Question

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how fast is the angle of elevation of the rocket increasing, as seen by an observer on the ground 5 kilometers from the launching pad

Answer 1

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Step-by-step explanation:

Geometrically speaking, the distance between the rocket and the observer (
`r`), measured in kilometers, can be represented by a right triangle:

`r = \sqrt{x^(2)+y^(2)}` (1)

Where:

`x` - Horizontal distance between the rocket and the observer, measured in kilometers.

`y` - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket (
`\theta`), measured in sexagesimal degrees, is defined by the following trigonometric relation:

`\tan \theta = (y)/(x)` (2)

If we know that
`x = 5\,km`, then the expression is:

`\tan \theta = (y)/(5)`

And the rate of change of this angle is determined by derivatives:

`\sec^(2)\theta \cdot \dot \theta = (1)/(5)\cdot \dot y`

`(\dot \theta)/(\cos^(2)\theta) = (\dot y)/(5)`

`(\dot \theta\cdot (25+y^(2)))/(25) = (\dot y)/(5)`

`\dot \theta = (5\cdot \dot y)/(25+y^(2))`

Where:

`\dot \theta` - Rate of change of the angle of elevation, measured in sexagesimal degrees.

`\dot y` - Vertical speed of the rocket, measured in kilometers per hour.

If we know that
`y = 4\,km` and
`\dot y = 400\,(km)/(h)`, then the rate of change of the angle of elevation is:

`\dot \theta = 48.780\,(\circ)/(s)`

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.