... use quotient rule
http://en.wikipedia.org/wiki/Quotient_rule

Answer 14

nice problem

Answer 15

you should get
\[\sec^2 \theta \frac{d \theta}{dx} = \frac{210 - 7x^2}{(x^2 +30)^2}\]

Answer 16

Ok yes caught up!

Answer 17

ok :)

now if we solve for d/dx and multiply by 3, we will have rate angle is changing
\[\frac{d \theta}{dt} = 3 \cos^2 \theta \frac{210-7x}{(x^2 +30)^2}\]
last part is finding cos^2 in terms of x

Answer 18

to do this imagine we have a triangle and use the tan relation

\[\cos^2 \theta = \frac{(x^2 +30)^2}{h^2}\]
find "h" using pythagorean thm

Answer 19

85.02 so 85

Answer 20

what is 85?

Answer 21

How we solve for h we know the 30ft not plug in for x

Answer 22

oh well no you should prob wait til very end to plug in x=30
get "h" in terms of x

something went wrong with your computing, it wouldnt be 85 even if you plug in x=30

Answer 23

H= sqrt (x^4+109x^2+900)
Please lol

Answer 24

right

Answer 25

Thank god lol

Answer 26

haha so now you have
\[\frac{d \theta}{dt} = 3 (\frac{210 - 7x^2}{x^4 +109 x^2 +900})\]
plug in x =30 and you're done

Answer 27

Should the bottom be still under square root

Answer 28

no because remember we wanted cos^2 which has "h^2" on bottom
the square cancels out the sqrt

Answer 29

Right haaha

Answer 30

U are the best !!