Question 448.
At what rate is the viewing angle changing when the observer is 50 feet from the tunnel if they are traveling at 6 feet per second? The picture is as under.
Answer 448.
θ
= arctan(25/50) - arctan(15/50)
= arctan(1/2 - 3/10) / (1 + 3/20)
= arctan (4/23)
If the observer were x ft. away,
θ
= arctan(25/x - 15/x)/(1 + 375/x^2)
= arctan [10x/(x^2 + 375)]
=> tanθ = 10x/(x^2 + 375)
=> sec^2 θ dθ/dt = [(x^2 + 375) * 10 - 10x * 2x]/(x^2 + 375)^2 dx/dt ... ( 1 )
sec^2 θ = 1 + tan^2 θ = 1 + 4/23 = 27/23
Plugging sec^2 θ = 27/23, dx/dt = 6 and x = 50 in ( 1 )
=> dθ/dt
= [(2500+375)*10 - 20*2500)] * 6/[(27/23) * (2500+375)^2] rad/sec
= (28750 - 50000) * 6 / [(27/23) * (8265625) rad/sec
= - (21250 * 23) * 6 / (27 * 8265625) rad/s
= - 0.01314 rad/s.
[Negative sign indicates that the angle is decreasing.]
Link to YA!
At what rate is the viewing angle changing when the observer is 50 feet from the tunnel if they are traveling at 6 feet per second? The picture is as under.
θ
= arctan(25/50) - arctan(15/50)
= arctan(1/2 - 3/10) / (1 + 3/20)
= arctan (4/23)
If the observer were x ft. away,
θ
= arctan(25/x - 15/x)/(1 + 375/x^2)
= arctan [10x/(x^2 + 375)]
=> tanθ = 10x/(x^2 + 375)
=> sec^2 θ dθ/dt = [(x^2 + 375) * 10 - 10x * 2x]/(x^2 + 375)^2 dx/dt ... ( 1 )
sec^2 θ = 1 + tan^2 θ = 1 + 4/23 = 27/23
Plugging sec^2 θ = 27/23, dx/dt = 6 and x = 50 in ( 1 )
=> dθ/dt
= [(2500+375)*10 - 20*2500)] * 6/[(27/23) * (2500+375)^2] rad/sec
= (28750 - 50000) * 6 / [(27/23) * (8265625) rad/sec
= - (21250 * 23) * 6 / (27 * 8265625) rad/s
= - 0.01314 rad/s.
[Negative sign indicates that the angle is decreasing.]
Link to YA!
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