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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Let's break down the problem step by step.

### Part a: Finding the angle $$ \theta $$

1. **Understanding the scenario**: You are 2000 feet away from the rocket, and as it ascends, you need to find the angle $$ \theta $$ that your camera makes with the ground.

2. **Using trigonometry**: The relationship between the height of the rocket $$ y $$ and the angle $$ \theta $$ can be described using the tangent function:
   $$
   \tan(\theta) = \frac{y}{2000}
   $$
   where $$ y $$ is the height of the rocket.

3. **Finding $$ \theta $$**:
   - When $$ y = 2000 $$:
     $$
     \tan(\theta) = \frac{2000}{2000} = 1 \implies \theta = \tan^{-1}(1) = \frac{\pi}{4} \text{ radians}
     $$
   - When $$ y = 4000 $$:
     $$
     \tan(\theta) = \frac{4000}{2000} = 2 \implies \theta = \tan^{-1}(2)
     $$
   - When $$ y = 26000 $$:
     $$
     \tan(\theta) = \frac{26000}{2000} = 13 \implies \theta = \tan^{-1}(13)
     $$

4. **General formula for $$ \theta $$**:
   $$
   \theta = \tan^{-1}\left(\frac{y}{2000}\right)
   $$

### Part b: Relating $$ \frac{d \theta}{d t} $$ to $$ \frac{d y}{d t} $$

1. **Differentiating the formula**: We need to find the relationship between the rates of change of $$ \theta $$ and $$ y $$. We can use implicit differentiation on the equation:
   $$
   \tan(\theta) = \frac{y}{2000}
   $$

2. **Differentiating both sides**:
   $$
   \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{2000} \frac{dy}{dt}
   $$

3. **Rearranging the equation**:
   $$
   \frac{d\theta}{dt} = \frac{1}{2000} \sec^2(\theta) \frac{dy}{dt}
   $$

4. **Using $$ y' $$ for $$ \frac{dy}{dt} $$**:
   $$
   \frac{d\theta}{dt} = \frac{1}{2000} \sec^2(\theta) y'
   $$

### Units for $$ \frac{d \theta}{d t} $$

- The angle $$ \theta $$ is measured in radians, and time $$ t $$ is measured in seconds. Therefore, the units for $$ \frac{d \theta}{d t} $$ are:
  $$
  \text{radians per second}
  $$

### Summary of Results

- The angle $$ \theta $$ when the rocket is at height $$ y $$ is given by:
  $$
  \theta = \tan^{-1}\left(\frac{y}{2000}\right)
  $$
- The relationship between the rates of change is:
  $$
  \frac{d\theta}{dt} = \frac{1}{2000} \sec^2(\theta) y'
  $$
- The units for $$ \frac{d \theta}{d t} $$ are radians per second.
When the rocket is at height 2000 feet, the angle $$ \theta $$ is $$ \frac{\pi}{4} $$ radians. At 4000 feet, $$ \theta $$ is approximately 1.107 radians, and at 26000 feet, $$ \theta $$ is about 1.42 radians. The general formula for $$ \theta $$ in terms of height $$ y $$ is: $$ \theta = \tan^{-1}\left(\frac{y}{2000}\right) $$ The rate of change of the angle $$ \frac{d\theta}{dt} $$ is related to the rate of change of height $$ y' $$ by: $$ \frac{d\theta}{dt} = \frac{1}{2000} \sec^2(\theta) y' $$ The units for $$ \frac{d \theta}{d t} $$ are radians per second.

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