A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle of elevation changing when the balloon is at an altitude of 2500 feet?

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Question:

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle of elevation changing when the balloon is at an altitude of 2500 feet?

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Answer:0.0031792338 rad/sExplanation:[tex]\theta[/tex] = Angle of elevationy = Height of balloonUsing trigonometry[tex]tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta[/tex]Differentiating with respect to t we get[tex]\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta[/tex]Now, with the base at 200 ft and height at 2500 ftThe hypotenuse is [tex]h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft[/tex]Now y = 2500 ft[tex]cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974[/tex][tex]\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s[/tex]The angle is changing at 0.0031792338 rad/s

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