Answer:
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.
Explanation:
Geometrically speaking, the distance between the rocket and the observer (
), measured in kilometers, can be represented by a right triangle:
(1)
Where:
- Horizontal distance between the rocket and the observer, measured in kilometers.
- Vertical distance between the rocket and the observer, measured in kilometers.
The angle of elevation of the rocket (
), measured in sexagesimal degrees, is defined by the following trigonometric relation:
(2)
If we know that
, then the expression is:
![\tan \theta = \frac{y}{5}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%20%5Cfrac%7By%7D%7B5%7D)
And the rate of change of this angle is determined by derivatives:
![\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y](https://tex.z-dn.net/?f=%5Csec%5E%7B2%7D%5Ctheta%20%5Ccdot%20%5Cdot%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B5%7D%5Ccdot%20%5Cdot%20y)
![\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdot%20%5Ctheta%7D%7B%5Ccos%5E%7B2%7D%5Ctheta%7D%20%3D%20%5Cfrac%7B%5Cdot%20y%7D%7B5%7D)
![\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdot%20%5Ctheta%5Ccdot%20%2825%2By%5E%7B2%7D%29%7D%7B25%7D%20%3D%20%5Cfrac%7B%5Cdot%20y%7D%7B5%7D)
![\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}](https://tex.z-dn.net/?f=%5Cdot%20%5Ctheta%20%3D%20%5Cfrac%7B5%5Ccdot%20%5Cdot%20y%7D%7B25%2By%5E%7B2%7D%7D)
Where:
- Rate of change of the angle of elevation, measured in sexagesimal degrees.
- Vertical speed of the rocket, measured in kilometers per hour.
If we know that
and
, then the rate of change of the angle of elevation is:
![\dot \theta = 48.780\,\frac{\circ}{s}](https://tex.z-dn.net/?f=%5Cdot%20%5Ctheta%20%3D%2048.780%5C%2C%5Cfrac%7B%5Ccirc%7D%7Bs%7D)
The angle of elevation of the rocket is increasing at a rate of 48.780º per second.