2000 A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by 0 = tan 2000 where x is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5947 feet apart. 0' = radians per foot. Round to five decimal places.
2000 A television camera at ground level is 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure. The angle of elevation of the camera can be found by 0 = tan 2000 where x is the height of the rocket. Find the rate of change of the angle of elevation after launch when the camera and the rocket are 5947 feet apart. 0' = radians per foot. Round to five decimal places.
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
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Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In this diagram:
- The rocket is shown taking off vertically.
- The horizontal distance from the rocket to the camera is 2000 feet.
- The height of the rocket above the ground is represented by \( x \).
- The angle of elevation from the camera to the rocket is represented by \( \theta \).
The angle of elevation \( \theta \) can be found using the arctangent function:
\[ \theta = \tan^{-1} \left(\frac{x}{2000}\right) \]
We are interested in finding the rate of change of the angle of elevation (\( \theta \)) after launch when the distance between the camera and the rocket is 5947 feet.
To find this rate, we use the formula for the derivative of the angle:
\[ \theta' = \textcolor{white}{placeholderspace} \text{radians per foot} \]
Round your answer to five decimal places for precision.
Note:
1. Due to the nature of inverse trigonometric functions, proper consideration of units and differentials is necessary.
2. In practical applications, you might need calculus techniques to derive this rate accurately.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0e718db9-a0c6-4ca7-b587-f8420530cbf9%2F25173db8-980d-46c1-8919-f08b774eab84%2Fuwce8g.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Rate of Change of the Angle of Elevation During a Rocket Launch
A television camera at ground level is positioned 2000 feet away from the launching pad of a space rocket that is set to take off vertically, as seen in the following figure:
In this diagram:
- The rocket is shown taking off vertically.
- The horizontal distance from the rocket to the camera is 2000 feet.
- The height of the rocket above the ground is represented by \( x \).
- The angle of elevation from the camera to the rocket is represented by \( \theta \).
The angle of elevation \( \theta \) can be found using the arctangent function:
\[ \theta = \tan^{-1} \left(\frac{x}{2000}\right) \]
We are interested in finding the rate of change of the angle of elevation (\( \theta \)) after launch when the distance between the camera and the rocket is 5947 feet.
To find this rate, we use the formula for the derivative of the angle:
\[ \theta' = \textcolor{white}{placeholderspace} \text{radians per foot} \]
Round your answer to five decimal places for precision.
Note:
1. Due to the nature of inverse trigonometric functions, proper consideration of units and differentials is necessary.
2. In practical applications, you might need calculus techniques to derive this rate accurately.
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