Ann and Bob are walking on straight streets that meet at right angles. Ann approaches the intersection at 2 m/sec and Bob moves away from the intersection at 1 m/sec. At what rate is the angle, theta, changing when Ann is 10 meters from the intersection and Bob is 20 meters from the intersection? Express your answer in degrees per second to the nearest degree.
Thank you in advance!
Thank you in advance!
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ok
assuming Ann's distance is y and Bob's distance is x , so :
dy/dt = - 2 m/s , and dx/dt = 1
so tan θ = y / x
differentiate w.r.to time :
sec^2 θ (dθ/dt) = (xdy/dt - ydx/dt) / x^2
given that y = 10 and x = 20 , u have cosθ = 20 / sqrt(500) = 2 / sqrt(5)
so secθ = sqrt(5) / 2
=> sec^2 θ = 5/4
hence dθ/dt = (4/5) [20(-2) - 10(1)] / 400
..................= (1/500) (- 50)
..................= - 0.1 rad/s
in degrees it's - 6 degrees / s to nearest degree <<<<<<<<<<<<<
can u check if that answer is correct ??
assuming Ann's distance is y and Bob's distance is x , so :
dy/dt = - 2 m/s , and dx/dt = 1
so tan θ = y / x
differentiate w.r.to time :
sec^2 θ (dθ/dt) = (xdy/dt - ydx/dt) / x^2
given that y = 10 and x = 20 , u have cosθ = 20 / sqrt(500) = 2 / sqrt(5)
so secθ = sqrt(5) / 2
=> sec^2 θ = 5/4
hence dθ/dt = (4/5) [20(-2) - 10(1)] / 400
..................= (1/500) (- 50)
..................= - 0.1 rad/s
in degrees it's - 6 degrees / s to nearest degree <<<<<<<<<<<<<
can u check if that answer is correct ??