   Chapter 2.8, Problem 43E

Chapter
Section
Textbook Problem

# A television camera is positioned 4000 ft from the base of a rocket launching pad. The angle of elevation of the camera has to change at the correct rate in order to keep the rocket in sight. Also, the mechanism for focusing the camera has to take into account the increasing distance from the camera to the rising rocket. Let’s assume the rocket rises vertically and its speed is 600 ft/s when it has risen 3000 ft.(a) How fast is the distance from the television camera to the rocket changing at that moment?(b) If the television camera is always kept aimed at the rocket, how fast is the camera’s angle of elevation changing at that same moment?

To determine

a)

To find:

How fast is the distance from the television camera to rocket changing?

Explanation

1. Concept:

We can plot the figure for the given situation and then apply the Pythagorean Theorem to write the equation. Then apply the differentiation to find the required rate

2. Formula:

i. Pythagorean Theorem: Hypoteneous2=Sum of squares of perpendicular sides

ii. Power rule of differentiation: ddxxn=n*xn-1

iii. Chain rule of differentiation: ddtfx=ddxfx*dxdt

3. Given:

A television camera is placed 4000 ft. from the base of the launching pad.

dhdt= 600 ft/s

4. Calculations:

By Pythagoras Theorem x2= 40002+h2

Differentiate it with respect to t using the power and chain rule

To determine

b)

To find:

How fast the camera’s angle of elevation is changing

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