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AP Calculus AB· difficulty 3/5

A balloon rises straight up at 5 ft/s5~\text{ft/s}. An observer is 100 ft100~\text{ft} from the launch point. When the balloon is 100 ft100~\text{ft} high, the line of sight changes angle (in radians) at

  • A

    1200 rad/s\tfrac{1}{200}~\text{rad/s}

  • B

    150 rad/s\tfrac{1}{50}~\text{rad/s}

  • C

    120 rad/s\tfrac{1}{20}~\text{rad/s}

  • D

    140 rad/s\tfrac{1}{40}~\text{rad/s}

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Explanation

tan⁡θ=h/100\tan\theta = h/100, sec⁡2θ dθ/dt=(1/100)dh/dt\sec^2\theta\,d\theta/dt = (1/100)dh/dt. At h=100h = 100, θ=45∘\theta = 45^\circ, sec⁡2=2\sec^2 = 2. So dθ/dt=(1/100)(5)/2=1/40 rad/sd\theta/dt = (1/100)(5)/2 = 1/40~\text{rad/s}.

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