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A rocket travels vertically at a speed of 1200 km/hr. The rocket is tracked through a telescope by an observer located 11 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after lift-off. (Round your answer to two decimal places.)
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my draw button is not working
Where to go from there?
|dw:1426992839744:dw|
\( \Large { \tan\theta = \frac{y}{11}\\ ~ \\ \\ \therefore implicit ~ derivative~ with~ respect ~ to ~ time \\ \sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{11} \cdot \frac{dy}{dt} } \)
we want to find \( \Large \frac{d\theta}{dt} \) , when \( \Large \frac{dy}{dt}=1200 \) and t = 3/60 hour
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What is d theta/dt?
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