Answer to Question #250810 in Calculus for Sam

Question

10. A picture 40 cm high is placed on a wall with its base 30 cm above the level of the eye of an observer. If the observer is approaching the wall at the rate of 40 cm/sec, how fast is the measure of the angle subtended at the observer's eye by the picture changing when the observer is 1 m from the wall?

Answer 1

$Angle \; θ = θ_2-θ_1 \\ θ = cot^{-1}(\frac{x}{70}) -cot^{-1}(\frac{x}{30}) \\ \frac{dθ }{dt} = \frac{\frac{-1}{70} \frac{dx}{dt}}{1 + \frac{x^2}{70^2}} - \frac{\frac{-1}{30} \frac{dx}{dt}}{1 + \frac{x^2}{30^2}} \\ \frac{dθ }{dt} = \frac{-dx/dt}{70}(\frac{4900}{4900+x^2} + \frac{dx-dt}{30}(\frac{900}{900+x^2}) \\ \frac{dx}{dt} = -40 \\ x = 100 \;cm \\ \frac{dθ }{dt} = \frac{-40}{70}\frac{4900}{4900+10000} + \frac{-40}{30}(\frac{900}{900+10000}) \\ \frac{dθ }{dt} = 4(\frac{700}{14900}) -4(\frac{300}{10900}) \\ \frac{dθ }{dt} = \frac{28}{149} - \frac{12}{109} = 0.07782 \;rad/sec$