You are 2000 feet away from a rocket on the Cape Canaveral launchpad filming a rocket launch. The rocket launches and flies upward. You pivot your camera upward, keeping the rocket in the camera's sights. a. Let \( \theta \) be the angle that your camera points relative to the ground. That is, when the rocket is on the ground and your camera points straight ahead towards it, the angle \( \theta \) is equal to 0 . As the rocket flies into the air, your camera pivots upward, and \( \theta \) increases. What is \( \theta \) when the rocket is at height 2000 ? \[ \theta=\frac{\pi}{4} \text { radians } \] \[ \theta=1.107 \text { radians } \] What is \( \theta \) when the rocket is at height 4000 when the rocket is at height 26000 ? \( \theta=1.42 \) radians Suppose the rocket is at some unknown height \( y \). Give a formula for \( \theta \) in terms of \( y \). \( \theta=1 \) radians b. Let \( t \) be the number of seconds since the rocket launched. Give a formula that relates \( \frac{d \theta}{d t} \) to \( \frac{d y}{d t} \). (In your answer, write \( y \) ' for \( \frac{d y}{d t} \).) What are the correct units to measure \( \frac{d \theta}{d t} \) ? What I
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The relationship between angles and heights can be a bit of a mathematical ride! To derive a formula for \( \theta \) in terms of \( y \), you can use the tangent function. Since your distance to the rocket remains consistent at 2000 feet, you can express the angle as \( \theta = \tan^{-1}\left(\frac{y}{2000}\right) \). As the rocket ascends, this formula allows you to capture the angle of your camera in a neat package. To relate the rate of change of the angle \( \frac{d\theta}{dt} \) to the rate of change of height \( \frac{dy}{dt} \), you would use the chain rule. You'd express it like this: \( \frac{d\theta}{dt} = \frac{1}{2000 + y^2} \cdot y' \), where \( y' \) is the height's rate of change. The units for \( \frac{d\theta}{dt} \) are typically radians per second, which captures the angular speed as the rocket climbs skywards!